Characterizing 2-crossing-critical graphs
Drago Bokal, Bogdan Oporowski, R. Bruce Richter, Gelasio Salazar
CCharacterizing 2-crossing-critical graphs
Drago BokalBogdan OporowskiR. Bruce RichterGelasio Salazar
Author address:
University of Maribor, Maribor SLOVENIA
E-mail address : [email protected] Louisiana State University, Baton Rouge U.S.A.
E-mail address : [email protected] University of Waterloo, Waterloo CANADA
E-mail address : [email protected] UA de San Luis Potosi, San Luis Potosi MEXICO
E-mail address : [email protected] a r X i v : . [ m a t h . C O ] D ec ontents List of Figures vChapter 1. Introduction 1Chapter 2. Description of 2-crossing-critical graphs with V Q G embeds with all spokes in M H -bridges 62Chapter 11. Every rim edge has a colour 73Chapter 12. Existence of a red edge and its structure 83Chapter 13. The next red edge and the tile structure 98Chapter 14. Graphs that are not 3-connected 11914.1. 2-critical graphs that are not 2-connected 11914.2. 2-connected 2-critical graphs that are not 3-connected 119Chapter 15. On 3-connected graphs that are not peripherally-4-connected 12615.1. A 3-cut with two non-planar sides 12615.2. 3-reducing to peripherally-4-connected graphs 12815.3. Planar 3-reductions 13215.4. Reducing to a basic 2-crossing-critical example 13915.5. Growing back from a given peripherally-4-connected graph 14315.6. Further reducing to internally-4-connected graphs 14415.7. The case of V -free 2-crossing-critical graphs 149Chapter 16. Finiteness of 3-connected 2-crossing-critical graphs with no V n V n -bridges are small 155 iiONTENTS iii bstract It is very well-known that there are precisely two minimal non-planar graphs: K and K , (degree 2 vertices being irrelevant in this context). In the languageof crossing numbers, these are the only 1-crossing-critical graphs: they each havecrossing number at least one, and every proper subgraph has crossing number lessthan one. In 1987, Kochol exhibited an infinite family of 3-connected, simple 2-crossing-critical graphs. In this work, we: (i) determine all the 3-connected 2-crossing-critical graphs that contain a subdivision of the M¨obius Ladder V ; (ii)show how to obtain all the not 3-connected 2-crossing-critical graphs from the 3-connected ones; (iii) show that there are only finitely many 3-connected 2-crossing-critical graphs not containing a subdivision of V ; and (iv) determine all the 3-connected 2-crossing-critical graphs that do not contain a subdivision of V . Mathematics Subject Classification.
Primary 05C10.
Key words and phrases. crossing number, crossing-critical graphs.Bokal acknowledges the support of NSERC and U. Waterloo for 2006-2007, Slovenian Re-search Agency basic research projects L7-5459, J6-3600, J1-2043, L1-9338, J1-6150, research pro-gramme P1-0297, and an international research grant GReGAS.Richter acknowledges the support of NSERC.Salazar acknowledges the support of CONACYT Grant 106432. iv ist of Figures R P . 103.2 The 2-crossing-critical 3-representative embeddings in R P . 113.3 Standard labellings of the representativity 2 embeddings of V . 135.1 The two possibilities for D i when j = i + 2. 275.2 The two possibilities for D . 285.3 The two possibilities for D . 296.1 The case e ∈ r i +4 r i +5 for ¯ Q i being a ( ¯ Q i ∪ M ¯ Q i )-prebox. Only two of thethree spokes are shown. 357.1 The subgraph K of G in R P . 427.2 The 1-drawing of K . 447.3 The 1-drawings D [( K − (cid:104) s (cid:105) ) ∪ P ] and D [( K − (cid:104) s (cid:105) ) ∪ P ]. 447.4 The 1-drawings D [( K − (cid:104) s (cid:105) ) ∪ P ] and D [( K − (cid:104) s (cid:105) ) ∪ P ]. 468.1 The two possibilities for D . 518.2 The two possibilities for D . 5111.1 The locations of e , f , w e , w f , H e , N , and K f . 7711.2 The spine and its constituent paths. 7912.1 One of several examples of a ∆. 8713.1 D [ H (cid:48) ] 10713.2 Definition 13.7. 11414.1 The 2-crossing-critical graphs that are not 2-connected. 12014.2 2-connected, not 3-connected, 2-crossing-critical graphs, 2 non-planarcleavage units 122 vi LIST OF FIGURES T, U )-configurations. 14515.2 The thick edge is a bear hug. The dotted edges tw and vz might besubdivided, and the dashed edge uw need not be present. If uw is notpresent, then { ux, uy } is a simultaneously deletable pair of bear hugs. 14615.3 When s = b , G i − is a subgraph of the illustrated planar graph. 148HAPTER 1 Introduction
For a positive integer k , a graph G is k -crossing-critical if the crossing numbercr( G ) is at least k , but every proper subgraph H of G has cr( H ) < k . In general,it is not true that a k -crossing-critical graph has crossing number exactly k . Forexample, any edge-transitive non-planar graph G satisfies cr( G − e ) < cr( G ), forany edge e of G , so every such graph is k -crossing-critical for any k satisfyingcr( G − e ) < k ≤ cr( G ). If G is the complete graph K n , then cr( K n ) − cr( K n − e )is of order n , so K n is k -crossing-critical for many different values of k .Insertion and suppression of vertices of degree 2 do not affect the crossingnumber of a graph, and a k -crossing-critical graph has no vertices of degree 1 andno component that is a cycle. Thus, if G is a k -crossing-critical graph, the graph G (cid:48) whose vertex set consists of the nodes of G (i.e., the vertices of degree differentfrom 2) and whose edges are the branches of G (i.e., the maximal paths all ofwhose internal vertices have degree 2 in G ) is also k -crossing-critical. Our interestis, therefore, in k -crossing-critical graphs with minimum degree at least 3.By Kuratowski’s Theorem, the only 1-crossing-critical graphs are K , and K .The classification of 2-crossing-critical graphs is currently not known. The earliestpublished remarks on this classification of which we are aware is by Bloom, Kennedy,and Quintas [ ], where they exhibit 21 such graphs. Kochol [ ] gives an infinitefamily of 3-connected, simple 2-crossing-critical graphs, answering a question ofˇSir´aˇn [ ] who gave, for each n ≥
3, an infinite family of 3-connected n -crossing-critical graphs. Richter [ ] shows there are just eight cubic 2-crossing-criticalgraphs.About 15 years ago, Oporowski gave several conference talks about showingthat every large peripherally-4-connected, 2-crossing-critical graph has a very par-ticular structure which was later denoted as ‘being composed of tiles’. The methodsuggested was to show that if a peripherally-4-connected, 2-crossing-critical graphhas a subdivision of a particular V k (that is, k is fixed), then it has the desiredstructure and that only finitely many peripherally-4-connected, 2-crossing-criticalgraphs do not have a subdivision of V k . (The graph V n is obtained from a 2 n -cycleby adding the n diagonals. Note that V is K and V is K , .)Approximately 10 years ago, it was proved by Ding, Oporowski, Thomas, andVertigan [ ] that, for any k , a large (as a function of k ) 3-connected, 2-crossing-critical graph necessarily has a subdivision of V k . It remains to show that havingthe V k -subdivision implies having the desired global structure. Their proof involvesfirst showing a statement about non-planar graphs that is of significant independentinterest: for every k , any large (as a function of k ) “almost 4-connected” non-planargraph contains a subdivision of one of four non-planar graphs whose sizes grow with k . One of the four graphs is V k . This theorem is then used for the crossing-criticalapplication mentioned above. Tiles have come to be a very fruitful tool in the study of crossing-critical graphs.Their fundamentals were laid out by Pinontoan and Richter [ ], and later theyturned out to be a key in Bokal’s solution of Salazar’s question regarding averagedegrees in crossing-critical graphs [
8, 28, 31 ]. These results all rely on the easeof establishing the crossing number of a sufficiently large tiled graph, and theygenerated considerable interest in the reverse question: what is the true structureof crossing-critical graphs? How far from a tiled graph can a large crossing-criticalgraph be? Hlinˇen´y’s result about bounded path-width of k -crossing-critical graphs[ ] establishes a rough structure, but is it possible that, for small values of k , tileswould describe the structure completely? It turns out that, for k = 2, the answer ispositive. A more detailed discussion of these and other matters relating to crossingnumbers can be found in the survey by Richter and Salazar [ ].Our goal in this work, not quite achieved, is to classify all 2-crossing-criticalgraphs. The bulk of our effort is devoted to showing that if G is a 3-connected2-crossing-critical graph that contains a subdivision of V , then G is one of acompletely described infinite family of 3-connected 2-crossing-critical graphs. Thesegraphs are all composed from 42 tiles. This takes up Chapters 3 – 13. This combineswith [ ] to prove that a “large” 3-connected 2-crossing-critical graph is a memberof this infinite family.The remainder of the classification would involve determining all 2-crossing-critical graphs that either are not 3-connected or are 3-connected and do not havea subdivision of V . In Chapter 14, we deal with the 2-crossing-critical graphsthat are not 3-connected: they are either one of a small number of known particularexamples, or they are 2-connected and easily obtained from 3-connected examples.There remains the problem of determining the 3-connected 2-crossing-criticalgraphs that do not contain a subdivision of V . In the first five sections of Chapter15, we explain how to completely determine all the 3-connected 2-crossing-criticalgraphs from peripherally-4-connected graphs that either have crossing number 1 orare themselves 2-crossing-critical. In the sixth and final subsection, we determinewhich peripherally-4-connected graphs do not contain a subdivision of V and eitherhave crossing number 1 or are themselves 2-crossing-critical. Combining the twoparts yields a definite (and practical) procedure for finding all the 3-connected 2-crossing-critical graphs that do not contain a subdivision of V . This leaves openthe problem of classifying those that contain a subdivision of V but do not have asubdivision of V . In Sections 16.1 and 16.2, we show that there are only finitelymany. (Although this follows from [11], the approach is different and it keeps ourwork self-contained.)There is hope for a complete description. In her master’s essay, Urrutia-Schroeder [ ] begins the determination of precisely these graphs and finds 326of them. Oporowski (personal communication) had previously determined 531 3-connected 2-crossing-critical graphs, of which 201 contain a subdivision of V butnot of V . Austin [ ] improves on Urrutia-Schroeder’s work, correcting a minorerror (only 214 of Urrutia-Schroeder’s graphs are actually 2-crossing-critical) andfinding several others, for a total of 312 examples. Only 8 of Oporowski’s examplesare not among the 312. A few have been determined by us as stepping stones in ourclassification of those that have a subdivision of V . We have hopes of completingthe classification. . INTRODUCTION 3 The principal facts that we prove in this work are summarized in the followingstatement.
Theorem . Let G be a 2-cros-sing-critical graph with minimum degree at least 3. Then either: • if G is 3-connected, then either G has a subdivision of V and a veryparticular tile structure or has at most 3 million vertices; or • G is not 3-connected and is one of particular examples; or • G is 2- but not 3-connected and is obtained from a 3-connected exampleby replacing digons by digonal paths. We remark again that vertices of degree 2 are uninteresting in the context ofcrossing-criticality, so we assume all graphs have minimum degree at least 3.Chapters 2–13 of this work contain the proof of the following, which is the maincontribution of this work. (The formal definitions required for the statement givenbelow are presented in Chapter 2.)
Theorem V ) . Let G be a 3-connected, 2-crossing-critical graph containing a subdivision of V . Then G is a twisted circularsequence ( T , T , . . . , T n ) of tiles, with each T i coming from a set of 42 possibilities. This is part of the first item in the statement of Theorem 1.1.Chapter 14 is devoted to 2-crossing-critical graphs that are not 3-connected.(We remind the reader of Tutte’s theory of cleavage units and introduce digonalpaths in Chapter 14.) The results there are summarized in the following.
Theorem . Let G be a 2-crossing-critical graph with minimum degree at least 3 that is not 3-connected.(1) If G is not 2-connected, then G is one of 13 graphs. (See Figure 14.1.)(2) If G is 2-connected and has two nonplanar cleavage units, then G is oneof 36 graphs. (See Figures 14.2 and 14.3.)(3) If G is 2-connected with at most one nonplanar cleavage unit, then G hasprecisely one nonplanar cleavage unit and is obtained from a 3-connected,2-crossing-critical graph by replacing pairs of parallel edges by digonalpaths. Chapter 15 shows how to reduce the determination of 3-connected 2-crossing-critical graphs to “peripherally-4-connected” 2-crossing-critical graphs. A graph G is peripherally-4-connected if G is 3-connected and, for every 3-cut X in G , anypartition of the components into nonnull subgraphs H and J has one of H and J being a single vertex. The main result here is the following. Theorem . Every 3-connected, 2-crossing-critical graph is obtained from aperipherally-4-connected, 2-crossing-critical graph by replacing each degree 3 vertexwith one of at most 20 different graphs, each having at most 6 vertices.
We combine this with Robertson’s characterization of V -free graphs to explainhow to determine all the 3-connected 2-crossing-critical graphs that do not havea subdivision of V . This requires a further reduction to “internally 4-connected”graphs.Chapter 16 shows that a 3-connected, 2-crossing-critical graph with a subdivi-sion of V but no subdivision of V has at most three million vertices. The generalresult we prove there is the following.
1. INTRODUCTION
Theorem . Suppose G is a 3-connected, 2-crossing-critical graph. Let n ≥ be such that G has a subdivision of V n but not of V n +1) . Then | V ( G ) | = O ( n ) . HAPTER 2
Description of 2-crossing-critical graphs with V V . As mentioned in the introduction, they are composed of tiles. Thisconcept was first formalized by Pinontoan and Richter [
27, 28 ] who studied largesequences of equal tiles. Bokal [ ] extended their results to sequences of arbitrarytiles, which are required in this section. In those results, “perfect” tiles were intro-duced to establish the crossing number of the constructed graphs. However, thisproperty required a lower bound on the number of the tiles that is just slightly toorestrictive to include all our graphs. As we are able to establish the lower bound onthe crossing number of all these graphs in a different way (Theorem 5.5), we sum-marize the concepts of [ ] without reference to “perfect” tiles. Where the readerfeels we are imprecise, please refer to [ ] for details. Definition . (1) A tile is a triple T = ( G, λ, ρ ), consisting of a graph G and two sequences λ and ρ of distinct vertices of G , with no vertex of G appearing in both λ and ρ .(2) A tile drawing is a drawing D of G in the unit square [0 , × [0 ,
1] forwhich the intersection of the boundary of the square with D [ G ] containsprecisely the images of the vertices of the left wall λ and the right wall ρ , and these are drawn in { } × [0 ,
1] and { } × [0 , y -coordinates of the vertices are increasing with respect to theirorders in the sequences λ and ρ .(3) The tile crossing number tcr( T ) of a tile T is the smallest number ofcrossings in a tile drawing of T .(4) The tile T is planar if tcr( T ) = 0.(5) A k -drawing of a graph or a k -tile-drawing of a tile is a drawing or tile-drawing, respectively, with at most k crossings.It is a central point for us that tiles may be “glued together” to form largertiles. We formalize this as follows. Definition . (1) The tiles T = ( G, λ, ρ ) and T (cid:48) = ( G (cid:48) , λ (cid:48) , ρ (cid:48) ) are compatible if | ρ | = | λ (cid:48) | .(2) A sequence ( T , T , . . . , T m ) of tiles is compatible if, for each i = 1 , , . . . , m , T i − is compatible with T i .(3) The join of compatible tiles ( G, λ, ρ ) and ( G (cid:48) , λ (cid:48) , ρ (cid:48) ) is the tile ( G, λ, ρ ) ⊗ ( G (cid:48) , λ (cid:48) , ρ (cid:48) ) whose graph is obtained from G and G (cid:48) by identifying thesequence ρ term by term with the sequence λ (cid:48) ; left wall is λ ; and rightwall is ρ (cid:48) .(4) As ⊗ is associative, the join ⊗T of a compatible sequence T = ( T , T , . . . ,T m ) of tiles is well-defined as T ⊗ T ⊗ · · · ⊗ T m . V Note that identifying wall vertices in a join may introduce either multiple edgesor vertices of degree two. If we are interested in 3-connected graphs, we maysuppress vertices of degree two, but we keep the multiple edges.We have the following simple observation.
Observation . Let ( T , T , . . . , T m ) be a compatible sequence T of tiles.Then tcr( ⊗T ) ≤ m (cid:88) i =0 tcr( T i ) . An important operation on tiles that we need converts a tile into a graph.
Definition . (1) A tile T is cyclically compatible if T is compatiblewith itself.(2) For a cyclically-compatible tile T , the cyclization of T is the graph ◦ T obtained by identifying the respective vertices of the left wall with theright wall. A cyclization of a cyclically-compatible sequence of tiles isdefined as ◦T = ◦ ( ⊗T ).The following useful observation is easy to prove. Typically, we will apply thisto the tile ⊗T obtained from a compatible sequence T of tiles. Lemma
8, 28 ]) . Let T be a cyclically compatible tile. Then cr( ◦ T ) ≤ tcr( T ) . We now describe various operations that turn one tile into another.
Definition . (1) For a sequence ω , ¯ ω denotes the reversed sequence.(2) • The right-inverted tile of a tile T = ( G, λ, ρ ) is the tile T (cid:108) = ( G, λ, ¯ ρ ); • the left-inverted tile is (cid:108) T = ( G, ¯ λ, ρ ); • the inverted tile is (cid:108) T (cid:108) = ( G, ¯ λ, ¯ ρ ); and • the reversed tile is T ↔ = ( G, ρ, λ ). ( T ↔ made an item.)(3) A tile T is k -degenerate if T is planar and, for every edge e of T ,tcr( T (cid:108) − e ) < k .Note that our k -degenerate tiles are not necessarily perfect, as opposed to thedefinition in [ ]. However, the following analogue of [ , Cor. 8] is still true. Lemma . Let T = ( T , . . . , T m ) , m ≥ , be a cyclically-compatible sequenceof k -degenerate tiles. Then ⊗ ( T ) is a k -degenerate tile. Proof.
By Lemma 2.5, ⊗T is planar. Let e be any edge of ⊗T . Let T i bethe tile of T containing e . Let T (cid:48) = ( T , . . . , T i − , T i (cid:108) − e, (cid:108) T i +1 (cid:108) , . . . , (cid:108) T m (cid:108) ), so ⊗T (cid:48) = ⊗T (cid:108) − e ; in particular, they have the same tile crossing number. As T i (cid:108) is k -degenerate, tcr( T i (cid:108) − e ) < k . Since all other tiles of T (cid:48) are planar, Lemma 2.5implies tcr( ⊗T (cid:108) − e ) ≤ tcr( T i (cid:108) − e ) < k . (cid:3) The following is an obvious corollary.
Corollary . Let T be a k -degenerate tile so that cr( ◦ ( T (cid:108) )) ≥ k . Then ◦ ( T (cid:108) ) is a k -crossing-critical graph. Definition . (1) T is a compatible sequence ( T , T , . . . , T m ), then: • the reversed sequence T ↔ is the sequence ( T ↔ m , T ↔ m − , . . . , T ↔ ); . DESCRIPTION OF 2-CROSSING-CRITICAL GRAPHS WITH V • the i -flip T i is the sequence ( T , . . . , T i (cid:108) , (cid:108) T i +1 , T i +2 . . . , T m ); and • the i -shift T i is the sequence ( T i , . . . , T m , T , . . . , T i +1 ).(2) Two sequences of tiles are equivalent if one can be obtained from the otherby a series of shifts, flips, and reversals.Note that the cyclizations of two equivalent sequences of tiles are the samegraph. Definition . The set S of tiles consists of those tiles obtained as combina-tions of two frames , illustrated in Figure 2.1, and 13 pictures , shown in Figure 2.2,in such a way, that a picture is inserted into a frame by identifying the two squares.A given picture may be inserted into a frame either with the given orientation orwith a 180 ◦ rotation (some examples are given in Figure 2.3). Figure 2.1.
The two frames.
Figure 2.2.
The thirteen pictures.We remark that each picture produces either two or four tiles in S ; see Figure2.3 Lemma . Let T be a tile in the set S . Then both T and (cid:108) T i (cid:108) are -degenerate. Proof.
Figure 2.4 shows that all the tiles are planar. The claim for T impliesthe result for (cid:108) T i (cid:108) , so it is enough to prove the result for an arbitrary T ∈ S . Let e be an arbitrary edge of T . We consider cases, depending on whether e is eitherdotted, thin solid, thick solid, thin dashed, or thick dashed in Figure 2.4. Usingthis classification, we argue that tcr( T − e ) <
2. DESCRIPTION OF 2-CROSSING-CRITICAL GRAPHS WITH V Figure 2.3.
Each picture produces either two or four tiles.
Figure 2.4.
The different kinds of edges in the pictures.If e is a dotted edge, then T − e has a wall with a single vertex and tcr( T (cid:108) − e ) =0. If e is a thin solid edge, then there is a 1-tile-drawing of T (cid:108) with two dottededges of T crossing each other.If e is a thick solid edge, then there is a unique thin dashed edge f adjacent to e , and there exists a 1-tile-drawing of T (cid:108) − e with f crossing the dotted edge noton the same horizontal side of T as f .If e is a thin dashed edge, then there is a unique thick dashed edge e (cid:48) suchthat e and e (cid:48) are in the same face of the exhibited planar drawing of T , as wellas a unique dotted edge f , that is not in the same horizontal side of T as e . Forsuch e and e (cid:48) , there exists a 1-tile-drawing of T (cid:108) − e with e (cid:48) crossing f , as well as a . DESCRIPTION OF 2-CROSSING-CRITICAL GRAPHS WITH V T (cid:108) − e (cid:48) with e crossing f . As each thick dashed edge correspondsto at least one thin dashed edge, this concludes the proof. (cid:3) We now define the set of graphs that is central to this work.
Definition . The set T ( S ) consists of all graphs of the form ◦ (( ⊗T ) (cid:108) ),where T is a sequence ( T , (cid:108) T (cid:108) , T , . . . , (cid:108) T (cid:108) m − , T m ) so that m ≥ i = 0 , , , . . . , m , T i ∈ S .The rim of an element of T ( S ) is the cycle R that consists of the top andbottom horizontal path in each frame (including the part that sticks out to eitherside) and, if there is a parallel pair in the frame, one of the two edges of the parallelpair.The following is an immediate consequence of Lemmas 2.7 and 2.11. Corollary . Let G ∈ T ( S ) . For every edge e of G , cr( G − e ) < . In Theorem 5.5, we complete the proof that each graph G in T ( S ) is 2-crossing-critical by proving there that cr( G ) ≥ Theorem . If G is a -connected -crossing-critical graph containing asubdivision of V , then G ∈ T ( S ) . This theorem is proved in the course of Chapters 3 – 13. We remark that notevery graph in T ( S ) contains a subdivision of V .HAPTER 3 Moving into the projective plane
It turns out that considering the relation of a 2-crossing-critical graph to itsembeddability in the projective plane is useful. This perspective was employed byRichter to determine all eight cubic 2-crossing-critical graphs [ ]. It is a trivialitythat, if G has a 1-drawing, then G embeds in the projective plane (put the crosscapon the crossing). Therefore, any graph G that does not embed in the projectiveplane has crossing number at least 2. Moreover, Archdeacon [
1, 2 ] proved thatit contains one of the 103 graphs that do not embed in the projective plane butevery proper subgraph does. Each obstruction for projective planar embedding hascrossing number at least 2. Of these, only the ones in Figure 3.1 are 3-connectedand 2-crossing-critical. (The non-projective planar graphs that are not 3-connectedare found by different means in Section 14.) These are the ones labelled — left toright, top to bottom — D17, E20, E22, E23, E26, F4, F5, F10, F12, F13, and G1in Glover, Huneke, and Wang [ ]. Figure 3.1.
The 3-connected, 2-crossing-critical graphs that donot embed in R P . Definition . Let G be a graph embedded in a (compact, connected) surfaceΣ. Then:(1) the representativity rep( G ) of G is the largest integer n so that every non-contractible, simple, closed curve in Σ intersects G in at least n points(this parameter is undefined when Σ is the sphere);(2) G is n -representative if n ≥ r ( G );
10. MOVING INTO THE PROJECTIVE PLANE 11 (3) G is embedded with representativity n if rep( G ) = n .Representativity is also known as face-width and gained notoriety in the GraphMinors project of Robertson and Seymour. We only require very elementary aspectsof this parameter; the reader is invited to consult [ ] or [ ] for further informationon representativity and Graph Minors.Barnette [ ] and Vitray [ ] independently proved that every 3-representativeembedding in the projective plane topologically contains one of the 15 graphs ([ ,Figure 2.2]). Vitray pointed out in a conference talk [ ] that each of these15 graphs has crossing number at least 2. Therefore, any graph that has a 3-representative embedding in the projective plane has crossing number at least 2.One immediate conclusion is that there are only finitely many 2-crossing-criticalgraphs that embed in R P and do not have a representativity at most 2 embeddingin R P , and, not only are there only finitely many of these, but they are all knownand are shown in Figure 3.2. Vitray went on to show that the only 2-crossing-critical graph whose crossing number is not equal to 2 is C (cid:50) C , whose crossingnumber is 3. Figure 3.2.
The 2-crossing-critical 3-representative embeddingsin R P .Since every graph that has an embedding in the projective plane with represen-tativity at most 1 is planar, it remains to explore those 2-crossing-critical graphsthat have an embedding in R P with representativity precisely 2. To cement someterminology and notation, we have the following. Definition . Let n ≥ V n is the M¨obius ladder consisting of: • the rim R of V n , which is a 2 n -cycle ( v , v , v , . . . , v n − , v ); and, • for i = 0 , , , . . . , n − the spoke v i v n + i .Suppose V n ∼ = H ⊆ G . (The notation L ∼ = H means that H is a subdivision of L .Thus, V n ∼ = H ⊆ G means H is a subgraph of G and is also a subdivision of V n .) • The H -nodes are the vertices of H corresponding to v , v , . . . , v n − in V n ; the H -nodes are also labelled v , v , . . . , v n − . • For i = 0 , , , . . . , n −
1, the H -rim branch r i is the path in H corre-sponding to the edge v i v i +1 of V n . • For i = 0 , , , . . . , n −
1, the H -spoke is the path s i in H correspondingto the edge v i v n + i in V n . • We also use H -rim and R for the cycle in H corresponding to the rim of V n .Whenever we discuss elements of a subdivision H of the M¨obius ladder V n ,we presume the indices are read appropriately. For the H -nodes v k and the H -rimbranches r k , the index k is to be read modulo 2 n . For the H -spokes s (cid:96) , the index (cid:96) is to be read modulo n . Thus, for example, s n = s and v n = v , while r n (cid:54) = r .Let G be a 2-crossing-critical graph embedded in R P with representativity 2.Let γ be a simple closed curve in R P meeting G in precisely the two points a and b . We further assume V n ∼ = H ⊆ G , with n ≥
3. Because G − a and G − b have1-representative embeddings in the projective plane, they are both planar. We notethat, for n ≥ V n is not planar; therefore, a, b ∈ H . Remark . Throughout this work, we abuse notation slightly. If K is anygraph and x is either a vertex or an edge of K , then we write x ∈ K , rather thanthe technically correct x ∈ V ( K ) or x ∈ E ( K ). We have taken care so that, in anyinstance, the reader will never be in doubt about whether x is a vertex or an edge.If n ≥
4, the deletion of a spoke of V n leaves a non-planar subgraph; thus,when n ≥
4, we conclude a, b ∈ R . If γ does not cross R at a , say, then deletingthe H -spoke incident with a (if there is one), and shifting γ away from a leaves asubdivision of K , in R P that meets the adjusted γ only at b . But then this K , has a 1-representative embedding in R P , showing K , is planar, a contradiction.Therefore, γ must cross R at a and b . As any two non-contractible curves cross anodd number of times, R is contractible and so bounds a closed disc D and a closedM¨obius strip M .Let P and Q be the two ab -subpaths of R , let α = γ ∩ D and β = γ ∩ M .(We alert the reader that the notations D , M , α , β , and γ will be reserved forthese objects.) Since each spoke is internally disjoint from γ , the spoke is eithercontained in D or contained in M . Since the spokes interlace on R , at most onecan be embedded in D .Moreover, observe that α divides D into two regions, one bounded by P ∪ α andthe other bounded by Q ∪ α . Thus, if a spoke — label it s — is embedded in D ,then s has both attachments in just one of P and Q , say P . In this case, P containseither all the H -nodes v , v , . . . , v n or all the H -nodes v n , v n +1 , . . . , v n − , v . Itfollows that, for n ≥
4, there are only two (up to relabelling) representativity 2embeddings of V n in the projective plane. See Figure 3.3. We remark that it ispossible that one or both of a and b might be an H -node.We introduce a notation that will be used extensively in this work. Definition . The set of 3-connected, 2-crossing-critical graphs is denoted M .It is a tedious (and unimportant) exercise to check the observation that noneof the graphs in M found among the obstructions to having a representativity 2embedding in R P has a subdivision of V . We record it in the following assertion. Theorem . Let G ∈ M and V ∼ = H ⊆ G . Then G has a representativity2 embedding in R P . We will also need information about 1-drawings of V n , for n ≥
4. These aresimilarly straightforward facts that can be proved by considering K , ’s in V n . . MOVING INTO THE PROJECTIVE PLANE 13 s bα βaab v v v v v v v v v v αβ bα βaab αv v v v v v v v v v β Figure 3.3.
Standard labellings of the representativity 2 embed-dings of V . Lemma . Let n ≥ and let D be a 1-drawing of V n . Then there is an i sothat r i crosses one of r i + n − , r i + n , and r i + n +1 . HAPTER 4
Bridges
The notion of a bridge of a subgraph of a graph is a valuable tool that allowsus to organize many aspects of this work. This section is devoted to their definitionand an elucidation of their properties that are relevant to us. Bridges are discussedat length in [ ] and, under the name J -components, in [ ]. Definition . Let G be a graph and let H be a subgraph of G .(1) For a set W of vertices of G , (cid:107) W (cid:107) consists of the subgraph of G withvertex set W and no edges.(2) An H -bridge in G is a subgraph B of G such that either B is an edge notin H , together with its ends, both of which are in H , or B is obtainedfrom a component K of G − V ( H ) by adding to K all the edges fromvertices in K to vertices in H , along with their ends in H .(3) For an H -bridge B in G , a vertex u of B is an attachment of B if u ∈ V ( H );att( B ) denotes the set of attachments of B .(4) If B is an H -bridge, then the nucleus Nuc( B ) of B is B − att( B ).(5) For u, v ∈ V ( G ), a uv -path P in G is H -avoiding if P ∩ H ⊆ (cid:107){ u, v }(cid:107) .(6) Let A and B be either subsets of V ( G ) or subgraphs of G . An AB -path isa path with an end in each of A and B but otherwise disjoint from A ∪ B .If, for example, A is the single vertex u , we write uB -path for { u } B -path.We will be especially interested in the bridges of a cycle. Definition . Let C be a cycle in a graph G and let B and B (cid:48) be distinct C -bridges.(1) The residual arcs of B in C are the B -bridges in C ∪ B ; if B has at leasttwo attachments, then these are the maximal B -avoiding subpaths of C .(2) The C -bridges B and B (cid:48) do not overlap if all the attachments of B are inthe same residual arc of B (cid:48) ; otherwise, they overlap .(3) The overlap diagram OD ( C ) of C has as its vertices the C -bridges; two C -bridges are adjacent in OD ( C ) precisely when they overlap.(4) The cycle C has bipartite overlap diagram , denoted BOD , if OD ( C ) isbipartite; otherwise, C has non-bipartite overlap diagram , denoted NBOD .The following is easy to see and well-known.
Lemma . Let C be a cycle in a graph G . The distinct C -bridges B and B (cid:48) overlap if and only if either:(1) there are attachments u, v of B and u (cid:48) , v (cid:48) of B (cid:48) so that the vertices u, u (cid:48) , v, v (cid:48) are distinct and occur in this order in C (in which case B and B (cid:48) are skew C -bridges ); or(2) att( B ) = att( B (cid:48) ) and | att( B ) | = 3 (in which case B and B (cid:48) are ).
14. BRIDGES 15
The following concept plays a central role through the next few sections of thiswork.
Definition . Let C be a cycle in a graph G and let B be a C -bridge. Then B is a planar C -bridge if C ∪ B is planar. Otherwise, B is a non-planar C -bridge .Note that there is a difference between C ∪ B being planar and, in some em-bedding of G in R P , C ∪ B being plane , that is, embedded in some closed disc in R P . If C ∪ B is plane, then B is planar, but the converse need not hold.We now present the major embedding and drawing results that we shall use.The theorem is due to Tutte, while the corollary is the form that we shall frequentlyuse. Theorem . [ , Theorems XI.48 and XI.49] Let G be a graph.(1) G is planar if and only if either G is a forest or there is a cycle C of G having BOD and all C -bridges planar.(2) G is planar if and only if, for every cycle C of G , C has BOD. For the corollary, we need the following important notion.
Definition . Let H be a subgraph of a graph G and let D be a drawing of G in the plane. Then H is clean in D if no edge of H is crossed in D . Corollary . Let G be a graph and let C be a cycle with BOD. If there isa C -bridge B so that every other C -bridge is planar and there is a 1-drawing of C ∪ B in which C is clean, then cr( G ) ≤ . Proof.
Let × denote the crossing in a 1-drawing D of C ∪ B in which C is clean.As C is not crossed in D , × is a crossing of two edges of B . Let G × denote the graphobtained from G by deleting those two edges and adding a new vertex adjacent tothe four ends of the deleted edges. Then C has BOD in G × and every C -bridgein G × is planar. By Theorem 4.5 (2), G × is planar. Any planar embedding of G × easily converts to a 1-drawing of G .We will also need the following result. Lemma . Let G be a graph, C a cycle in G , B a setof non-overlapping C -bridges. Let P and Q be disjoint paths in C , with V ( C ) = V ( P ∪ Q ) . Suppose that each B ∈ B has at least one attachment in each of P and Q . Let P B and Q B be the minimal subpaths of P and Q , respectively, containing P ∩ B and Q ∩ B , respectively. Then:(1) the { P B } and { Q B } are pairwise internally disjoint and there is an order-ing ( B , . . . , B k ) of B so that P = P B . . . P B . . . P B i . . . P B k and Q = Q B . . . Q B . . . Q B i . . . Q B k ; and(2) if, for each B, B (cid:48) ∈ B , att( B ) (cid:54) = att( B (cid:48) ) , the order is unique up to inver-sion. Proof.
Suppose
B, B (cid:48) ∈ B are such that P B and P B (cid:48) have a common edge e .Then B and B (cid:48) have attachments x , x , x (cid:48) , x (cid:48) in both components of P − e andattachments x, x (cid:48) in Q . If |{ x , x (cid:48) , x , x (cid:48) , x, x (cid:48) }| = 3, then they have 3 commonattachments and so overlap, a contradiction. Otherwise, some y ∈ { x (cid:48) , x (cid:48) , x (cid:48) } isnot in { x , x , x } . Then y is in one residual arc A of x , x , x in C and not both ofthe other two of { x (cid:48) , x (cid:48) , x (cid:48) } are in A . So again B, B (cid:48) overlap, a contradiction fromwhich we conclude P B and P B (cid:48) are internally disjoint.Let C = P − R QR . Suppose B, B (cid:48) ∈ B are such that P = . . . P B . . . P B (cid:48) . . . and Q = . . . Q B (cid:48) . . . Q B . . . . We claim that either P B = P B (cid:48) or Q B = Q B (cid:48) . If not,then there is an attachment u P of one of B and B (cid:48) in P that is not an attachmentof the other and likewise an attachment u Q of one of B and B (cid:48) in Q that is notan attachment of the other. Note that u P and u Q are not attachments of the sameone of B and B (cid:48) , as otherwise the orderings in P and Q imply B and B (cid:48) overlap.For the sake of definiteness, we assume u P ∈ att( B ), so that u Q ∈ att( B (cid:48) ).Let w P ∈ att( B (cid:48) ) ∩ P and let w Q ∈ att( B ) ∩ Q . The ordering of B and B (cid:48) in P and Q imply that, in C , these vertices appear in the cyclic order w P , u P , u Q , w Q .Since u P , u Q , w P , w Q are all different, we conclude that B and B (cid:48) overlap on C , acontradiction.It follows that, by symmetry, we may assume P B = P B (cid:48) . As P B and P B (cid:48) areinternally disjoint, they are just a vertex. So if P = . . . P B . . . P B (cid:48) . . . and Q = . . . Q B (cid:48) . . . Q B . . . , we may exchange P B and P (cid:48) B , to see that P = . . . P B (cid:48) . . . P B . . . and Q = . . . Q B (cid:48) . . . Q B . . . . We conclude there is an ordering of B as claimed.Let ( B , . . . , B k ) and ( B π (1) , . . . , B π ( k ) ) be distinct orderings so that P = P B , . . . , P B k , P = P B π (1) , . . . , P B π ( k ) , Q = Q B . . . Q B k and Q = Q B π (1) , . . . , Q B π ( k ) .There exist i < j so that π ( i ) > π ( j ). We may choose the labelling ( P versus Q )so that the preceding argument implies that P B i = P B j = u . If Q B i = Q B j , then Q B i = Q B j = w and att( B i ) = att( B j ), which is (2). Therefore, we may assumethere is an attachment y of one of B i and B j that is not an attachment of theother. Let z be an attachment of the other. Since Q is either ( Q , y, Q , z, Q ) or( Q − , z, Q − , y, Q − ), the only possibility is that π is the inversion ( k, k − , . . . , Quads have BOD
There are two main results in this section. One is to show that each graph inthe set T ( S ) is 2-crossing-critical and the other, rather more challenging and centralto the characterization of 3-connected 2-crossing-critical graphs with a subdivisionof V , is to show that all H -quads and some H -hyperquads have BOD. We startwith the definition of quads and hyperquads. Definition . Let G be a graph and V ∼ = H ⊆ G .(1) For a path P and distinct vertices u and v in P , [ uP v ] denotes the uv -subpath of P , while [ uP v (cid:105) denotes [ uP v ] − v , (cid:104) uP v ] is [ uP v ] − u , and (cid:104) uP v (cid:105) is (cid:104) uP v ] − v .(2) When concatenating a uv -path P with a vw -path Q , we may write either P Q or [ uP vQw ]. If u = w and P and Q are internally disjoint, thenboth P Q and [ uP vQu ] are cycles. The reader may have to choose theappropriate direction of traversal of either P or Q in order to make theconcatenation meaningful.(3) If L is a subgraph of G and P is a path in G , then L − (cid:104) P (cid:105) is obtainedfrom L by deleting all the edges and interior vertices of P . (In particular,this includes the case P has length 1, in which case L − (cid:104) P (cid:105) is just L lessone edge.)(4) For i = 0 , , , ,
4, the H -quad Q i is the cycle r i s i +1 r i +5 s i .(5) For i = 0 , , , ,
4, the H -hyperquad Q i is the cycle ( Q i − ∪ Q i ) − (cid:104) s i (cid:105) .(6) The M¨obius bridge of Q i is the Q i -bridge M Q i in G such that H ⊆ Q i ∪ M Q i .(7) The M¨obius bridge of Q i is the Q i -bridge M Q i in G for which ( H − (cid:104) s i (cid:105) ) ⊆ Q i ∪ M Q i .The following notions will help our analysis. Definition . Let G be a graph, V n ∼ = H ⊆ G , n ≥
3, and let K be asubgraph of G . Then:(1) a claw is a subdivision of K , with centre the vertex of degree 3 and talons the vertices of degree 1;(2) an { x, y, z } -claw is a claw with talons x , y , and z ;(3) an open H -claw is the subgraph of H obtained from a claw in H consistingof the three H -branches incident with an H -node, which is the centre ofthe open H -claw, but with the three talons deleted;(4) K is H -close if K ∩ H is contained either in a closed H -branch or in aopen H -claw.(5) A cycle C in K is a K -prebox if, for each edge e of C , K − e is not planar.The following is elementary but not trivial.
178 5. QUADS HAVE BOD
Lemma . Let C be an H -close cycle, for some H ∼ = V . Then C is a ( C ∪ H ) -prebox. Proof.
For e ∈ E ( C ), if e / ∈ H , then evidently ( C ∪ H ) − e contains H , which isa V ; therefore ( C ∪ H ) − e is not planar. So suppose e ∈ H . Since C is H -close, C ∩ H is contained in either a closed H -branch b or an open H -claw Y . There isan H -avoiding path P in C − e having ends in both components of either b − e or Y − e . In the former case, ( H − e ) ∪ P , and hence ( C ∪ H ) − e , contains a V . Inthe latter case, ( Y − e ) ∪ P contains a different claw that has the same talons as Y , so again ( H − e ) ∪ P , and ( C ∪ H ) − e , contains a V . Lemma . Let K be a graph and C a cycle of K . If C is a K -prebox, then,in any 1-drawing of K , C is clean. Proof.
Let D be a 1-drawing of K and let e be any edge of C . Since K − e isnot planar, D ( K − e ) has a crossing. It must be the only crossing of D ( K ) and,therefore, e is not crossed in D ( K ).We can now show that any of the tiled graphs described in Section 13 in facthave crossing number 2, thereby completing the proof that they are all 2-crossing-critical. Theorem . If G ∈ T ( S ) , then G ∈ M . Proof.
By Lemmas 2.7 and 2.11 and Corollary 2.8, we know that if K is a propersubgraph of G , then cr( K ) ≤
1. Thus, it suffices to prove that cr( G ) ≥ G is a ∆ -base if it is one of these edges.A ∆ -cycle is a face-bounding cycle in the natural projective planar embedding of G containing precisely one ∆-base. Recall that the rim R of G is described inDefinition 2.12.There are at least three ∆-cycles contained in G and any two are totally disjoint.From each ∆-cycle we choose either of its RR -paths (by definition, these are R -avoiding) as a “spoke”, and, with R as the rim, we find 8 different subdivisionsof V . There are two of these that are edge-disjoint on the spokes, so if D is a1-drawing of G , the crossing must involve two edges of R . Claim . If e is a rim edge in one of the 13 pictures, then e is in an H (cid:48) -closecycle C e , for some H (cid:48) ∼ = V in G .The point of this is that Lemmas 5.3 and 5.4 imply that C e is clean in D .This is also obviously true for the other edges of the rim that are in digons. Theconclusion is that we know the two crossing edges must be from among the ∆-bases.We shall show below that no two of these can cross in a 1-drawing of G , the desiredcontradiction. Proof of Claim 1.
Let e be in edge in the rim R of G that is in the picture T ,let r be the component of T ∩ R containing e , and let r (cid:48) be the other componentof T ∩ R . There is a unique cycle in T − r (cid:48) containing e ; this is the cycle C e . Let e (cid:48) be the one of the two ∆-bases incident with T that has an end in r . Choose the RR -subpath of the e (cid:48) -containing ∆-cycle that is disjoint from r . For any other twoof the ∆-cycles, choose arbitrarily one of the RR -subpaths. These three “spokes”, . QUADS HAVE BOD 19 together with R , constitute a subdivision H (cid:48) of V for which C e is H (cid:48) -close, asrequired.The proof is completed by showing that no two ∆-bases can cross in a 1-drawingof G . If there are at least five tiles, then it is easy to find a subdivision of V sothat the two ∆-bases are on disjoint H -quads and therefore cannot be crossed ina 1-drawing of G . Thus, we may assume there are precisely three tiles and thecrossing ∆-bases e and e are, therefore, in consecutive ∆-cycles.Let T be the picture incident with both e and e . Choose a subdivision H (cid:48) of V containing R but so that T ∩ H (cid:48) = T ∩ R . There is a unique 1-drawing D of H (cid:48) with e and e being the crossing pair. For i = 1 ,
2, let the H (cid:48) -branch containing e i be b i . The end u i of e i that is in T is in the interior of b i .The vertices u and u are two of the four attachments of T in G . Let w and w be the other two, labelled so that w is in the same component of T ∩ R as u .It follows that w is in the same component of T ∩ R as u . In T , there is a uniquepair of totally disjoint R -avoiding u w - and u w -paths P and P , respectively.The crossing in D is of e with e , so [ u b w ] and [ u b w ] are both not crossedin D . Therefore, D [ P ] and D [ P ] are both in the same face F of D .Since the two paths P and P are totally disjoint(text deleted), D [ P ] and D [ P ] are disjoint arcs in F ; the contradiction arises from the fact that their endsalternate in the boundary of F , showing there must be a second crossing.One important by-product of cleanliness is that it frequently shows a cycle hasBOD. Lemma . Let C be a cycle in a graph G . Let D be a 1-drawing of G inwhich C is clean. If there is a non-planar C -bridge, then C has BOD and exactlyone non-planar bridge. Proof.
Let B be a non-planar C -bridge. Then D [ C ∪ B ] has a crossing, and, since C is clean in D , the crossing does not involve an edge of C . Therefore, it involvestwo edges of B . This is the only crossing of D , so inserting a vertex at this crossingturns D into a planar embedding of a graph G × . As C is still a cycle of G × , C has BOD in G × and all C -bridges in G × are planar. But OD G × ( C ) is the same as OD G ( C ) and all C -bridges other than B are the same in G and G × .We shall routinely make use of the following notions. Definition . Let G be a connected graph and let H be a subgraph of G .Then:(1) H is the subgraph of G induced by E ( G ) \ E ( H ); and(2) if G is embedded in R P , then an H -face is a face of the induced embed-ding of H in R P .We will often use this when B is a C -bridge, for some cycle C in a graph G ,in which case B is the union of C and all C -bridges other than B . The followingtwo lemmas are useful examples. Lemma . Let G be a graph embedded in R P with representativity 2 and let γ be a non-contractible curve in R P so that G ∩ γ = { a, b } . Let C be a contractiblecycle in G and let B be a C -bridge so that Nuc( B ) ∩{ a, b } (cid:54) = ∅ . Then B is planar. Proof.
This is straightforward: B = G − Nuc( B ) ⊆ G − ( { a, b } ∩ Nuc B ) and thelatter has a representativity at most 1 embedding in R P . Therefore it is planar.The following result, when combined with the (not yet proved) fact that H -quads and some H -hyperquads have BOD, yields the fact, often used in the sectionsto follow, that deleting some edge results in a 1-drawing in which a particular H -quad or H -hyperquad must be crossed. Lemma . Let G be a graph with cr( G ) ≥ and let C be a cycle in G . If C has BOD in G , then, for any planar C -bridge B , C is crossed in any 1-drawing of B . Proof.
Suppose there is a 1-drawing D of B with C clean. Since C has BODand G is not planar, there is a non-planar C -bridge B (cid:48) . Because C is clean, anycrossing in D [ C ∪ B (cid:48) ] involves two edges of B (cid:48) . The only crossing in D involves twoedges of B (cid:48) , so every other C -bridge in B is planar. Since B is planar, it followsfrom Corollary 4.7 that cr( G ) ≤
1, a contradiction.We remark that M Q is a non-planar Q -bridge whenever Q is an H -quad or H -hyperquad. Corollary . Let G ∈ M and V ∼ = H ⊆ G . If the H -quad Q i and H -hyperquad Q j are disjoint, Q j has BOD, and there is a planar Q j -bridge B , then Q i has BOD and there is precisely one non-planar Q i -bridge. Proof.
Let B be a planar Q j -bridge. Because G is 2-crossing-critical, there is a1-drawing D of B . By Lemma 5.9, Q j is crossed in D . Note that H − (cid:104) s j (cid:105) ⊆ B . In any 1-drawing of H − (cid:104) s j (cid:105) in which Q j is crossed, the crossing is between r j − ∪ r j − ∪ r j ∪ r j +1 and r n + j − ∪ r n + j − ∪ r n + j ∪ r n + j +1 . Since Q i is edge-disjointfrom these crossing rim segments, Q i is clean in D .The two graphs OD G ( Q i ) and OD B ( Q i ) are isomorphic: the Q i -bridges inboth G and B are the same, except M Q i in G becomes M Q i − Nuc( B ) in B andthey have the same attachments. Since Q i is clean in D , OD B ( Q i ) is bipartite.Furthermore, the crossing in D is between two edges of Q j , so D shows that every Q i -bridge other than M Q i is planar.We next introduce boxes, which are cycles that, it turns out, cannot exist ina 2-crossing-critical graph G . On several occasions in the subsequent sections, weprove a result by showing that otherwise G has a box. Definition . Let C be a cycle in a graph G . Then C is a box in G if C has BOD in G and there is a planar C -bridge B so that C is a B -prebox. Lemma . Let G ∈ M . Then G has no box. Proof.
Suppose C is a box in G . Then C has BOD and there is a planar C -bridge B so that C is a B -prebox. As B is a proper subgraph of G , there is a 1-drawing D of B . By Lemma 5.4, D [ C ] is clean. This contradicts Lemma 5.9.We can now determine the complete structure of a 2-connected H -close sub-graph. . QUADS HAVE BOD 21 Lemma . Let G ∈ M and V n ∼ = H ⊆ G with n ≥ . If K is a 2-connected H -close subgraph of G , then K is a cycle. Proof. If K ∩ H consists of at least two vertices, then we include in K the minimalconnected subgraph of the H -branch or open H -claw containing K ∩ H . Since K is H -close, there is a K -bridge M K in G so that H ⊆ K ∪ M K . Let e be an edgeof any H -spoke totally disjoint from K . Note that M K − e is a K -bridge in G − e and that M K has the same attachments in G as M K − e has in G − e .Since K is 2-connected, every edge of K is in an H -close cycle contained in K . Thus, for any 1-drawing D of G − e , Lemmas 5.3 and 5.4 imply that D [ K ] isclean. There is a face F of D [ K ] containing D [ M K − e ]. As D [ K ] is clean and K is 2-connected, F is bounded by a cycle C of K .Lemma 5.3 implies the cycle C is a ( C ∪ H )-prebox. If K is not just C , thenthere is a C -bridge B contained on the side of D [ C ] disjoint from M K . Evidently B is a planar C -bridge.Lemma 5.6 implies C has BOD. Since C is a ( C ∪ H )-prebox, C is a B -prebox.We conclude that C is a box, contradicting Lemma 5.12. This shows that K = C .The second of the following two corollaries is used several times later in thiswork. We recall from Definition 4.1 that, for a set W of vertices, (cid:107) W (cid:107) is thesubgraph with vertex set W and no edges. Corollary . Let G ∈ M , let V n ∼ = H ⊆ G with n ≥ , let B be an H -bridge.(1) If x, y ∈ att( B ) are such that (cid:107){ x, y }(cid:107) is H -close, then there is a unique H -avoiding xy -path in G .(2) There do not exist vertices x, y, z ∈ att( B ) so that (cid:107){ x, y, z }(cid:107) is H -close. Proof.
Suppose P and P are distinct H -avoiding xy -paths. There is either aclosed H -branch or an open H -claw containing an xy -path; this subgraph of H contains a unique xy -path P . Then P ∪ P ∪ P is a 2-connected H -close subgraphof G and so, by Lemma 5.13, is a cycle. But it contains three distinct xy -paths, acontradiction.For the second point, suppose by way of contradiction that such x, y, z exist.Let Y be an { x, y, z } -claw in B . There is a minimal connected subgraph Z of H contained either in a closed H -branch or in an open H -claw and containing x , y , and z . We note that Z is either a path or an { x, y, z } -claw. Thus, Y ∪ Z is2-connected and is H -close. It is a cycle by Lemma 5.13, but the centre of Y hasdegree 3 in Y ∪ Z , a contradiction. Corollary . Let G ∈ M , let V ∼ = H ⊆ G , and let B be a Q -local H -bridge, for some H -quad Q . If s is an H -spoke and r is an H -rim branch, bothcontained in Q , then | att( B ) ∩ s | ≤ and | att( B ) ∩ ( Q − [ r ]) | ≤ . Proof.
The first claim follows immediately from Corollary 5.14. For the second,suppose there are three such attachments x , y , and z . Corollary 5.14 implies theyare not all in the other H -rim branch r (cid:48) of Q , so at least one of x , y , and z is inthe interior of some H -spoke of Q .Suppose first that some H -spoke s in Q is such that (cid:104) s (cid:105) ∩ { x, y, z } = ∅ . Thenlet H (cid:48) = H − (cid:104) s (cid:105) , let B (cid:48) be the H (cid:48) -bridge containing B , and let r (cid:48) and s (cid:48) be the two H -branches in Q other than r and s . Then x , y , and z are all attachments of B (cid:48) and they are all in the same open H (cid:48) -claw containing ( r (cid:48) ∪ s (cid:48) ) − r , contradictingCorollary 5.14.Otherwise, we may suppose both H -spokes s and s (cid:48) in Q have one of x , y , and z in their interiors. We may suppose s has no other one of x , y and z . Choosethe labelling so that x ∈ (cid:104) s (cid:105) . Let r (cid:48) be the H -rim branch in Q other than r andagain let H (cid:48) = H − (cid:104) s (cid:105) and B (cid:48) be the H (cid:48) -bridge containing B . Then y and z areattachments of B (cid:48) , as is the H -node in s ∩ r (cid:48) . But now these three attachments of B (cid:48) contradict Corollary 5.14.We want to find cycles having BOD in our G ∈ M that is embedded withrepresentativity 2 in the projective plane. The following will be helpful. Lemma . Let G be a graph embedded in R P and let C be a contractiblecycle in G . Suppose B is a C -bridge so that C ∪ B has no non-contractible cyclesand let F be the C -face containing B . If B (cid:48) is another C -bridge embedded in F ,then B and B (cid:48) do not overlap on C . Proof.
Let x and y be any distinct attachments of B and let P be a C -avoiding xy -path in B . Then C ∪ P has three cycles, all contractible by hypothesis. Weclaim that one bounds a closed disc ∆ so that C ∪ P ⊆ ∆. If P is contained in thedisc ∆ bounded by C , then we are done. In the remaining case, let C (cid:48) be one ofthese cycles containing P . If the closed disc ∆ (cid:48) bounded by C (cid:48) contains C , then weare done. Otherwise, ∆ ∩ ∆ (cid:48) is a path in C and then ∆ ∪ ∆ (cid:48) is the desired closeddisc.As no other C -bridge in F can have attachments in the interiors of both thetwo xy -subpaths of C and, therefore, there is no C -bridge embedded in F that isskew (see Lemma 4.3 (1)) to B .Likewise, if x, y, z are three distinct attachments of B , then there is a disc ∆ (cid:48) containing the union of C with a C -avoiding { x, y, z } -claw in B . This disc showsthat no other C -bridge embedded in F can have all of x, y, z as attachments and,therefore, no C -bridge embedded in F is 3-equivalent (see Lemma 4.3 (2)) to B .The following is an immediate consequence of Lemma 5.16 and the fact that C has only two faces. Corollary . Let G be a graph embedded in R P and let C be a cycle of G bounding a closed disc in R P . If at most one C -bridge B is such that C ∪ B contains a non-contractible cycle, then C has BOD and, for every other C -bridge B (cid:48) , C ∪ B (cid:48) is planar. The following result is surprisingly useful in later sections.
Lemma . Let G ∈ M and suppose G is embedded with representativity 2in the projective plane. Let γ be a non-contractible curve in the projective plane sothat | γ ∩ G | = 2 and let C be a cycle of G so that γ ∩ C = ∅ . If there is a non-planar C -bridge B , then γ ∩ G ⊆ B , C has BOD, and, for every other C -bridge B (cid:48) , C ∪ B (cid:48) is planar. Proof.
Let a and b be the two points in γ ∩ G . We note that G − a and G − b areplanar, as they have representativity 1 embeddings in R P . Thus, if, for example, a / ∈ B , then C ∪ B ⊆ G − a and so C ∪ B is planar, a contradiction. . QUADS HAVE BOD 23 If B (cid:48) is any other C -bridge, then a, b / ∈ C ∪ B (cid:48) and, therefore, C ∪ B (cid:48) is disjointfrom γ . Since any non-contractible cycle must intersect γ , C ∪ B (cid:48) has no non-contractible cycles. The result is now an immediate consequence of Corollary 5.17.Here is a simple result that we occasionally use. Lemma . Suppose G ∈ M and V n ∼ = H ⊆ G , with n ≥ . Let B be an H -bridge.(1) Then | att( B ) | ≥ .(2) If | att( B ) | = 2 , then B is isomorphic to K .(3) If | att( B ) | = 3 , then B is isomorphic to K , . Proof.
Note that att( B ) = B ∩ B and G = B ∪ B . If | att( B ) | ≤
1, then G is not2-connected. If | att( B ) | = 2 and Nuc( B ) has a vertex, then G is not 3-connected.Now suppose | att( B ) | = 3 and B is not isomorphic to K , . Let Y be an att( B )-claw contained in B . As B ∪ Y is a proper subgraph of G , it has a 1-drawing D ; Y is clean in D , as H must be self-crossed. On the other hand, if s is an H -spokedisjoint from B , there is a 1-drawing D of G − (cid:104) s (cid:105) . Again, the crossing in D involves two edges of H − (cid:104) s (cid:105) , so B is clean. We can substitute D [ B ] for D [ Y ] toconvert D into a 1-drawing of G , a contradiction.The following lemma is the last substantial one we need before proving thatevery H -quad has BOD. Lemma . Let G be a graph that is embedded in R P and let C be a cycleof G . Let B be a C -bridge so that Nuc( B ) contains a non-contractible cycle. Then C is contractible, C has BOD, and every C -bridge other than B is planar. Proof.
Let N be a non-contractible cycle in Nuc( B ) and let B (cid:48) be a C -bridgedifferent from B . Then C ∪ B (cid:48) is disjoint from N . Since any two non-contractiblecycles in R P intersect, C ∪ B (cid:48) does not contain a non-contractible cycle. Clearlythis implies C is contractible and the remaining items are an immediate consequenceof Corollary 5.17.We prove below that every H -quad has BOD and that at least two hyperquadshave BOD. A standard labelling of the embedded V will help make the details ofthe statement comprehensible. We have seen that, up to relabelling, there are tworepresentativity 2 embeddings of V in R P . There is a simple non-contractiblecurve γ in R P meeting G in two points a and b . These are both in the rim R of H and either none or one of the H -spokes is outside the M¨obius band M boundedby R . Let α and β be the two ab -subarcs of γ , labelled so that β ⊆ M . Definition . Let G be a graph and let V ∼ = H ⊆ G . If G is embeddedin R P so that one H -spoke is not in M , then H has an exposed spoke and theexposed spoke is the H -spoke not in M .In this case, the standard labelling is chosen so that the exposed spoke is s and so that v , v , v , v , v , v are all incident with one of the two faces of H ∪ γ incident with s .The faces of H ∪ γ are bounded by the cycles:(1) [ a, r , v ] s [ v , r , b, α, a ]; (2) r r r r r s ;(3) [ a, r , v ] r s [ v , r , b, β, a ];(4) Q , Q , Q ;(5) r [ v , r , b, β, a, r , v ] s ; and(6) [ b, r , v ] r r r [ v , r , a, α, b ].This case is illustrated in the diagram to the left in Figure 3.3.In the case all the H -spokes are in M , the labelling of H may be chosen so thatthe faces of H ∪ γ are bounded by:(1) [ a, r , v ] r r r r [ v , r , b, α, a ];(2) [ a, r , v , s , v , r , b, β, a ];(3) Q , Q , Q , Q ;(4) [ v , r , b, β, a, r , v , s , v ]; and(5) [ b, r , v ] r r r r [ v , r , a, α, b ].This case is illustrated in the diagram to the right in Figure 3.3.We need one more technical lemma before the main result of this section. Lemma . Let G ∈ M , let V ∼ = H ⊆ G , and let i, j ∈ { , , , , } be suchthat Q i and Q j have precisely one H -spoke in common. If Q i has BOD and s i isin a planar Q i -bridge, then ( M Q j ) is planar. Proof.
Let e be any edge of s i and let D be a 1-drawing of G − e . By Lemma 5.9, Q i is crossed in D . Thus, the crossing of D involves an edge of M Q j , showing that( M Q j ) is planar.The following is the main result of this section. Theorem . Let G ∈ M and V ∼ = H ⊆ G . Let G be embedded withrepresentativity 2 in the projective plane, with the standard labelling. Then:(1) each H -quad Q of G has BOD and exactly one non-planar bridge;(2) Q has BOD;(3) for each i ∈ { , , , } , ( M Q i ) is planar;(4) if there is an exposed spoke, then Q has BOD;(5) if there is no exposed spoke, then at least one of Q and Q has BOD.(6) if there is no exposed spoke and Q does not have BOD, then there is a Q -bridge B different from M Q so that B ⊆ D and either:(a) a = v and B has an attachment at a , an attachment in r r , and att( B ) ⊆ { a } ∪ r r ; or(b) b = v and B has an attachment at b , an attachment in r r , and att( B ) ⊆ { b } ∪ r r . (The analogous statement holds for Q in placeof Q .) The following definitions will be useful throughout the remainder of this work.
Definition . Let G be a graph embedded in R P and let C be a cycle of G bounding a closed disc ∆ in R P . A C -bridge B is C -interior if B is containedin ∆ and C -exterior otherwise. Proof of Theorem 5.23.
We distinguish two cases.
Case 1: H has an exposed spoke. . QUADS HAVE BOD 25 We adopt the standard labelling, so s is the exposed spoke. We note that Q is disjoint from G ∩ γ and, therefore, Lemma 5.18 implies Q has BOD and preciselyone non-planar bridge, which is part of (1).The arguments for Q , Q , Q , Q are all analogous and so we do Q . Since s is exposed, the cycle [ a, r , v ] s r s [ v , r , a ] is not contractible and is disjointfrom Q . Lemma 5.20 shows Q has BOD and precisely one non-planar bridge,proving (2) and (4). We have also proved (3) for j = 3 and (1) for Q and Q .To complete the proof of (1) in Case 1, it remains to deal with Q and Q . Thesetwo cases are symmetric and so it suffices to prove Q has BOD and only one non-planar bridge. We note that Q is completely disjoint from Q and we have shownthat Q has BOD. Let B be the Q -bridge containing s . As Q is contractible and B is Q -interior, we conclude that B is planar. Therefore, Corollary 5.10 implies Q has BOD, and each Q -bridge except M Q is planar, as required for (1).For (3), it remains to prove that, for j ∈ { , , } , ( M Q j ) is planar. We applyLemma 5.22: for j = 0 or 4, we take i = 2; for j = 1, we take i = 3. In all cases,the result follows. Case 2: H has no exposed spoke. Lemma 5.18 shows Q , Q , and Q all have BOD and just one non-planarbridge. This proves (2) and part of (1). We use this in Corollary 5.10 to see that Q has BOD and just one non-planar bridge, another part of (1). Also, taking i = 2and j ∈ { , } in Lemma 5.22, we see that ( M Q j ) is planar, part of (3).If Q has BOD, then Corollary 5.10 implies Q has BOD, so in order to show Q has BOD, we may assume Q has NBOD. There is an analogous situation for Q and Q . We first prove (6) for Q ; we will use this to prove both Q has BODand (5).If v (cid:54) = b and v (cid:54) = a , then Lemma 5.18 shows that Q has BOD and exactlyone non-planar bridge. So suppose either (or both) v = b or v = a . If every Q -bridge other than M Q has only contractible cycles, then Q has BOD by Corollary5.17. Thus, some Q -bridge B other than M Q is such that Q ∪ B contains anon-contractible cycle. Evidently, B is Q -exterior. If B ⊆ M , then again Q ∪ B has only contractible cycles. Thus, B ⊆ D .Any Q -exterior bridge B contained in the face of H ∪ γ bounded by[ a, r , v ] r r r [ v , r , b, α, a ]has all its attachments in { a } ∪ r r . Note that B is planar; moreover, if a is notan attachment, then Q ∪ B has no non-contractible cycle and, therefore, does notoverlap any other Q -exterior bridge. We have the analogous conclusions if B iscontained in the face of H ∪ γ bounded by [ b, r , v ] r r r [ v , r , a, α, b ].We conclude that either B has a as an attachment and also has an attachmentin r r or, symmetrically, B has b as an attachment and also has an attachment in r r . This proves (6).We now prove (5). If { v , v } ∩ { a, b } = ∅ , then Q has BOD and just onenon-planar bridge; likewise if { v , v } ∩ { a, b } = ∅ , then Q has BOD and just onenon-planar bridge. Up to symmetry, the only other possibility is that v = a and v = b .Now suppose that Q also has NBOD. Then (6) implies that there must be, upto symmetry, a Q -bridge B different from M Q having attachments at a and in r r . Likewise, there is an H -bridge B different from M Q having attachments at b and in r r . As B cannot have an attachment at b , B (cid:54) = B . Considering theembedding of G in R P , we see that both B and B must be embedded in the faceof H ∪ γ incident with [ b, r , v ] r r r r [ v , r , a, α, b ]. If B , say, has an attachmentother than a and v , then the H -avoiding path in B from b to any attachment in r r crosses B , a contradiction. So att( B ) = { a, v } , att( B ) = { b, v } , and, byLemma 5.19, both B and B are just edges.Now recall that Q has BOD and, letting B be the Q -bridge containing s ,Lemma 5.9 implies Q is crossed in a 1-drawing D of B . The crossing must bebetween the paths r r r r and r r r r .There are two maximal uncrossed subpaths of R in D and we know that v and v are on one uncrossed segment, say S , of R , while v and v are on S . Supposefirst that v is on S . Then the cycle [ v , B , v ] r r r s r separates v from v in D , yielding the contradiction that s is crossed in D . On the other hand, if v is on S , then the same cycle separates v from v , yielding the contradiction that s is crossed in D .We conclude that not both Q and Q can have NBOD which is (5). Bysymmetry, we may assume Q has BOD. Then Lemma 5.22 shows ( M Q ) is planar.Furthermore, Corollary 5.10 implies Q has BOD and precisely one non-planarbridge.What remains is to prove that Q has BOD and precisely one non-planar bridgeand that there is precisely one non-planar Q -bridge. Recall that symmetry impliesthis will show the same things for Q and Q , completing the proofs of (1) and (3).From (6), we may assume that v = a and that there is a Q -bridge B attachingat a and in r r . Let w be any attachment of B in r r , let P be an H -avoiding v w -path in B , and let Q be the subpath of r r joining w to v . Then thecycle [ v , P, w, Q, v , s , v ] is non-contractible in R P and is disjoint from Q . ByLemma 5.20, Q has BOD and has just one non-planar bridge.As for Q , we consider two cases. If Q has BOD, then Lemma 5.22 implies( M Q ) is planar. If Q has NBOD, then (6) implies either v = a or v = b . Inboth cases, Nuc( M Q ) ∩ { a, b } (cid:54) = ∅ , so Lemma 5.8 implies ( M Q ) is planar, asrequired.The following technical corollary of Theorem 5.23 and Lemmas 5.6 and 5.9 willbe used in a few different places later. Corollary . Let G ∈ M and V ∼ = H ⊆ G . With indices read modulo ,suppose, i ∈ { , , , , } is such that Q i has BOD and, where { j, k } = { i +2 , i +3 } ,suppose further that Q j has NBOD. Then s i is in a planar Q i -bridge B i and Q k hasBOD. Moreover, if e i is any edge of B i and D i is a 1-drawing of G − e i , then either r i − r i crosses whichever of r i +3 and r i +6 is in Q j or r i +4 r i +5 crosses whicheverof r i − and r i +1 is in Q j . The two possibilities for D i in the case j = i + 2 are illustrated in Figure 5.1. Proof of Corollary 5.25.
By way of contradiction, suppose s i is not in a planar Q i -bridge. We observe that s must be exposed, as otherwise we have the contra-diction that, for every (cid:96) ∈ { , , , , } , s (cid:96) is in a planar Q (cid:96) -bridge. It follows that,for (cid:96) ∈ { , } , s (cid:96) is in a planar Q (cid:96) -bridge. Thus, i / ∈ { , } . . QUADS HAVE BOD 27 v i +1 v i +2 v i +3 v i +4 v i +6 v i +7 v i +8 v i +9 v i +1 v i +6 v i +4 v i +9 v i +8 v i +3 v i +2 v i +7 Figure 5.1.
The two possibilities for D i when j = i + 2.Let (cid:96) ∈ { , } be such that i and (cid:96) are not consecutive in the cyclic order(0 , , , , e (cid:96) be the edge of s (cid:96) incident with v (cid:96) and let D (cid:96) be a 1-drawing of G − e (cid:96) . By Lemma 5.9, Q (cid:96) is crossed in D (cid:96) .If Q (cid:96) is self-crossed in D (cid:96) , then D (cid:96) shows that the Q i -bridge containing s i isplanar. Thus, we have that Q (cid:96) is not self-crossed in D (cid:96) . One of s (cid:96) − and s (cid:96) +1 isexposed in D (cid:96) . If this exposed spoke is not also in Q i , then again s i is in a planar Q i -bridge; therefore, we must have that the exposed spoke is in Q i . For the sakeof definiteness, we assume that s (cid:96) − is exposed, which implies that (cid:96) = i + 2.As the only non-planar Q i -bridge is M Q i , we must have an H -avoiding path P from the interior of s i to the interior of one of r (cid:96) − r (cid:96) r (cid:96) +1 and r (cid:96) +4 r (cid:96) +5 r (cid:96) +6 . Thedrawing D (cid:96) restricts the possibility to the interior of one of r (cid:96) − r (cid:96) and r (cid:96) +4 r (cid:96) +5 .But now the embedding in R P implies i = 0. This implies j ∈ { , } ; however,neither Q nor Q has NBOD. Therefore, s i is in a planar Q i -bridge.Because M Q j − e i and M Q j have the same attachments, OD G − e i ( Q j ) and OD G ( Q j ) are isomorphic. As the latter is not bipartite, neither is the former. ByLemma 5.6, Q j is not clean in D i . Thus, either r j − r j or r j +4 r j +5 is crossed in D . These are edge-disjoint from Q i .Lemma 5.9 implies that Q i is also crossed in D i . Since Q i is crossed and, fromthe preceding paragraph, something outside of Q i is crossed, either r i − r i crosses r i +3 ∪ r i +6 or r i +4 r i +5 crosses r i − ∪ r i +1 , as required.Since Q always has BOD, Corollary 5.25 implies at least one of Q and Q has BOD. Together with the fact that, in all cases, at least one of Q and Q hasBOD, we conclude that at least three of the H -hyperquads have BOD.The last result in this section will be useful early in the next section. Corollary . Let G ∈ M and let V ∼ = H ⊆ G and suppose G has arepresentativity 2 embedding in the projective plane, with the standard labelling.Suppose, for some i , B is an H -bridge having an attachment in both (cid:104) r i − s i − (cid:105) and (cid:104) r n + i s i +1 (cid:105) . (1) if i (cid:54) = 0 , then B ⊆ D .(2) If i = 0 , then either Q has NBOD or B consists only of the edge v v . Proof.
For (1), we may assume B ⊆ M . The two representativity 2 embeddingsof V in R P show that B can only be embedded in a face bounded by either[ a, r , v ] r s [ v , b, β, a ] or [ b, β, a, r , v ] s r [ v , r , b ] and that s is necessarily ex-posed in R P . Notice that i = 0 in both cases, proving (1).Now assume i = 0 and suppose Q has BOD. From Theorem 5.23, we knowthat Q also has BOD. For j ∈ { , } , let e j be the edge of s j incident with v j andlet D j be a 1-drawing of G − e j . Because s j is in a Q j -interior bridge, from Lemma5.9, we know that Q j is crossed in D j .If Q is clean in D j , then no face of D j is incident with vertices in both (cid:104) r s (cid:105) and (cid:104) s r (cid:105) . Therefore, D j [ B ] cannot be crossing-free in D j , a contradiction. Thus, Q is crossed in D j . The two possibilities for D are shown in Figure 5.2, while thetwo possibilities for D are shown in Figure 5.3. v v v v v v v v v v v v v v v v Figure 5.2.
The two possibilities for D .Let P be an H -avoiding path in G joining a vertex in each of (cid:104) r s (cid:105) and (cid:104) s r (cid:105) . The left-hand version of D has no face incident with both these paths,and so we must have the right-hand version of D . Thus, D implies P has oneend in (cid:104) v , r , v ] and one end in (cid:104) v , s , v ]. The right-hand version of D has noface incident with these paths, so it must be the left-hand version of D . The onlypossibility there for the ends of P are v and v , as claimed. . QUADS HAVE BOD 29 v v v v v v v v v v v v v v v v Figure 5.3.
The two possibilities for D .HAPTER 6 Green cycles
In this section, we begin our study of the rim edges of H . Ultimately, we willpartition them into three types: “green”, “yellow”, and “red”, and it will be thered ones that we focus on to find the desired tile structure. In this section, however,we begin with the study of green edges. We shall show that the cycles C we labelgreen and yellow cannot be crossed in any 1-drawing of H ∪ C . Definition . An edge e of a non-planar graph G is red in G if G − e isplanar.We will eventually prove that every edge of R is either in a green cycle, or in ayellow cycle, or red. The main result in this section, one of the three main steps ofthe entire proof, is that no edge of R is in two green cycles. Definition . Suppose G is a graph and V ∼ = H ⊆ G . Suppose furtherthat G is embedded in R P with representativity 2 and that M is the M¨obius bandbounded by the H -rim R .(1) A cycle C in G is H -green if C is the composition P P P P of four paths,such that:(a) P ⊆ R and P has length at least 1;(b) P P P is R -avoiding;(c) P ∪ P ⊆ H ;(d) P is H -avoiding (and, therefore, is either trivial or contained in an H -bridge); and(e) either(i) P contains at most 3 H -nodes or(ii) P is exceptional , that is, for some i ∈ { , , , . . . , } and in-dices read modulo 10, P = r i r i +1 r i +2 . (2) An edge of R is H -green if it is in an H -green cycle.(3) A vertex v of R is H -green if both edges of R incident with v are in thesame H -green cycle.There is a natural symmetry between P and P : if C is an H -green cycle,consisting of the composition P P P P as in Definition 6.2, then P − P − P − P − is another H -green cycle. Thus P − and P can both be considered to be P . Asthe orientations of the individual P i will not be of any importance (except in asmuch as they are required to make C a cycle), we may say P and P are symmetric.Note that the exceptional case 1(e)ii is the only one in which P has 4 H -nodes. Lemma . Suppose G is a graph and V ∼ = H ⊆ G . Let C be any H -greencycle expressed as the composition P P P P as in Definition 6.2.
30. GREEN CYCLES 31 (1) If i ∈ { , } , then P i has an end in R and is either trivial or contained inan H -spoke.(2) The path P is not trivial.(3) If P and P are both non-trivial, then they are contained in different H -spokes. Proof. (1)
For sake of definiteness, we assume i = 2. If P is not trivial, thenthere is an edge e in P . From the definition, e is in H but not in R . Therefore,there is a spoke s containing e . If P has a vertex u not in s , then P is a pathcontained in H and containing e and u . This implies that one end of s , a vertex of R , is internal to P , contradicting the fact that P P P is R -avoiding. So P ⊆ s ,as required. Since P ⊆ R and P has an end in common with P , P has an endin R . (2) Suppose P is trivial. Then P P is an R -avoiding path joining the ends of P . Each of P and P is either trivial or in a spoke and, since P P is R -avoiding,either both are trivial or P P is contained in a single spoke. If both are trivial,then P is the cycle P P P P , which is impossible, since P is properly containedin the cycle R . Each of P and P has an end in R (or is trivial) and P P hasboth ends in common with P , so P P is the entire spoke. But then P containssix H -nodes, a contradiction. (3) For j = 2 , P j is non-trivial by hypothesis. Therefore, (1) shows it iscontained in an H -spoke s . As it has a vertex in common with P , P j has a vertexin R . This vertex is an H -node incident with s . If P and P are contained in thesame spoke s , then, as in the proof of (2), they contain different H -nodes. But then P contains six H -nodes, contradicting Definition 6.2.There is a small technical point that must be dealt with before we can success-fully analyze the relation of an H -green cycle to the embedding of G in R P . Definition . Let Π be a representativity 2 embedding of a graph G in R P and let V ∼ = H ⊆ G . Then Π is H -friendly if, for each H -green cycle C of G andany non-contractible simple closed curve γ in R P meeting Π( G ) in precisely twopoints, Π[ C ] is contained in the closure of some face of Π[ H ] ∪ γ . Lemma . Suppose G ∈ M and V ∼ = H ⊆ G . Let Π be any representativity2 embedding of G in R P , let γ be a non-contractible simple closed curve in R P meeting Π( G ) in precisely two points, and let C be an H -green cycle in G . Give H the standard labelling relative to γ .(1) Either Π[ C ] is contained in the closure of some face of Π[ H ] ∪ γ or v v isan edge of G embedded in M and C = r r r [ v , v v , v ] . In particular,if Π[ H ] ⊆ M , then Π is H -friendly.(2) If Π is not H -friendly, then there is an H -friendly embedding of G in R P obtained from Π by reembedding only v v .(3) In particular, there is an H -friendly embedding of G in R P . Proof.
Suppose Π[ C ] is not contained in the closure of any face of Π[ H ] ∪ γ and let P P P P be the decomposition of C as in Definition 6.2. As P is ( H ∪ γ )-avoidingand non-trivial by Lemma 6.3 (2), there is an ( H ∪ γ )-face F containing P . Notethat, if P is not trivial, then Lemma 6.3 (1) asserts it is contained in an H -spoke s and it contains an end of P , so P is contained in the boundary of F . Likewisefor P . We assume by way of contradiction that P (cid:54)⊆ cl( F ). Claim . Then:(1) P = r r r ;(2) s is exposed;(3) either a = v or b = v ; and(4) if F ⊆ D , then both v = b and v = a . Proof.
We first consider the case F ⊆ D . Both ends of P are contained inone of the ab -subpaths of R . If P is not contained in the boundary of F , thenit must contain the other complete ab -subpath of R . As each of these has at least4 H -nodes, the only possibility is that it is precisely 4 H -nodes. In this case, P must be exceptional and s must be exposed. In particular, P = r r r and P has ends v and v . The paths P and P are both trivial. Moreover, as P is notincident with F , we must have v = b and v = a .In the other case, F ⊆ M . If F is contained in the interior of an H -quad,then P joins two vertices in the same quad and is not contained in the quad.In this case, P must have at least 5 H -nodes, which is impossible. Therefore, F is not contained in the interior of an H -quad, and so is bounded by one of[ a, r , v ] r s [ v , r , b, β, a ] and [ a, r , v ] s r [ v , r , b, β, a ]. (Recall β = γ ∩ M .)Notice that s is exposed.These cases are symmetric; for sake of definiteness, we presume F is boundedby [ a, r , v ] r s [ v , r , b, β, a ]. The path P has at most 4 H -nodes and joins twovertices on Q . If P ⊆ Q , then Π[ C ] is contained in the closure of one of thetwo (Π[ H ] ∪ γ )-faces whose boundary is contained in Π[ Q ] ∪ γ ; thus, P (cid:54)⊆ Q .Therefore, P has at least 4 H -nodes; by definition it has at most 4, so P hasprecisely 4 H -nodes. In particular, P can only be r r r and v = a . (cid:3) Because s is exposed, Theorem 5.23 implies that both Q and Q have BOD.Let e be any edge in s and let D be a 1-drawing of G − e . Since Q has BOD,Lemma 5.9 shows Q is crossed in D , so r r r r crosses r r r r . This impliesthat neither s nor s is exposed in D and, therefore, P cannot be in the same( H − (cid:104) s (cid:105) )-bridge as s .Let B and B be the ( H − (cid:104) s (cid:105) )-bridges containing s and P , respectively.These evidently overlap on Q and they both overlap M Q − e (in G − e ). Therefore, Q has NBOD. Since M Q − e is a non-planar Q -bridge in G − e , Lemma 5.6 impliesthat Q is not clean in D .As Q and Q have only s in common and both are crossed in D , s mustbe exposed in D . It follows that D [ P ] is in the face of D [ H − (cid:104) s (cid:105) ] bounded by s r r r r r .The same arguments apply with Q in place of Q , showing that D [ P ] is inthe face of D [ H − (cid:104) s (cid:105) ] bounded by s r r r r r . These two drawings implythat att( B ) ⊆ r r r .If F ⊆ D , then F is bounded by r s r [ v , α, v ] (recall α = γ ∩ D ). Thus,att( B ) = { v , v } and Lemma 5.19 implies that P is just the edge v v . In thiscase, Claim 1 implies P can obviously be embedded in the other face of H ∪ γ contained in D and incident with both v and v .If F ⊆ M , then F is bounded by either[ a, r , v ] r s [ v , r , b, β, a ] or [ a, r , v ] s r [ v , r , b, β, a ] . . GREEN CYCLES 33 Again, this implies that att( B ) ⊆ { v , v } , so P is just the edge v v . In this case,Claim 1 implies only that either v = b or v = a . Again these cases are symmetric,so we assume v = a .We remark that if v ∈ A ∩ B , then ( v ) is an AB -path and this is the only pathcontaining v that is an AB -path. We now return to the proof.We wish to reembed v v in the ( H ∪ γ )-face incident with v , v , v , and v .We need only verify that there is no H -avoiding [ b, r , v (cid:105) (cid:104) v , r , v , r , v , r , v ]-path. But such a path would have to appear in D , where it can only also be in theface of D [ H − (cid:104) s (cid:105) ] bounded by s r r r r r . But then it crosses v v in D , acontradiction completing the proof.We are now prepared for our analysis of H -green cycles. Lemma . Let G ∈ M , V ∼ = H ⊆ G , and let Π be an H -friendly embeddingof G in R P . Let C be an H -green cycle expressed as the composition P P P P as in Definition 6.2. Then:(1) P is contained in one of the two ab -subpaths of R ;(2) if C ⊆ M and s is any H -spoke contained in M that is totally disjointfrom C , then C is a ( C ∪ ( H − (cid:104) s (cid:105) )) -prebox;(3) if C is not contained in M and s is any H -spoke contained in M havingone end in the interior of P , then C is a ( C ∪ ( H − (cid:104) s (cid:105) )) -prebox;(4) there is a C -bridge M C so that H ⊆ C ∪ M C ;(5) C is contractible, C has BOD, and all C -bridges other than M C are planar;(6) C is a ( C ∪ H ) -prebox;(7) M C is the unique C -bridge (that is, there are no planar C -bridges);(8) C bounds a face of Π ;(9) there are at most two H -nodes in the interior of P ; and(10) in any 1-drawing of H ∪ C , C is clean. Proof.
Because Π is H -friendly, there is a face F of ( H ∪ γ ) whose closure contains C . (1) This is an immediate consequence of Definition 6.4, as the boundary ∂ ofany face of H ∪ γ has each component of ∂ ∩ R contained in one of the ab -subpathsof R . (2) and (3) Note that H − (cid:104) s (cid:105) contains a subdivision of V . In particular, if e is an edge of C not in R , then H −(cid:104) s (cid:105) is a non-planar subgraph of ( C ∪ ( H −(cid:104) s (cid:105) )) − e ,as required. If e ∈ C is in R , then we claim the cycle R (cid:48) = ( R − (cid:104) P (cid:105) ) ∪ P P P isthe rim of a V . We see this in the two cases. Case 1: (2)
In this case, there are three H -spokes t , t , t other than s containedin M . Each t i has an end v i in R − (cid:104) P (cid:105) and a maximal R (cid:48) -avoiding subpath t (cid:48) i containing v i . It is straightforward to verify that R (cid:48) ∪ t (cid:48) ∪ t (cid:48) ∪ t (cid:48) is a subdivisionof V , as required. Case 2: (3)
In the exceptional case P = r i r i +1 r i +2 , s is different from all of s i , s i +3 , and s i +4 , so R (cid:48) ∪ s i ∪ s i +3 ∪ s i +4 is the required V . (Note that one of s i and s i +3 can be the exposed spoke and part of that spoke might be in either P or P ,but whatever part is not in P ∪ P makes the third spoke.)In the remaining case, there are two H -spokes s i and s i +1 that are completelydisjoint from C . Any other H -spoke s (cid:48) , different from s , s i , and s i +1 , and containedin M , will connect to R (cid:48) to make a third spoke, either because both its ends are in R (cid:48) or because one end is in R (cid:48) and the other end is in P and one of the paths in P − e joins the other end of s to a vertex in R (cid:48) . (4) Let M C be the C -bridge containing the ab -subpath Q of R that is P -avoiding. We claim H ⊆ C ∪ M C . Observe that the maximal P -avoiding subpath Q (cid:48) of R containing Q is contained in M C and, therefore, R ⊆ C ∪ M C . Note thatevery H -spoke has at least one end in Q (cid:48) that is not in P and, therefore, that endis in Nuc( M C ). Thus, if P is not contained in M , it is obvious that H ⊆ C ∪ M C .So suppose P is contained in M . The H -spokes other than those that contain P and P are obviously in M C , and the ones containing P and P are in the unionof M C and C . (5) If either P has at most 3 H -nodes, or s is not exposed, or P is neither r r r nor r r r , then there is an H -spoke s contained in M and totally disjointfrom C . The spoke s combines with the one of the two subpaths of R joining theends of s that is disjoint from P to give a non-contractible cycle disjoint from C .The claim now follows immediately from Lemma 5.20.We now treat the case s is exposed and P is either r r r or r r r . In thiscase, F is a face of H ∪ γ contained in D . Let B (cid:48) be a C -bridge other than M C . If B (cid:48) ⊆ cl( F ), then C ∪ B (cid:48) ⊆ cl( F ) and cl( F ) is a closed disc in R P . Therefore, C ∪ B (cid:48) has no non-contractible cycles in R P . Otherwise, B (cid:48) is contained in the closure ofone of the H -faces bounded by Q or Q or Q . For each i ∈ { , , } , let F i be the H -face bounded by Q i . Then cl( F i ) ∩ cl( F ) is a path and, therefore, cl( F i ) ∪ cl( F )is a closed disc containing C ∪ B (cid:48) and again C ∪ B (cid:48) has no non-contractible cycles.The result now follows from Corollary 5.17. (6) In the case P ⊆ M , at most the H -spokes containing P and P meet C . There are at least two others contained in M that are disjoint from C ; let s be one of these. By (2), for any edge e of C , ( C ∪ ( H − (cid:104) s (cid:105) )) − e is not planar, so( C ∪ H ) − e is not planar.Now suppose P ⊆ D . If some H -spoke s contained in M has an end in theinterior of P , then (3) implies that, for any edge e of C , ( C ∪ ( H − (cid:104) s (cid:105) )) − e is notplanar, so ( C ∪ H ) − e is not planar.In the alternative, no H -spoke contained in M has an end in the interior of P .If e is not in P , then H ∩ M , which is a V or V , is contained in ( C ∪ H ) − e , sowe may assume e ∈ P . But then ( R − (cid:104) P (cid:105) ) ∪ P P P and the H -spokes containedin M make a V or V , showing ( C ∪ H ) − e is not planar. (7) Observe that (5) shows any other C -bridge is planar and that C has BOD.If B is any other C -bridge, then C is a B -prebox by (6) and, therefore, is, bydefinition, a box, contradicting Lemma 5.12. (8) This is an immediate consequence of the facts that C is contractible (5)and there is only one C -bridge (7). (9) Suppose by way of contradiction that v i − , v i , v i +1 are internal to P .Notice that P is not exceptional. We claim that Q i is a box, contradicting Lemma5.12.For s ∈ { s i − , s i , s i +1 } , s is contained in one of the two faces of R (i.e., theM¨obius band M and the disc D ). By (8), C is the boundary of some face F of G . Clearly F and s are in different R -faces, so one is in M and the other is in D . Therefore, all of s i − , s i , and s i +1 are contained in the same one of M and D .Since D contains at most one H -spoke, it must be that all three are contained in M . Clearly, this implies F ⊆ D and, therefore, P P P ⊆ D . . GREEN CYCLES 35 There is another H -spoke s contained in M that is totally disjoint from Q i .As P P P ⊆ D , R ∪ P P P ∪ s contains a non-contractible cycle including both P P P and s that is totally disjoint from Q i . Thus, Lemma 5.20 implies Q i hasBOD and all Q i -bridges except M Q i are planar.We claim Q i is a ( Q i ∪ M Q i )-prebox. Note that Q i ∪ M Q i contains H − (cid:104) s i (cid:105) andso the deletion of any edge in s i − ∪ s i +1 leaves a V . By (3), C is a C ∪ ( H − (cid:104) s i (cid:105) )-prebox, so the deletion of any edge e in r i − ∪ r i leaves a non-planar subgraph in( C − e ) ∪ ( H − (cid:104) s i (cid:105) ), which is contained in ( Q i − e ) ∪ M Q i . That is, if e ∈ r i − ∪ r i ,then ( Q i − e ) ∪ M Q i is not planar.We must also consider an edge in r i +4 ∪ r i +5 (these indices are read modulo10). Let R (cid:48) be the cycle made up of the following four parts: the two paths in R − (cid:104) P (cid:105) − (cid:104) r i +4 r i +5 (cid:105) , P P P , and s i − r i − r i s i +1 . To get the V , add to R (cid:48) both H -spokes totally disjoint from P and either of the two R (cid:48) -avoiding subpaths of P whose ends are in R (cid:48) . Thus, if e ∈ r i +4 r i +5 , then ( Q i − e ) ∪ M Q i is not planar,completing the proof that Q i is a ( Q i ∪ M Q i )-prebox. (See Figure 6.1.) r i +4 r i +6 r i +5 r i − r i r i +1 C D Figure 6.1.
The case e ∈ r i +4 r i +5 for ¯ Q i being a ( ¯ Q i ∪ M ¯ Q i )-prebox. Only two of the three spokes are shown.Since the Q i -bridge B containing s i is contained in the closed disc in R P bounded by Q i , B is planar and, therefore, Q i is a box, the desired contradiction. (10) Let D be a 1-drawing of H ∪ C . Let P P P P be the decomposition of C into paths as in Definition 6.2, so P ⊆ R and P is H -avoiding. If C is crossedin D , then it is P that is crossed, while P P P , being R -avoiding, is not crossedin D . We claim that there is an H -spoke v i v i +5 disjoint from C that is not exposedin D . The existence of s and the fact that C is crossed in D shows that no face of R ∪ s is incident with both ends of P and, therefore, P P P must cross R ∪ s in D , the desired contradiction.To prove the claim, we consider two cases. If P has at most 3 H -nodes, thenthis is obvious, since only one H -spoke can be exposed. In the alternative, P isexceptional, say P = r i r i +1 r i +2 . As the spoke exposed in D is incident with anend of the H -rim branch that is crossed, we see that s i +4 is not the exposed spokeand is disjoint from P , as required.The next result is the main result of this section and the first of three mainsteps along the way to obtaining the classification of 3-connected, 2-crossing-criticalgraphs having a subdivision of V . The other two major steps are, for G ∈ M
326 6. GREEN CYCLES containing a subdivision H of V : (i) G has a representativity 2 embedding in R P so that H ⊆ M ; and (ii) G contains a subdivision of V with additionalproperties (that we call “tidiness”). It is this tidy V for which the partition ofthe edges of the rim into the red, yellow, and green edges that allows us to find thedecomposition into tiles. Theorem . If G ∈ M and V ∼ = H ⊆ G , then no two H -green cycles havean edge of R in common. Proof.
Suppose e ∈ R is in distinct H -green cycles. By Lemma 6.5 (3), there isan H -friendly embedding Π of G in R P . By Lemma 6.6 (8), any H -green cyclebounds a face of Π[ G ]. As e is in R and R is the boundary of both the (closed)M¨obius band M and the (closed) disc D , one of these faces, call it F M , is containedin M , while the other, call it F D , is contained in D . For n ∈ { M , D } , let C n be thegreen cycle bounding F n and let P n P n P n P n be the path decomposition of C n asin Definition 6.2; in particular, P n ⊆ R and P n is H -avoiding.Note P D P D P D is disjoint from M (except for its ends) and P M P M P M iscontained in M . Thus, C D ∩ C M = P D ∩ P M . Lemma 6.6 (9) implies that, for n ∈ { M , D } , P n has at most 4 H -nodes. We conclude that P D ∪ P M is not all of R , and so C D ∩ C M is a path. Therefore, there is a unique cycle C in C D ∪ C M not containing e and, furthermore, C bounds a closed disc in R P having e in itsinterior.On the other hand, Lemma 6.6 (1) shows there is an ab -subpath A of R thatcontains P D . Since e ∈ P D ∩ P M , it is also the case that P M ⊆ A . Let A bethe other ab -subpath of R , so that A is ( C D ∪ C M )-avoiding. In particular, thereis a C -bridge M C containing A . By Lemma 6.6 (7), for n ∈ { M , D } , A is in theunique C n -bridge M C n . Since M C n (and therefore A ) is not contained in the face of G bounded by C n , we conclude that A is not in the disc bounded by C . Therefore, M C is different from the C -bridge B C containing e . Claim . For each H -spoke s , some H -node incident with s is not in C M ∪ C D . Proof.
By Lemma 6.6 (9), there exists an i so that P D ⊆ r i r i +1 r i +2 . Inparticular, e is in r i ∪ r i +1 ∪ r i +2 . Thus, P M has an edge in at least one of r i , r i +1 , and r i +2 .Lemma 6.6 (8) implies that C M bounds a face of G . Therefore, C M is containedin the closure cl( F ) of a face F of Π[ H ] and F ⊆ M . Thus, P M is contained inone of the two components of cl( F ) ∩ R . Since such a component is contained inconsecutive H -rim branches, if P M contains an edge in r j , then P M is containedin either r j − r j or r j r j +1 . From the preceding paragraph, P M is contained in oneof r i − r i , r i r i +1 , r i +1 r i +2 , and r i +2 r i +3 .We conclude that P D ∪ P M is contained in either r i − r i r i +1 r i +2 or r i r i +1 r i +2 r i +3 showing that no H -spoke has both ends in P D ∪ P M . (cid:3) Claim . (1) H ⊆ C ∪ M C ∪ B C .(2) If s is an H -spoke contained in M disjoint from C M , then ( C ∪ M C ) − (cid:104) s (cid:105) is not planar. Proof.
For (1), we note that it is clear that R ⊆ C ∪ M C ∪ B C . Now let s bean H -spoke. Suppose first that s ⊆ M . By Claim 1, there is an H -node v incidentwith s and not in C M ∪ C D . If s ∩ C M is at most an end of s , then it is evident that s ⊆ M C . If s ∩ C M is more than just an end of s , then s consists of a C M -avoiding . GREEN CYCLES 37 subpath s (cid:48) joining v to a vertex in C M , together with the path C M ∩ s (which is byLemma 6.3 (1)) either P M or P M ). But then it is again evident that s ⊆ C ∪ M C .Otherwise, s is exposed, in which case we have the same argument, but replacing C M with C D , completing the proof of (1).For (2), a V is found whose rim is ( R − (cid:10) P M (cid:11) ) ∪ P M P M P M . The spokes arecontained in the three other spokes in M , namely they are the parts that are notin P M ∪ P M . (cid:3) Claim . C has BOD. Proof.
Let S be the set of H -spokes contained in M and disjoint from C M .As C M meets at most two H -spokes in M , | S | ≥
2. If some s ∈ S is also disjointfrom C D , then R ∪ s contains a non-contractible cycle disjoint from C , in whichcase Lemma 5.20 shows C has BOD, as claimed.So we may assume that no element of S is also disjoint from C D . Let s be anyelement of S ; then s ∩ C D is a vertex v of P D . Let e be the edge of s incidentwith v . In order to show that C has BOD, we will show that: (i) the overlapdiagrams OD G − e ( C ) and OD G ( C ) are the same; and (ii) OD G − e ( C ) is bipartite.For (i), note that C D bounds a face in R P and that (cid:104) s (cid:105) is in the boundary of two( H ∪ γ )-faces. Thus, there can be no C -bridge that overlaps M C in G because ofits attachment at v . That is, OD G − e ( C ) and OD G ( C ) are the same.For (ii), Lemma 6.6 (2) applied to C M and (3) applied to C D , combined withLemma 5.4, shows C D and C M are both clean in D e . Therefore, C is clean in D e .By Claim 2 (2), ( C ∪ M C ) − e is not planar, so Lemma 5.6 shows C has BOD in G − e . Therefore, C has BOD in G . (cid:3) Claim . C is a C ∪ H -prebox. Proof.
Note that C D ∪ C M ⊆ C ∪ H . If e ∈ C , then let i ∈ { M , D } besuch that e ∈ C i . Lemma 6.6 (6) says that C i is a ( C i ∪ H )-prebox and, therefore,( C i ∪ H ) − e is not planar. Since ( C i ∪ H ) − e ⊆ ( C ∪ H ) − e , we conclude that C is a ( C ∪ H )-prebox. (cid:3) Claim . G = C ∪ M C ∪ B C . Proof.
By way of contradiction, suppose there is another C -bridge B (cid:48) . Let F be the ( H ∪ γ )-face containing B (cid:48) . Then C ∪ B (cid:48) is contained in the closed discthat is the union of the closure of F and the disc bounded by C , showing B (cid:48) isplanar. By Claim 4 and the fact that C ∪ H ⊆ B (cid:48) , Lemma 5.4 says that C is cleanin a 1-drawing of B (cid:48) , of which there is at least one, since G is 2-crossing-critical.This yields a 1-drawing of C ∪ M C with C clean. By Claim 3, C has BOD, B C is planar because it is contained in the closed disc bounded by C , and above weshowed that every other C -bridge is planar; Corollary 4.7 implies cr ( G ) ≤
1, acontradiction. (cid:3)
We are now on the look-out for a box in G ; it is not true that C is necessarilyone. Our next claim gives a sufficient condition under which we can find some boxand the following two claims show that, in all other cases, C is a box. Claim . Suppose all of the following:(1) there is an i so that P M is in a Q i -local H -bridge;(2) P M contains v i and is a non-trivial subpath of s i ; and (3) v i +2 is in the interior of P D .Then G has a box. Proof.
We note that (2) implies s i ⊆ M . Subclaim . Both s i +1 and s i +2 are contained in M . Proof.
Suppose first that s i +2 is exposed. Then (3) implies P D and P D areboth trivial. That is, C D = P D P D . But P D is H -avoiding and overlaps s i +2 on R (because P D has at most four H -nodes, only two of which can be in the interior of P D ). Thus, P D and s i +2 cross in R P , a contradiction. Therefore, s i +2 ⊆ M .Next, suppose s i +1 is exposed. Then, by symmetry, we may assume i = 4 or i = 9. In either case, P M and P D are in different ab -subpaths of R and so do nothave an edge in common, a contradiction. Hence s i +1 is also contained in M . (cid:3) Let u be the common end of P M and P M and let w be the common end of P M and P M . By (2), u ∈ (cid:104) s i (cid:105) and, by (1) and (2), w ∈ r i . Observe that the edge e common to C M and C D is in [ v i , r i , w ].Let C (cid:48) be the cycle (cid:2) v i +5 , s i , u, P M P M , w, r i , v i +1 (cid:3) r i +1 s i +2 r i +6 r i +5 . We notethat there are two obvious C (cid:48) -bridges: the C (cid:48) -interior bridge B C (cid:48) containing theedge of s i +1 incident with v i +6 ; and the C (cid:48) -exterior bridge M C (cid:48) for which H −(cid:104) s i +1 (cid:105) ⊆ C (cid:48) ∪ M C (cid:48) . To show C (cid:48) is a box, it suffices to show that C (cid:48) has BOD and C (cid:48) is a ( C (cid:48) ∪ M C (cid:48) )-prebox.Notice that v i +2 is in the interior of P D by hypothesis and v i +1 is in the interiorof P D because e ∈ r i . Lemma 6.6 (9) implies that the only H -nodes in the interiorof P D are v i +1 and v i +2 . In particular, v i and v i +3 are in R − (cid:10) P D (cid:11) , as are all theends of s i +3 and s i +4 .To see that C (cid:48) has BOD, we produce a non-contractible cycle in Nuc( M C (cid:48) ).Lemma 5.20 then implies C (cid:48) has BOD and precisely one non-planar bridge. Westart with the two paths P D P D P D and s i +4 , and easily complete the required cycleusing two paths in R , one containing r i +3 and the other containing r i +9 .It remains to show that C (cid:48) is a ( C (cid:48) ∪ M C (cid:48) )-prebox. Since V ∼ = H − (cid:104) s i +1 (cid:105) ⊆ C (cid:48) ∪ M C (cid:48) , it is obvious that, if e ∈ C (cid:48) and e / ∈ R , then ( C (cid:48) ∪ M C (cid:48) ) − e contains a V and so is not planar. So suppose e ∈ C (cid:48) and e ∈ R . There are two cases.If e ∈ r i r i +1 , then take ( R − (cid:10) P D (cid:11) ) ∪ P D P D P D as the rim. We choose asspokes s i , s i +3 , and s i +4 .If e ∈ r i +5 r i +6 , then the rim consists of the two paths P D P D P D and C (cid:48) −(cid:104) r i +5 r i +6 (cid:105) , together with the two subpaths of R joining them, one containing v i +3 , v i +4 , and v i +5 , and the other containing v i +7 , v i +8 , v i +9 , and v i . In this case, thespokes are s i +3 , s i +4 , and P M . (cid:3) In the remaining case, we show that C is a box. The following simple observa-tions get us started, the first being the essential ingredient. Claim . Either:(1) there is an i so that • P M is in a Q i -local H -bridge; • s i contains an edge of C M ; and • v i +2 is in the interior of P D ;or (symmetrically)(2) there is an i so that . GREEN CYCLES 39 • P M is in a Q i -local H -bridge; • s i +1 contains an edge of C M ; and • v i − is in the interior of P D ;or(3) there are three H -spokes not having an edge in C M and not having anincident vertex in the interior of P D . Proof.
Lemma 6.6 (9) implies there are at most two H -nodes in the interiorof P D . Therefore, if no H -spoke contains an edge of C M , then (3) holds. So wemay suppose C M has an edge in some H -spoke.Suppose first that s is exposed, C M has an edge in s and e is in either[ a, r , v , r , v ] or [ b, r , v ]. Therefore, P D has one end in either [ a, r , v , r , v (cid:105) or [ b, r , v (cid:105) . Lemma 6.6 (9) implies at most two H -nodes can be in the interior of P D , so no end of s can be in the interior of P D . We conclude that s , s and s are the required three spokes yielding (3).Symmetry treats the same case on the other side.In the remaining case, P M is contained in a Q i -local H -bridge and both s i and s i +1 are contained in M . The edge e is in either r i or r i +5 . If the only H -nodes inthe interior of P D are incident with either s i or s i +1 , then the other three H -spokessuffice for (3).Thus, by symmetry we may assume an end of s i +2 is in the interior of P D .This implies that an end of s i +1 is also in the interior of P D . Lemma 6.6 (9) showsthese are the only H -nodes in the interior of P D . If s i does not contain an edge of C M , then the three spokes other than s i +1 and s i +2 suffice for (3), while if s i doescontain an edge of C M , then we have (1). (cid:3) Claims 6 and 7 show we need only consider the third possibility in Claim 7 tofind a box.
Claim . If there are three H -spokes not having any edge in C M and nothaving an incident H -node in (cid:10) P D (cid:11) , then C is a box. Proof.
By Claim 3 and the fact that B C is a planar C -bridge, it suffices toshow C is a ( C ∪ M C )-prebox. For each e ∈ C , we show that ( C ∪ M C ) − e containsa V .We note that 3-connection and the fact that C M and C D both bound facesimplies C M ∩ C D is just e and its ends. That is, B C consists of just e and itsends. Thus, Claim 5 implies that G − e = C ∪ M C . In particular, every spoke isin C ∪ M C .Let w be any H -node that is not in C . There are two wC -paths in R − e ;let them be R x with end x ∈ C and R y with end y ∈ C . Thus, R consists of the C -avoiding path R x ∪ R y , a subpath of C , the edge e , and another subpath of C .The cycle C consists of two xy -paths; let them be N D containing P D P D P D and N M containing P M P M P M . We note that N D ⊆ D and N M ⊆ M . Subclaim . Let s be an H -spoke with no edge in C M and not having anincident H -node in (cid:10) P D (cid:11) .(1) If s ⊆ M , then s ∩ C is either empty, x , or y .(2) If s ⊆ D (that is, s = s is exposed), then s ∩ C contains at most one of v and v . Proof.
For (1), the alternative is that s contains a vertex u in (cid:10) N M (cid:11) . Byhypothesis, s has no edge in C M and, therefore, s has no edge in C . Being in N M ,the vertex u is either in R or in P M P M P M .Suppose that u is in P M P M P M . If u is in P M , then, since P M is H -avoiding, u is an end of P M , and so is in P M ∪ P M . Thus, if u is in P M P M P M , then u isin P M ∪ P M . Since both P M and P M are contained in H , are R -avoiding, andneither has an edge of s , the one containing u is trivial and u is in R .Thus, in every case u is in R and so is an H -node. It follows that one of (cid:2) x, N M , u (cid:3) and (cid:2) u, N M , y (cid:3) contains P M P M P M and the other is contained in R .We choose the labelling so that (cid:2) x, N M , u (cid:3) ⊆ R .As we follow R − e from w to x and continue to u along N M , we see there is anedge of C incident with x and not in R . That it is in N D implies it is in P D P D P D .All the vertices in (cid:2) x, N M , u (cid:11) are incident with two rim edges in what we have justtraversed. In particular, e is not incident with any of these vertices and, therefore, (cid:2) x, N M , u (cid:3) is contained in C D . More precisely, (cid:2) x, N M , u (cid:3) is contained in P D . Aswe continue along R past u , we either find e is incident with u or the other edgeof C incident with u is in R . In either case, u is in (cid:10) P D (cid:11) , a contradiction.For (2), suppose v and v are both in C . Then P M ∪ P D contains both v and v . By Definition 6.2 (1e), v and v are not both in the same one of P M and P D ,so one is in P M and the other is in P D . By symmetry, we may assume v is in P M .Because Π is H -friendly, P M is contained in either [ a, r , v , r , v ] or, if a = v , r (these being the only two faces of Π[ H ] ∪ γ in M that can be incident with v ).Recall that e is in both P M and P D . If P M ⊆ [ a, r , v , r , v ], then e is in ei-ther r or r and P D is, by Definition 6.2 (1e), contained in either [ a, r , v ] r r [ v ,r , v (cid:105) or r r r [ v , r , v (cid:105) , and v is not in C . If P M ⊆ r , then e is in r , so P D is contained in r r r [ v , r , v (cid:105) , and again v is not in C . (cid:3) The case e ∈ N D is easy: the rim of the V is ( R − (cid:10) P M (cid:11) ) ∪ P M P M P M andwe choose as spokes any three of the H -spokes that are contained in M . (If oneintersects C M , then only the part of the spoke that is C M -avoiding will be theactual spoke of the V .)If e ∈ N M , then the rim R (cid:48) of the V is ( R − (cid:10) P D (cid:11) ) ∪ P D P D P D and the spokesare the three H -spokes from the hypothesis. If all three hypothesized H -spokesare contained in M , then it is evident from Subclaim 2 (1) that we have indeeddescribed a V in ( C ∪ M C ) − e .So suppose that one of the H -spokes in the hypothesis is the exposed spoke s .From Subclaim 2 (2), either s is disjoint from C or precisely one H -node incidentwith s is in C . We may choose the labelling so that v is not in C .If v is not in C , then s is disjoint from C . Subclaim 2 (1) shows the othertwo hypothesized H -spokes meet C in at most x or y ; it is now obvious that thethree hypothesized H -spokes combine with R (cid:48) to make a V .Finally, suppose v is in C . Because C D is H -green, P D ⊆ r r r [ v , r , b ].In particular, s is disjoint from C M . If s has no edge in C M , then R (cid:48) ∪ s ∪ s ,together with the portion of s from v to C D is a V avoiding N M . If s has anedge in C M , then C M is in the Π[ H ]-face bounded by Q . In this case, we mayreplace s with s r to obtained the desired V . (cid:3) Evidently, Claims 6, 7, and 8 show that G has a box, contradicting Lemma5.12.HAPTER 7 Exposed spoke withadditional attachment not in Q V -bridges. This will be used in the next sectionwhen we get our second major step by showing that there is a representativity 2embedding of G in R P for which all the H -spokes are contained in the M¨obiusband. Theorem . Suppose G ∈ M and V ∼ = H ⊆ G . Let Π be an H -friendlyembedding of G in R P , with the standard labelling. Then there is no H -bridgehaving attachments in both (cid:104) s (cid:105) and (cid:104) r r r (cid:105) . At one point in the proof of this theorem, we need the following lemma. Mostof it is used again several times.
Lemma . Let G be a graph and let V ∼ = H ⊆ G . Let P be an H -avoiding pathin G joining distinct vertices x and y of R and let P (cid:48) be one of the two xy -subpathsof R . Let D be a 1-drawing of H ∪ P .(1) If P (cid:48) has at most two H -nodes or, for some i , P (cid:48) = r i r i +1 , then P (cid:48) is notcrossed in D .(2) If there are only the two H -nodes v i , v i +1 in the interior of P (cid:48) and P (cid:48) hasat most one other H -node, then r i +4 is not crossed in D .(3) Suppose r i r i +1 ⊆ P (cid:48) , P (cid:48) (cid:54)⊆ r i r i +1 , but P (cid:48) ⊆ r i r i +1 [ v i +2 , r i +2 , v i +3 (cid:105) .(a) Then r i r i +1 is not crossed in D .(b) If P (cid:48) is crossed in D , then s i +3 is exposed in D and P (cid:48) ∩ r i +2 crosses r i − . Proof.
Let x and y be the ends of P and let R (cid:48) = ( R − (cid:104) P (cid:48) (cid:105) ) ∪ P . For (1) and (2),we find three spokes to add to R (cid:48) to find a subdivision of V disjoint from P (cid:48) — orat least some part of P (cid:48) . The part of P (cid:48) disjoint from the V cannot be crossed inany 1-drawing of H .For (1), if P (cid:48) contains at most one H -node, then this is easy: any three H -spokes not having an end in P (cid:48) will suffice. If P (cid:48) = r i r i +1 , then the three H -spokes s i , s i +2 , and s i +3 suffice.In the remaining case, P (cid:48) has precisely two H -nodes. We may express P (cid:48) inthe form P (cid:48) = [ x, r j − , v j ] r j [ v j +1 , r j +1 , y ] , where either of [ x, r j − , v j ] and [ v j +1 , r j +1 , y ] might be a single vertex. In thiscase, the spokes are s j +2 , s j +3 and s j +1 [ v j +1 , r j +1 , y ], showing that [ x, r j − , v j ] r j is not crossed in D , while replacing s j +1 [ v j +1 , r j +1 , y ] with [ x, r j − , v j ] s j shows[ v j +1 , r j +1 , y ] is not crossed in D . This completes the proof of (1).
412 7. EXPOSED SPOKE WITH ADDITIONAL ATTACHMENT NOT IN Q For (2), replace R (cid:48) with ( R (cid:48) − (cid:104) r i +4 (cid:105) ) ∪ ( s i r i s i +1 ). We now need three spokes.If there is a third H -node in P (cid:48) , then symmetry allows us to assume it is v i − . Ineither case, we choose s i − , [ v i +1 , r i +1 , y ], and s i +2 as the three spokes for the V .This V avoids r i +4 , showing it is not crossed in D .For (3), x = v i and the hypotheses imply that y ∈ (cid:104) r i +2 (cid:105) . For (3a), we mayuse the spokes s i , s i +2 [ v i +2 , r i +2 , y ], and s i +3 to see that r i r i +1 is not crossed in D , as required.For (3b), suppose P (cid:48) is crossed in D . Part (3a) shows that it must be P (cid:48) ∩ r i +2 that is crossed and (2) shows that r i +5 = r i − is not crossed in D . We need onlyshow that r i − is also not crossed in D . If it were, then [ v i +2 , r i +2 , y ] crosses r i − .But then the cycle r i +3 r i +4 r i − r i − s i − separates v i = x from y in D , showingthat P is also crossed in D , a contradiction. Proof of Theorem 7.1.
This is obvious if no spoke is exposed in Π, so we maysuppose s is exposed. Claim . There is no H -avoiding (cid:104) s (cid:105) (cid:104) v , r , v ]- or (cid:104) s (cid:105) [ v , r , v (cid:105) -path. Proof.
By symmetry, it suffices to prove only one. By way of contradiction,we suppose that there is an H -avoiding path P from x ∈ (cid:104) s (cid:105) to y ∈ (cid:104) v , r , v ].Let e ∈ s and consider a 1-drawing D of G − e . By Lemma 5.9 and Theorem5.23 (4), we know that Q is crossed in D . This implies that r r r r crosses r r r r . This already implies neither s nor s is exposed in D . Furthermore,the crossing is of two edges in R and, since P is H -avoiding, we conclude that D [ P ]is not crossed in D . Therefore, the end of P in (cid:104) v , r , v ] must occur in the intervalof r r r r between the crossing and v ; that is, the crossing must involve an edgeof r . In particular, r r r r is not crossed in D .Since Q is crossed in D and r is crossed in D , the other crossing edge is in r r . Thus it is in r r r . It follows that s is exposed in D . Thus, the cycle r r s r r s separates x from y in D , showing P is crossed in D , a contradiction. (cid:3) It follows from Claim 1 that, if there is an H -avoiding path P joining x ∈ (cid:104) s (cid:105) to y ∈ (cid:104) r r r (cid:105) , then y ∈ (cid:104) r (cid:105) . Let K = H ∪ P . See Figure 7.1. v xyab bav v v v P v v v v v Figure 7.1.
The subgraph K of G in R P . . EXPOSED SPOKE WITH ADDITIONAL ATTACHMENT NOT IN Q Let J and J be the two cycles r r [ v , r , y, P , x, s , v ] and r r [ v , r , y, P ,x, s , v ], respectively. Claim . The cycles J and J both bound faces of G in R P . Proof.
These cycles are both H -green, so this is just Lemma 6.6 (8). (cid:3) The following claim completes the determination of the ( H ∩ M )-bridge con-taining s . Claim . The ( H − (cid:104) s (cid:105) )-bridge containing s is s ∪ P . Proof.
Suppose not and let B be the ( H − (cid:104) s (cid:105) )-bridge containing s . ThenLemma 5.19 implies that B has an attachment z other than v , y , and v . ByClaim 2, z ∈ [ a, r , v (cid:105) ∪ (cid:104) v , r , b ]; by symmetry we may assume the former. Let P be a K -avoiding z (cid:104) s (cid:105) -path.Suppose z = v . Let e be the edge of s incident with v . We show thatcr(( K ∪ P ) − e ) ≥
2. As this is a proper subgraph of G , we contradict the fact that G is 2-crossing-critical. In P ∪ ( s − e ) ∪ P , there is a claw Y with talons z = v , y and v . We show cr(( H − (cid:104) s (cid:105) ) ∪ Y ) ≥ D is a 1-drawing of ( H −(cid:104) s (cid:105) ) ∪ Y . As H −(cid:104) s (cid:105) ∼ = V , Lemma 7.2 (1) implies that (using the labelling from H ) [ y, r , v ] r r is not crossed in D , while (2) of the same lemma implies neither r nor r is crossedin D . Part (3a) implies r r r is not crossed, while (3b) implies (since r is notcrossed) that [ v , r , y ] is not crossed. The only remaining possibilities for crossed( H − (cid:104) s (cid:105) )-rim branches are r and r . But no 1-drawing of H − (cid:104) s (cid:105) has these tworim-branches crossed, the desired contradiction.So z (cid:54) = v . But then we may replace s with the zv -path s (cid:48) in P ∪ s andreplace P with the ys (cid:48) -path in P ∪ s to get a new subdivision H (cid:48) of V . Wenotice that Lemma 6.5 (1) implies that Π is H (cid:48) -friendly. However, the analogue J (cid:48) of J does not bound a face, contradicting Claim 2. (cid:3) Claim . There is a unique 1-drawing of K . In this 1-drawing, s is exposed.The 1-drawing of K is illustrated in Figure 7.2. Proof. If D is a 1-drawing of K , then Claim 2 and Lemma 6.6 (10) implyneither J nor J is crossed in D . It follows that none of r , r , r , r , and r iscrossed in D . Lemma 3.6 implies r cannot be crossed in D , so Q is clean in D .Therefore, s must be in a face of D [ R ∪ Q ] incident with r . This is only possibleif s is exposed, which determines D . (cid:3) For j ∈ { , } , let D j be a 1-drawing of G − (cid:104) s j (cid:105) . Claim . The crossing in D [( H − s ) ∪ P ] is of r with [ y, r , v ]. Likewise,the crossing in D [( H − s ) ∪ P ] is of r with [ v , r , y ].The 1-drawings of Claim 5 are illustrated in Figure 7.3. Proof.
We treat the case j = 2; the case j = 3 is very similar. By Theorem5.23 (2), Q has BOD, so Lemma 5.9 implies Q is crossed in D . This implies that s is not exposed in D . The H -avoiding path P joins x ∈ (cid:104) s (cid:105) to y ∈ (cid:104) r (cid:105) , so y must be on a face incident with s . It follows that Q must be crossed in D . This Q v v v v v v v v v v Figure 7.2.
The 1-drawing of K . x v v v v v v v yv v v yv v v v v xv v v v v Figure 7.3.
The 1-drawings D [( K − (cid:104) s (cid:105) ) ∪ P ] and D [( K −(cid:104) s (cid:105) ) ∪ P ].implies that s is exposed. We deduce that either r crosses r ∪ r or r crosses r ∪ r . In the latter case, D [ P ] must cross D [ H − s ], a contradiction, so it mustbe the former.As D [ P ] is not crossed, y occurs between v and the crossing in r ∪ r , asrequired. (cid:3) The following claims help us obtain the structure of ( M Q ) ; we will use thisto find a 1-drawing of G , which is the final contradiction. Claim . Suppose B is a Q -bridge having an attachment in each of r and r . Then B is one of M Q , v v , v v , and v v . Proof.
We note that s ∪ P ⊆ M Q . Either B = M Q , or, in the drawing D , B is in a face of D [( H − s ) ∪ P ] incident with both r and r . There are only . EXPOSED SPOKE WITH ADDITIONAL ATTACHMENT NOT IN Q two such faces, namely F , bounded by Q , and F (cid:48) , the other face incident with r . Whichever face B is in, its attachments are in the intersection of Q with theboundary of the containing face. Thus, if B is in F , then att( B ) ⊆ r s r . In thiscase, the only possibility for an attachment in r is v , so v ∈ att( B ). If, on theother hand, B is in F (cid:48) , then att( B ) ⊆ r r s . In this case, v ∈ att( B ). Similarly, D shows either B = M Q , or att( B ) ⊆ r s r and v ∈ att( B ), or att( B ) ⊆ s r r and v ∈ att( B ). Comparing these possibilities, we conclude that one ofthe following four cases holds for att( B ): att( B ) = { v , v } ; att( B ) = { v , v } ; v , v ∈ att( B ) and att( B ) ⊆ r ∪ s ; and v , v ∈ att( B ) and att( B ) ⊆ r ∪ s .We claim v v is not an H -bridge. For if it were, let D be a 1-drawing of G − v v . Then s ∪ P is not crossed in D and Claim 3 says the ( H − (cid:104) s (cid:105) )-bridgecontaining s is s ∪ P . In particular, s consists of the two edges v x and xv , and x has degree 3 in G . Thus, we can draw v v alongside s , yielding a 1-drawing of G , a contradiction.We must show that, if v , v ∈ att( B ) and att( B ) ⊆ r ∪ s , then B = v v .Likewise, if v , v ∈ att( B ) and att( B ) ⊆ r ∪ s , then B = v v . We consider theformer case, the latter being completely analogous. Corollary 5.15 shows that B can have at most one other attachment. Lemma 5.19 shows that either B = v v or B is a claw with talons v , v , and z ∈ (cid:104) v , r , v , s , v (cid:105) . Since we are trying toshow B = v v , we assume the latter. Let e be the edge of B incident with z and let D be a 1-drawing of G − e . Since K ⊆ G − e , D extends the 1-drawing illustratedin Figure 7.2. We modify D to obtain a 1-drawing of G , which is impossible.Observe that B − z is an H -avoiding v v -path P (having length 2); there isonly one place D [ P ] can occur in Figure 7.2. Notice that B is a Q -local H -bridgeand, furthermore, P overlaps M Q .Theorem 5.23 shows Q has BOD in G ; let ( B , M ) be the bipartition of OD ( Q ), with B ∈ B . Then M Q ∈ M . Every Q -bridge is drawn in D , withthe exception that we have B − e in place of B .Because we cannot add e back into D to get a 1-drawing of G , there must bean H -avoiding path P (cid:48) in G − e joining the two components of [ v , r , v , s , v ] − z so that D [ P (cid:48) ] is on the same side — henceforth, the inside — of D [ Q ] as P . Let B (cid:48) be the Q -bridge containing P (cid:48) . If B (cid:48) has just v and v as attachments, thenlet D be a 1-drawing of G − v v . As we did above for v v , we can add v v alongside P to recover a 1-drawing of G . Therefore, B (cid:48) does not have just v and v as attachments.It follows that B (cid:48) overlaps B , so it is in M . Therefore, it does not overlap M Q ;in particular, it cannot have an attachment in both [ v , s , v (cid:105) and [ v , r , v (cid:105) . Weconclude that, for some q ∈ { r , s } ; and (ii) att( B (cid:48) ) ⊆ q . Let q (cid:48) be such that { q, q (cid:48) } = { r , s } .Let B , B , . . . , B k be a path in OD ( Q ) − { M Q , B } so that B (cid:48) = B . Subclaim . For i = 1 , , . . . , k , att( B i ) ⊆ q . Proof.
Above, we chose q to contain att( B (cid:48) ), which is the case i = 1. Noticethat B , B , . . . are all on the same side of D [ Q ] as B (cid:48) and P , while B , B ,. . . are all on the other side of D [ Q ]. The former are all in M , while the latter arein B . Let i be least so that B i has an attachment outside q . Then it also has anattachment in (cid:104) q (cid:105) (in order to overlap B i − ).If B i is inside D [ Q ], then B i does not overlap M Q , so it has no attachmentin q (cid:48) − q . As B i cannot cross P in D , att( B i ) ⊆ q , a contradiction. Q If B i is outside D [ Q ], then either att( B i ) ⊆ s , so q = s and we are done, oratt( B i ) ⊆ r ∪ [ v , s , x ], so, in particular, q = r . Furthermore, B i does not overlap B . Therefore, B i has no attachment in (cid:104) v , s , x ], so att( B i ) ⊆ r . (cid:3) Let L be the component of OD ( Q ) − { M Q , B } containing B (cid:48) . We can flipthe Q -bridges in L so that they exchange sides of D [ Q ], yielding a new 1-drawingof G − e with fewer Q -bridges in M on the same side of D [ Q ] as P . Inductively,this shows there is a 1-drawing D (cid:48) of G − e in which all Q -bridges in the face of D (cid:48) [ K ∪ P ] bounded by r s P are in B . As none of these overlaps B , we may add e into D (cid:48) to obtain a 1-drawing of G , a contradiction. (cid:3) Let e be the edge in r that is crossed in D and let e be the edge in r thatis crossed in D . For i = 5 ,
9, let u i be the end of e i nearer to v i in r i and let w i bethe other end of e i . See Figure 7.4. We highlight some relevant “cut” properties ofthese edges in the next three claims. yv v v v v xv v v v v x v v v v v v v yv v v u w u w Figure 7.4.
The 1-drawings D [( K − (cid:104) s (cid:105) ) ∪ P ] and D [( K −(cid:104) s (cid:105) ) ∪ P ]. Claim . Any r -avoiding (cid:104) s r ] (cid:104) r s ]-path in ( M Q ) contains e . In par-ticular, there are not two edge-disjoint r -avoiding (cid:104) s r ] (cid:104) r s ]-paths in ( M Q ) . Proof.
Suppose P is a r -avoiding (cid:104) s r ] (cid:104) r s ]-path. Let e be any edge of s and let D be any 1-drawing of G − e . By Claim 5, D [( H − (cid:104) s (cid:105) ) ∪ P ] is illustratedin Figure 7.3. But here we see that the cycle C = [ v , s , x ] P [ y, r , v ] s r r separates (cid:104) s r ] and (cid:104) r s ]. Note that C consists of r and a Q -avoiding v v -path in M Q . Therefore, P is disjoint from C , and so it must cross C in D . Asthis can only happen at the crossing in D , it must be that the edge of r crossedin D is in P . (cid:3) Analogously, deleting e ∈ s provides a proof of the following claim. Claim . Any r -avoiding [ s r (cid:105) [ r s (cid:105) -path in ( M Q ) contains e . In par-ticular, there are not two edge-disjoint r -avoiding [ s r (cid:105) [ r s (cid:105) -paths in ( M Q ) . (cid:50) . EXPOSED SPOKE WITH ADDITIONAL ATTACHMENT NOT IN Q The final claim is a central point about M Q . Claim . Let P and P be the two paths of Q − { e , e } . Then there is no P P -path in ( M Q ) − { e , e , v v } . Proof.
Assume that there is a P P -path P in ( M Q ) −{ e , e } . For i = 1 , z i be the end of P in P i .Suppose first that z is in (cid:104) s r (cid:105) . If z is in [ v , r , w ], then P [ z , r , v ] is an r -avoiding (cid:104) s r ] (cid:104) r s ]-path in ( M Q ) that also avoids e , contradicting Claim 7.If z is not in [ v , r , w ], then there is an r -avoiding [ s r (cid:105) [ r s (cid:105) -path in ( M Q ) that also avoids e , contradicting Claim 8. Therefore, z is in P − (cid:104) s r (cid:105) ; that is z is in [ v , r , u ] ∪ [ v , r , u ]. Symmetrically, z is in [ w , r , v ] ∪ [ w , r , v ].If z is in [ v , r , u ], then Claim 7 implies z is not in [ w , r , v ]. Therefore, z is in [ w , r , v ]. By Claim 6, P is one of v v , v v , and v v . Clearly, neither z nor z is v and neither is v , so none of these outcomes is possible.Therefore, z is in [ v , r , u ]. Claim 8 implies z is not in [ w , r , v ]. ByClaim 6, the only possibility is that z = v and z = v and P is just the edge v v , as required. (cid:3) We will show that there is an embedding Π (cid:48) of G in R P and a non-contractiblesimple closed curve γ (cid:48) in R P so that γ (cid:48) ∩ G consists of one point in each of theinteriors of Π (cid:48) [ e ] and Π (cid:48) [ e ]. Standard surgery then implies that cr( G ) ≤ ]).Consider the two faces of Π[ K ] incident with both e and e . Let F Q be theone bounded by Q . Let F (cid:48) be the other; it is bounded by the cycle s r r r r r ,which we call C (cid:48) . Both Q and C (cid:48) contain both e and e . What we would like toprove is that, for each such face F with boundary C , there is no K -avoiding pathcontained in F and having an end in each of the two components of C − { e , e } .Although not necessarily true for Π, it is true for an embedding obtained from Πby possibly re-embedding the edges v v and v v .Let us begin with the possible re-embeddings. We deal with v v ; the argumentfor v v is completely analogous. If v v is not embedded in F (cid:48) , then do nothingwith it. Otherwise, it is embedded in F (cid:48) and we claim we can re-embed it in F Q .The embedding Π shows that v v is contained in one of the two faces of K ∪ γ into which F (cid:48) is split. Therefore, v and v must be on the same ab -subpath of R .This implies that either v = a or v = b , or both. In order not to be able to embed v v in F Q , there must be a Q -avoiding path P contained in F Q joining (cid:104) r s (cid:105) to (cid:104) r r s r (cid:105) .We first consider where D [ P ] can be. There are only two possibilities: it iseither in the face of D [ K − (cid:104) s (cid:105) ] bounded by [ v , r , × , r , v ] s r ; or in the faceincident with both r and s . The latter cannot occur, as v v is also in that faceand they overlap on the boundary of this face. So it must be the former.However, in this case, both v v and P are in the face of D [ K − (cid:104) s (cid:105) ] boundedby Q , and they overlap on Q , the final contradiction that shows that P does notexist, so we can re-embed v v in F Q . Let Π (cid:48) be the embedding of G obtained byany such re-embeddings of v v and v v . Q The faces F Q and F (cid:48) of Π[ K ] are also faces of Π (cid:48) [ K ] with the same boundaries;we will continue to use these names for them, while Q and C (cid:48) are still theirboundaries.We now show that there is no K -avoiding path in F Q joining the two paths P and P of Q − { e , e } . Such a path is necessarily in ( M Q ) . By Claim 9, sucha path is necessarily v v . But Π is H -friendly, so v v is not embedded in M andso, in particular, is not embedded in F Q . Thus, v v is also not in this face of Π (cid:48) ,whence there is no P P -path in F Q , as required.Now consider the possibility of a K -avoiding path in F (cid:48) having its ends in eachof the two paths in C (cid:48) − { e , e } . Such a path is in a C (cid:48) -bridge B embedded in F (cid:48) .By Claim 3, B has no attachment in (cid:104) s (cid:105) . Thus, B has an attachment either in[ v , r , w ] or in [ v , r , u ].We claim it must also have an attachment in (cid:104) r r r (cid:105) . If not, then all itsattachments are in[ v , r , w ] ∪ [ v , r , u ] ∪ [ w , r , v ] ∪ [ v , r , u ] . But then B is a Q -bridge. If it has an attachment in both r and r , then Claim 6implies B is one of v v , v v , and v v . The first two are not embedded in the Π (cid:48) -face F (cid:48) and the last does not have attachments in both components of C (cid:48) − { e , e } .In the alternative, either att( B ) ⊆ r or att( B ) ⊆ r , and then we contradict eitherClaim 7 or Claim 8.So B has an attachment in (cid:104) r r r (cid:105) . If B has an attachment in [ v , r , w ],then D [ B ] must have a crossing, which is not possible. If B has an attachmentin [ v , r , u ], then D [ B ] must have a crossing, which is not possible. Therefore,there is no such B , as claimed.For each of the faces F Q and F (cid:48) of Π (cid:48) and any points x and y in the interiors ofΠ (cid:48) [ e ] and Π (cid:48) [ e ], the preceding paragraphs show that there is a G -avoiding simple xy -arc in the face. The union of these two arcs is a simple closed curve γ (cid:48) in G thatmeets Π (cid:48) [ G ] in just the two points x and y .In a neighbourhood of x , there are points of e on both sides of γ (cid:48) . If γ (cid:48) werecontractible in R P , then { e , e } would be an edge-cut of size 2 in the 3-connectedgraph G , which is impossible. So γ (cid:48) is non-contractible. But this is also impossible,as it meets G precisely in x and y , showing that G has a 1-drawing, the finalcontradiction.HAPTER 8 G embeds with all spokes in M In this section, we prove that if G ∈ M and V ∼ = H ⊆ G , then G has arepresentativity 2 embedding in R P with H ⊆ M . This is an important step as itprovides the embedding structure we need to find the tiles.It turns out that we need something stronger than H ⊆ M . We must also showthat, in addition to H ⊆ M , the representativity 2 embedding of G is such that M Q is the only Q -local H -bridge B for which Q ∪ B contains a non-contractiblecycle. (We remind the reader that Q is special. Each H -quad bounds a face ofΠ[ H ]. In the standard labelling, the only one of these five faces that contains anarc of γ is the one bounded by Q .) Theorem . Suppose G ∈ M and V ∼ = H ⊆ G . Then G has a representa-tivity 2 embedding Π in R P so that, with the standard labelling:(1) s is not exposed in Π , that is, Π[ H ] ⊆ M ; and,(2) if B is a Q -local H -bridge other than M Q , then Π[ Q ∪ B ] has no non-contractible cycle. In principle, these two arguments are consecutive: we first show we can arrange H ⊆ M , and then deal with the Q -bridges. However, the arguments are essentiallythe same. Therefore, we shall have parallel statements and arguments, one forgetting the five H -spokes in M and one for getting such an embedding with Q nicely behaved. (If we knew that G had an embedding with H not contained in M ,then we could do both simultaneously.) Definition . A friendly, standard quadruple , denoted (( G, H, Π , γ )), con-sists of G ∈ M , V ∼ = H ⊆ G , an H -friendly embedding Π of G , and a non-contractible, simple closed curve γ meeting Π[ G ] in precisely two points, used asthe reference for giving H the standard labelling relative to Π. We abbreviatefriendly, standard quadruple as fsq .Observe that Theorem 3.5 implies G has a representativity 2 embedding in R P .Lemma 6.5 (3) implies G has an H -friendly embedding Π. Any non-contractiblesimple closed curve γ in R P meeting G in precisely two points yields a standardlabelling of H relative to Π and γ . Summarizing, we have the following observation. Lemma . If G ∈ M and V ∼ = H ⊆ G , then there is an fsq (( G, H, Π , γ )) . Let Q ∗ be Q if s is exposed in Π and let Q ∗ be Q if s is not exposedin Π, that is, if Π[ H ] ⊆ M . Our first step is to show that OD ( Q ∗ ) is (nearly)bipartite. Theorem 5.23 (1) implies OD ( Q ) is bipartite. For Q ∗ = Q , this ismore involved. In the following statement, v v and v v are meant to be possible Q -bridges consisting of a single edge joining the two indicated vertices. They neednot exist in G .
490 8. G EMBEDS WITH ALL SPOKES IN M Lemma . Let (( G, H, Π , γ )) be an fsq. If s is exposed in Π , then OD ( Q ) −{ v v , v v } is bipartite. The following observations will be needed throughout the proof of Theorem 8.1and, in particular, the proof of Lemma 8.4.
Definition . Let ((
G, H, Π , γ )) be an fsq and let Q ∗ be either Q (if s isexposed) or Q (otherwise). Then N — a function of (( G, H, Π , γ )) — denotes theset of Q ∗ -bridges B other than M Q ∗ for which Π[ Q ∗ ∪ B ] has a non-contractiblecycle. In the case Q ∗ = Q , any of v v and v v that occurs in G is a Q -bridge B for which Π[ Q ∪ B ] has a non-contractible cycle, and we do not include these in N . We remark that, if s is exposed in Π, then Theorem 7.1 implies the ( H ∩ M )-bridge B containing s is distinct from M Q . In this case, B ∈ N . If s is notexposed in Π, then Q ∗ = Q . If N = ∅ , then Π satisfies the conclusions of Theorem8.1. Therefore, in this case, we may assume N (cid:54) = ∅ .Before we can prove Lemma 8.4, we need some results common to both cases.An easy corollary of the following lemma will be used to deal with the maincase in the proof of Lemma 8.4. Lemma . Let D be a 1-drawing of V (with the usual labelling) in which Q is crossed. Then:(1) Q bounds a face of D ; and(2) if Q is crossed in D , then either r crosses r or r crosses r . Proof. As Q is crossed in D , either r crosses r r r in D or r crosses r r r in D . This already shows that Q bounds a face of D .As Q is crossed in D , either r r or r r is crossed in D . Compare each ofthese with the possible crossing of Q . In the former case, r crosses r , while inthe latter case r crosses r .The following is the simple corollary that we will use. Corollary . Let G ∈ M and V ∼ = H ⊆ G . Let D be a 1-drawing of G − (cid:104) s (cid:105) . Then:(1) Q bounds a face of D [ H − s ] ; and(2) if Q is crossed in D , then either r r crosses r or r r crosses r (seeFigure 8.1 for the possibilities for D [ H − (cid:104) s (cid:105) ]) .Likewise, if D is a 1-drawing of G −(cid:104) s (cid:105) in which Q is crossed, thenthe two possibilities for D [ H − (cid:104) s (cid:105) ] are illustrated in Figure 8.2. Proof.
Theorem 5.23 implies Q has BOD. Lemma 5.9 implies Q is crossed in D . The results now follow immediately from Lemma 8.6.Let r ∗ denote r ∪ r in the case Q ∗ = Q and r in the case Q ∗ = Q . We alsolet r ∗ +5 denote the other component of Q ∗ ∩ R . Lemma . Let (( G, H, Π , γ )) be an fsq. If B ∈ N , then Π[ B ] ⊆ D , att( B ) ⊆ r ∗ ∪ r ∗ +5 , and B has an attachment in each of r ∗ and r ∗ +5 . Proof.
If Π[ B ] ⊆ M , then Π[ Q ∗ ∪ B ] is contained in a closed disc and, therefore,has only contractible cycles, a contradiction. Thus, Π[ B ] ⊆ D . It now follows . G EMBEDS WITH ALL SPOKES IN M v v v v v v v v v v v v v v v v Figure 8.1.
The two possibilities for D . v v v v v v v v v v v v v v v v Figure 8.2.
The two possibilities for D .that att( B ) is contained in the intersection of Q with the boundary of D ; that is,att( B ) ⊆ r ∗ ∪ r ∗ +5 .Suppose by way of contradiction that att( B ) ⊆ r ∗ . Let ¯ r ∗ be a minimal subpathof r ∗ containing att( B ). Then there is a non-contractible cycle C contained in B ∪ ¯ r ∗ .Let F be the closed (Π[ H ] ∪ γ )-face containing Π[ B ]. Then F contains Π[ B ∪ ¯ r ∗ ],so the non-contractible cycle Π[ C ] is contained in the closed disc F , a contradiction.So att( B ) is not contained in r ∗ and, likewise, it is not contained in r ∗ +5 .Let (( G, H, Π , γ )) be an fsq, with s exposed in Π. Suppose D is a 1-drawingof G − (cid:104) s (cid:105) in which Q is crossed. Corollary 8.7 implies that D [ H − (cid:104) s (cid:105) ] is oneof the two drawings illustrated in Figure 8.1. The outside of D [ Q ] is the face of D [ Q ] containing D [ s ]. The inside is the other face of D [ Q ]. Likewise, if D is a 1-drawing of G − (cid:104) s (cid:105) in which Q is crossed, then the outside of D [ Q ] is theface of D [ Q ] containing D [ s ]. Lemma . Let (( G, H, Π , γ )) be an fsq, with s exposed in Π . For i = 2 , , let D i be a 1-drawing of G − (cid:104) s i (cid:105) in which Q is crossed. Suppose B is a Q -bridge in N . (1) If D [ B ] is outside of D [ Q ] , then B ∈ { v v , v v } .(2) If D [ B ] is outside of D [ Q ] , then B ∈ { v v , v v } . G EMBEDS WITH ALL SPOKES IN M Proof.
We prove (1); (2) is completely analogous. We remark that B (cid:54) = B as D [ s ] is inside D [ Q ]. Lemma 8.8 shows that either: (i) att( B ) ⊆ [ b, r , v ] ∪ [ v , r , a ] and B has attachments in both [ b, r , v ] and [ v , r , a ]; or (ii) att( B ) ⊆ [ a, r , v ] r ∪ r [ v , r , b ] and B has attachments in both [ a, r , v ] r and r [ v , r , b ].Suppose first that D is the left-hand possibility illustrated in Figure 8.1. Con-sidering D , we see that v is one attachment of B and the others are in r r .Now consider the possibilities for D [ B ]. We see that D [ B ] can be outside D [ Q ] in only one of the two possible D ’s, namely the right-hand one, and thenonly if att( B ) = { v , v } . But in this case B is just the edge v v , which is not in N . So D [ B ] is inside D [ Q ]. It now follows from this and the previous paragraphsthat att( B ) ⊆ { v } ∪ r .Putting this information into Π, we see that the only possibility for B , whichis embedded in D and not in M , is that B = v v .In the case D is the right-hand possibility in Figure 8.1, D shows thatatt( B ) ⊆ { v } ∪ r r . Since v v / ∈ N , B (cid:54) = v v , so D [ B ] is not outside D [ Q ].Therefore, D shows att( B ) ⊆ { v } ∪ r .Again we recall that B is embedded in D in R P . If B is embedded in theface bounded by [ a, r , v , s , v , r , b, α, a ], then b = v and the only other possi-ble attachment for B is v , as required. If B is embedded in the face boundedby [ b, r , v ] r r r [ v , r , a, α, b ], then a = v and again this is the only possibleattachment other than v , as required.Let N be the graph (cid:91) B ∈N B .
Lemma . Let (( G, H, Π , γ )) be an fsq. Then there are not disjoint ( N ∩ r ∗ )( N ∩ r ∗ +5 ) -paths in N . In particular, if Q ∗ = Q and |N | ≥ , then either every B ∈ N has only v as an attachment in r r or every B ∈ N has only v as anattachment in r r . Proof.
Suppose by way of contradiction that P and P are disjoint r ∗ r ∗ +5 -pathsin N , with, for j = 1 , P j having the end p j in r ∗ and the end q j in r ∗ +5 . Choosethe labelling so that, in r ∗ , p is closer to v than p is. There are three possibilitiesfor how P and P are embedded by Π: both in the (closed) disc contained in D bounded by [ a, r , v ] r r r r r [ v , r , b ] α (recall that α = γ ∩ D ); both in thedisc in D bounded by [ b, r , v ] r r r [ v , r , a ] α ; or one in each of these discs. Inall cases, we conclude that q is closer in r ∗ +5 to v than q is. Summarizing, wehave the following. Fact 1
Any two disjoint r ∗ r ∗ +5 -paths in N overlap on Q ∗ . For Q ∗ = Q we are done: Corollary 8.7 implies D [ Q ] bounds a face of D [ H − (cid:104) s (cid:105) ]. Both P and P have ends in both r ∗ and r ∗ +5 , so both must be inside D [ Q ], yielding the contradiction that they cross in D [ Q ].Now suppose Q ∗ = Q . For i = 2 , D i [ Q ] is not self-crossing; thus Fact 1implies that D i [ P ] and D i [ P ] are on different sides of D i [ Q ]. If Q is clean in D i , then we have a contradiction, as no face of D i [ H − (cid:104) s i (cid:105) ] is incident with both r ∗ and r ∗ +5 except the ones bounded by Q and Q .Thus, Q is crossed in D i . By Lemma 8.9, the one that is outside is one of v v , v v , v v , and v v . We treat in detail that this one is v v , as the othercases are completely analogous. It is in D that v v is outside D [ Q ]. . G EMBEDS WITH ALL SPOKES IN M Because q is closer to v than q is, q cannot be v ; it follows that it is P that is v v . Lemma 8.9 also implies that P , that is v v , is not outside D [ Q ]and, therefore, it is inside D [ Q ]. Thus, P is outside D [ Q ]. By Lemma 8.9, P is one of v v and v v . By choice of the labelling, it cannot be that v is anend of P , so P = v v , which is not disjoint from P = v v , a contradiction. Weconclude that there are not such disjoint paths.For the “in particular”, there is a cut vertex u of N separating N ∩ ( r r ) and N ∩ ( r r ) in N , as claimed. As s is a ([ r r ]) ([ r r ])-path in N , we deduce u ∈ s . If B is not the only member of N , then any other element B of N sharesthe vertex u with B , so u is an attachment of both. But u ∈ s implies u ∈ { v , v } .As a final preparatory remark, we have the following. Lemma . Let (( G, H, Π , γ )) be an fsq. Let B and B (cid:48) be distinct elements of N . Then:(1) B and B (cid:48) do not overlap on Q ∗ ; and(2) either B overlaps M Q ∗ on Q ∗ or Q ∗ = Q and B is either v v or v v . Proof.
In the case Q ∗ = Q , Corollary 8.7 and Lemma 8.8 imply B and B (cid:48) areboth drawn inside the face of D [ H − (cid:104) s (cid:105) ] bounded by Q and, therefore, they donot overlap, yielding (1) for Q .For Q ∗ = Q , if both B and B (cid:48) are in the same face of either D [ Q ] or D [ Q ],then they obviously do not overlap on Q . Thus, we may assume one is outside D [ Q ] and the other is inside D [ Q ] and that one is outside D [ Q ] and the otheris inside D [ Q ].By Lemma 8.9, the one outside D [ Q ] is either v v or v v , while the oneoutside D [ Q ] is either v v or v v . Thus, we may assume B ∈ { v v , v v } and B (cid:48) ∈ { v v , v v } . But none of the four possibilities is an overlapping pair, whichis (1) for Q .As for overlapping M Q ∗ , we suppose first that B has an attachment x in theinterior of one of r ∗ and r ∗ +5 . (The “in particular” part of Lemma 8.10 implies thisis always the case when Q ∗ = Q .) In this case, it is a simple exercise to see that x , together with any attachment of B in the other one of r ∗ and r ∗ +5 , are skew toat least one of the pairs of diagonally opposite corners of Q ∗ (in the case of Q these pairs are { v , v } and { v , v } ; for Q , they are { v , v } and { v , v } ). Thus, B overlaps M Q ∗ .In the remaining case, Q ∗ = Q and att( B ) ⊆ { v , v , v , v } . If both v and v are attachments, then B is again skew to M Q ∗ ; the same happens if both v and v are attachments. The only remaining cases are: att( B ) = { v , v } and { v , v } ,as claimed.The next result contains the essence of the proof of Lemma 8.4. Lemma . Let (( G, H, Π , γ )) be an fsq. Suppose B ∈ N , B k = M Q ∗ , and B , B , . . . , B k is an induced cycle in OD ( Q ∗ ) . Then either(1) Q ∗ = Q , k = 3 , and B ∈ { v v , v v } or(2) k is even and B k − ∈ N ∪ { v v , v v } . Proof. Case 1. k is odd. G EMBEDS WITH ALL SPOKES IN M Theorem 5.23 implies OD ( Q ) is bipartite. Therefore, Q ∗ = Q and s isexposed in Π.For i = 2 ,
3, let e i be the edge of s i incident with v i and let D i be a 1-drawingof G − e i . Theorem 5.23 implies Q i has BOD; Lemma 5.9 implies Q i is crossed in D i . If, for some i ∈ { , } , Q is clean in D i , then Lemma 5.6 implies Q has BOD,yielding the contradiction that k is even. Therefore, Q is crossed in both D and D . Claim . If some B i is either v v or v v , then i = 2 and k = 3. Proof.
Since both v v and v v overlap M Q , neither is in N , B is in N ,and the cycle is induced, it must be that i = k −
1. For sake of definiteness, wesuppose B k − = v v ; the alternative is treated completely analogously.Because B k − = v v , we deduce that D is the left-hand one of the twodrawings in Figure 8.1, while D is the right-hand drawing in Figure 8.2; in bothdrawings, B k − is outside Q .We note that B overlaps v v , so if B is B , then k = 3, as claimed. Other-wise, B ∈ N \ { B } . By Lemma 8.10, either the only attachment of B in r r is v or the only attachment of B in r r is v . For sake of definiteness, we assumethe former; the latter is completely analogous. In order not to overlap v v , theonly attachment for B in r r is v . Therefore, either k = 3 and we are done, or B is just the edge v v . We show that B = v v is not possible.Suppose that B = v v . Because we know D , we see that D [ B ] = D [ v v ]is inside D [ Q ], while D [ B k − ] = D [ v v ] is outside. In D , both are outside.But this is impossible, as B , B , B , . . . , B k − , B k − alternate sides of Q in both D and D .We conclude that B = v v is impossible and therefore k = 3, as claimed. (cid:3) It remains to show that no other possibility can occur with k odd. So supposeno B i is either v v or v v . Suppose some B i other than B is in N . As B i overlaps M Q and the cycle B , B , . . . , B k is induced, Lemma 8.11 implies i = k −
1. Thesame lemma implies k ≥
5. Therefore, Lemma 5.16 implies B , B , . . . , B k − , B k − alternate sides of Π[ Q ]. Since k is odd, B and B k − are on different sides ofΠ[ Q ], contradicting the fact that both are in N . Hence no other B i is in N .By Lemma 8.9, for at least one i ∈ { , } , D i [ B ] is inside D i [ Q ]. For thesake of definiteness, we consider the case i = 2 and D is the left-hand drawing of H − (cid:104) s (cid:105) in Figure 8.1; the remaining cases are completely analogous. Thus, either B is B or B is either a Q - or a Q -bridge.Since k is odd, B k − is on the other side of D [ Q ] from B . Therefore, B k − is outside D [ Q ]. In order to understand how B k − can overlap M Q in D , weanalyze D [ M Q ].Let e be the edge of M Q that is crossed in D . The end w of e outside D [ Q ]is in Nuc( M Q ). If the other end u of e is not in Nuc( M Q ), then u = v and[ × , r , v ] is the only part of M Q inside D [ Q ]. Otherwise, Nuc( M Q ) − { e , e } isnot connected. Since Nuc( M Q ) − e is connected, Nuc( M Q ) −{ e , e } consists of thecomponent inside D [ Q ] and the component O outside. In particular, M Q −{ e , e } consists of two Q -bridges in G − { e , e } . Let I be the one contained inside D [ Q ]and let O be the one outside. All attachments of M Q are attachments of either I . G EMBEDS WITH ALL SPOKES IN M or O , and possibly both. In the case u = v , we take I to be the portion of e from × to v .We observe that D shows that, except for one end of e , all the attachments of I are in Q . On the other hand, Theorem 7.1 implies that M Q , and, therefore I ,has no attachment in (cid:104) s (cid:105) . The embedding Π shows that I has no attachment in (cid:104) r (cid:105) : otherwise, I is not just [ × , e , v ] and u (cid:54) = v . Thus, the simple closed curve s r r r s [ v , r , a ] α [ b, r , v ] bounds a closed disc in R P separating u from (cid:104) r (cid:105) and is disjoint from Nuc( I ) ∪ (cid:104) r (cid:105) . Unless v = a , the same simple closed curveseparates u from v ; thus, if v is an attachment of I , then a = v .Because B k − is outside D [ Q ] and att( B k − ) ⊆ Q , there are four candidatesfor the face of D [ H − (cid:104) s (cid:105) ] that contains B k − . The one bounded by Q is notpossible: if B k − were in that face, it would not overlap M Q , as all the M Q attachments there would be in s and, therefore, all in O and not in I ; both B k − and O being outside D [ Q ] shows they do not overlap.The face of D [ H − (cid:104) s (cid:105) ] incident with [ × , r , v ] is not a possibility for B k − for exactly the same reason: the only attachment of I there can be v and v is notpart of a pair of attachments of M Q that are skew to two attachments of B i − ,which are all contained in [ × , r , v ].The face of D [ H −(cid:104) s (cid:105) ] incident with r r is also not a possibility for B k − . Tosee this, v is the only possible attachment of I in the boundary of this face. Thus, v is an attachment of I and B k − must have attachments in each of [ v , r , v (cid:105) and (cid:104) v , r , × ]. However, in Π we must have a = v and then there is no way to embed B k − .Therefore, B k − is in the face of D [ H − (cid:104) s (cid:105) ] incident with r s .By way of contradiction, suppose B k − is outside D [ Q ]. Identical argumentsas those just above show that B k − is in the face of D [ H −(cid:104) s (cid:105) ] incident with r s .Because the previous paragraph shows att( B k − ) ⊆ r r s , it cannot overlap M Q using an attachment of the portion of M Q that is inside D [ Q ] and, therefore, itcannot overlap M Q at all, a contradiction. Therefore, B k − is inside D [ Q ]. Thisimplies B k − is either a Q - or Q -bridge.If B k − is a Q -bridge, then att( B k − ) ⊆ r (because of D ). Letting ¯ r denotethe minimal subpath of r containing att( B k − ), D shows that no attachment of I is in (cid:104) ¯ r (cid:105) and, because O and B k − do not overlap (in D ), O also has no attachmentin (cid:104) ¯ r (cid:105) . Consequently, B k − does not overlap M Q , a contradiction. Therefore, B k − is a Q -bridge.Because B k − is inside D [ Q ], has no attachments in s , and overlaps B k − as Q -bridges, we see that B k − is also a Q -bridge. Continuing back, we see thateach of B k − , . . . , B is a Q -bridge and that B is outside D [ Q ]. By Lemma8.9, B is either v v or v v . But neither of these overlaps B . This contradictionshows that, except for the case described in Claim 1, k is even. Case 2. k is even. For each i = 2 , , . . . , k − B i ∪ Q ∗ has no non-contractible cycle in R P . Thus,Lemma 5.16 implies B and B k − are on the same side of Q ∗ in R P ; since B is Q ∗ -exterior, we have that B k − is Q ∗ -exterior. If Π[ Q ∗ ∪ B k − ] has no non-contractiblecycle, then Lemma 5.16 shows that it cannot overlap M Q ∗ , a contradiction. In thecase Q ∗ = Q , this implies that B k − is in N , while if Q ∗ = Q , then B k − is in N ∪ { v v , v v } . G EMBEDS WITH ALL SPOKES IN M Proof of Lemma 8.4.
We show that any odd cycle C in OD ( Q ) contains either v v or v v . Theorem 5.23 (3) implies that OD ( Q ) − M Q is bipartite. Therefore, C contains M Q . Lemma 8.12 shows that any odd cycle in OD ( Q ) containing M Q and an element of N has length 3 and contains one of v v and v v , as required.Thus, we may suppose C avoids N ∪{ v v , v v } ; let C = ( B , B , . . . , B k , M Q ).For each i = 1 , , . . . , k , Π[ B i ∪ Q ] has no non-contractible cycles in R P . Lemma5.16 implies B i and B i +1 are on different sides of Π[ Q ]. From this, parity impliesthat B and B k are on opposite sides of Π[ Q ]. On the other hand, they are bothon the side of Π[ Q ] not containing M Q , a contradiction.We are now prepared for the proof of Theorem 8.1. Proof of Theorem 8.1.
By Theorem 3.5, G has a representativity 2 embeddingΠ in R P . For (1), if no spoke is exposed in Π, then we are done; thus, with thestandard labelling, we may suppose that s is exposed in Π. From Theorem 7.1,we know that the Q -bridge B containing s is different from M Q . From Lemma8.4, we know that OD ( Q ) − { v v , v v } is bipartite and from Theorem 5.23 (3),we know that ( M Q ) is planar.We need to modify Π so that the set N (Definition 8.5) becomes empty. Westart with terminology that will be useful for the next claims. Definition . Let L be a graph. A path ( v , v , . . . , v k ) in L is chordlessin L if there is no edge v i v j of L that is not in P except possibly v v k .The following is a simple consequence of Lemma 8.12. Claim . (1) If Q ∗ = Q , then every N M Q -path in OD ( Q ) of lengthat least two contains one of v v and v v .(2) If Q ∗ = Q , then every chordless N M Q -path in OD ( Q ) of length atleast two has length exactly two, one end is either v v or v v , and thatend does not overlap M Q . Proof.
Suppose first that Q ∗ = Q . Let P be any N M Q -path in OD ( Q )that has length at least 2. We may assume P is chordless: otherwise there is ashorter N M Q -path P (cid:48) of length at least 2 and V ( P (cid:48) ) ⊆ V ( P ); if P (cid:48) contains either v v or v v , then so does P . By Lemma 8.11 (2), the ends of P are adjacent in OD ( Q ). Thus, P together with this edge of OD ( Q ) makes an induced cycle. Asthis cycle has only one vertex in N , Lemma 8.12 implies the cycle has length 3 andcontains one of v v and v v .Now suppose that Q ∗ = Q and P = ( B , B , . . . , B k , M Q ) is a chordless N M Q -path in OD ( Q ) of length at least 2. Then B ∈ N . Since P is chordless and B k / ∈ N , Lemma 8.12 (2) implies B does not overlap M Q . Now Lemma 8.11 (2)implies B is either v v or v v . Thus, B is skew to B . Since att( B ) ⊆ att( M Q ), B is also skew to M Q . Since P is chordless, k = 2, as required. (cid:3) If Q ∗ = Q , then set M to be the set { M Q , v v , v v } , while if Q ∗ = Q , thenset M to be the set { M Q , v v , v v } . In either case, let M − = M \ { M Q ∗ } .Let N + be the set of Q ∗ -bridges B so that there is an N B -path in OD ( Q ∗ )that is disjoint from M . The next lemma shows that N + consists of the members . G EMBEDS WITH ALL SPOKES IN M of N , which have attachments in both r ∗ and r ∗ +5 , and other Q ∗ -bridges B thatsimply extend out along either r ∗ or r ∗ +5 . This structure is what will allow us tofind natural “breaking points” a (cid:48) and b (cid:48) in r ∗ and r ∗ +5 , respectively, to allow us to“flip” the members of N into M , yielding the embedding with H ⊆ M and N = ∅ . Claim . If B ∈ N + , then att( B ) ⊆ r ∗ ∪ r ∗ +5 . Furthermore, if B ∈ N + \ N ,then either att( B ) ⊆ r ∗ or att( B ) ⊆ r ∗ +5 . Proof.
Let P be a shortest N B -path in OD ( Q ∗ ) that is disjoint from M . Weproceed by induction on the length of P .If B ∈ N , then the result follows from Lemma 8.8. Otherwise, B / ∈ N . Theneighbour B (cid:48) of B in P is closer to N than B is, so att( B (cid:48) ) ⊆ r ∗ ∪ r ∗ +5 .If B overlaps M Q ∗ , then P extends to a chordless N M Q ∗ -path in OD ( Q ∗ ) −M − of length at least 2. This contradicts Claim 1, showing B does not overlap M Q ∗ .Suppose by way of contradiction that B has an attachment x in the interiorof some H -spoke s contained in Q ∗ . As B overlaps B (cid:48) and att( B (cid:48) ) ⊆ r ∗ ∪ r ∗ +5 ,not all attachments of B can be in [ s ]. But any attachment y of B in Q ∗ − [ s ]combines with x to show that B is skew to the ends of s and, therefore, overlaps M Q ∗ . Therefore, att( B ) ⊆ r ∗ ∪ r ∗ +5 .Next suppose that B has an attachment in (cid:104) r ∗ (cid:105) . If B also has an attachmentin Q ∗ − [ r ∗ ], then B overlaps M Q ∗ (the two identified attachments of B are skewto the two ends of r ∗ ). Thus, if B has an attachment in (cid:104) r ∗ (cid:105) , then att( B ) ⊆ r ∗ .Likewise, if B has an attachment in (cid:10) r ∗ +5 (cid:11) , then att( B ) ⊆ r ∗ +5 .If B has an attachment in each of r ∗ and r ∗ +5 , then the preceding paragraphshows that att( B ) consists of some of the four H -nodes that comprise the ends of r ∗ and r ∗ +5 . Because B overlaps B (cid:48) , att( B ) cannot be just the two ends of one ofthe two H -spokes in Q ∗ . In the remaining case, B is skew to M Q ∗ , a contradiction.Thus, either att( B ) ⊆ r ∗ or att( B ) ⊆ r ∗ +5 . (cid:3) Let OD − ( Q ) = OD ( Q ) − { v v , v v } and let OD − ( Q ) = OD ( Q ). ByLemma 8.4 or Theorem 5.23 (1), OD − ( Q ∗ ) is bipartite; let ( S, T ) be a bipartitionof OD − ( Q ∗ ), with M Q ∗ ∈ T . We briefly treat separately the cases Q ∗ = Q and Q ∗ = Q .For the former, every element of N overlaps M Q and so N ⊆ S . There isan embedding Φ of ( G − { v v , v v } ) − Nuc( M Q ) in the plane so that all the Q -bridges in N are on the same side of Φ[ Q ].In the case of Q ∗ = Q , N \ { v v , v v } ⊆ S . There is an embedding Φ of G − Nuc( M Q ) in the plane so that all the Q -bridges in N \ { v v , v v } are on thesame side of Φ[ Q ]. Any of v v and v v that is also in S can also be embeddedon that same side of Φ[ Q ].Among the attachments of the elements of N + , let a be the one in r ∗ nearest v and let a be the one in r ∗ +5 nearest v . Claim . No Q ∗ -bridge not in M is skew to { a , a } . Proof.
It is clear that, in the case Q ∗ = Q , neither v v nor v v is skew to { a , a } . We show that a Q ∗ -bridge not in M that is skew to { a , a } must overlapsome Q ∗ -bridge in N + ; this implies the contradiction that it is in N + .By the Ordering Lemma 4.8, the elements of N ∩ S occur in order on Q ∗ inΦ. Thus, there is one element B (cid:48) of N ∩ S that has both an attachment nearestto v (relative to r ∗ ) and an attachment nearest to v (relative to r ∗ +5 ). Let x (cid:48) G EMBEDS WITH ALL SPOKES IN M and y (cid:48) be the attachments of B (cid:48) nearest v in r ∗ and v in r ∗ +5 , respectively. Inthe case Q ∗ = Q , B is a candidate for B (cid:48) , so, even in this case, we have that x (cid:48) ∈ [ v , r , v ] and y (cid:48) ∈ [ v , r , v ].Suppose by way of contradiction that some Q ∗ -bridge B (cid:48)(cid:48) not in M has at-tachments x (cid:48)(cid:48) and y (cid:48)(cid:48) in the two components of Q ∗ − { a , a } . We note that, when Q ∗ = Q , B (cid:48)(cid:48) (cid:54) = v v and B (cid:48)(cid:48) (cid:54) = v v .If one of x (cid:48)(cid:48) and y (cid:48)(cid:48) is in the component of Q ∗ − { x (cid:48) , y (cid:48) } that is disjoint from s − { x (cid:48) , y (cid:48) } , then B (cid:48)(cid:48) overlaps B (cid:48) . Since B (cid:48) ∈ N , Lemma 8.11 implies B (cid:48)(cid:48) / ∈ N and, therefore, B (cid:48)(cid:48) ∈ N + . But this contradicts the definition of either a or a and, therefore, both x (cid:48)(cid:48) and y (cid:48)(cid:48) are contained in the component of Q ∗ − { x (cid:48) , y (cid:48) } thatcontains s − { x (cid:48) , y (cid:48) } . In particular, we may assume y (cid:48)(cid:48) ∈ (cid:104) a , r , x (cid:48) ] ∪ (cid:104) a , r , y (cid:48) ].For the sake of definiteness, we assume y (cid:48)(cid:48) ∈ (cid:104) a , r , y (cid:48) ].Some Q ∗ -bridge B + in N + has a as an attachment; since y (cid:48)(cid:48) is in (cid:104) a , r , y (cid:48) ], y (cid:48) (cid:54) = a and, therefore, B + is not in N . There is a shortest path P = ( B (cid:48) , B , . . . , B n )in OD − ( Q ∗ ) − M Q ∗ from B (cid:48) to some element B n of N + so that B n has an attach-ment y n in [ a , r , y (cid:48)(cid:48) (cid:105) ; choose y n so that it is as close to a in [ a , r , y (cid:48)(cid:48) (cid:105) as possible.The Q ∗ -bridge B n − is in N + and so, by minimality of n , does not have anattachment in [ a , r , y (cid:48)(cid:48) (cid:105) . Since B n overlaps B n − , there is an attachment z n of B n in (cid:104) y (cid:48)(cid:48) , r , x (cid:48) ]. Since B (cid:48)(cid:48) is skew to { a , a } , there is an attachment z (cid:48)(cid:48) of B (cid:48)(cid:48) in (cid:104) a , r , v ] s [ v , r , a (cid:105) . But now z n , y (cid:48)(cid:48) , y n , and z (cid:48)(cid:48) show B (cid:48)(cid:48) overlaps B n . Since B (cid:48)(cid:48) / ∈ M , B (cid:48)(cid:48) is in N + . But this contradicts the definition of a or a . (cid:3) The following is immediate from Claim 3.
Claim . Each Q ∗ -bridge not in M has all its attachments in one of the two a a -subpaths of Q ∗ . (cid:50) The proof now bifurcates into the two cases. We consider first the case Q ∗ = Q and that s is exposed in Π. The following is immediate from Claim 4. Claim . The planar embedding Φ of ( G − { v v , v v } ) − Nuc( M Q ) hasthe property that there is a simple closed curve in the plane that meets Φ[( G −{ v v , v v } ) − Nuc( M Q ∗ )] precisely at a and a . (cid:50) We are now prepared to describe a representativity 2 embedding of G in R P so that all H -spokes are in M .Let Ψ be an embedding of H in R P so that all H -spokes are contained inthe M¨obius band M Ψ bounded by Ψ[ R ] and let γ Ψ be a non-contractible, simple,closed curve that meets H in precisely the points a and a . The claim is that thisembedding extends to an embedding of G so that γ Ψ meets G only at a and a .Claim 4 implies that we can add all the Q -bridges other than v v , v v , and M Q to Ψ so that there is no additional intersection with γ Ψ . It remains to showthat we may also add the at most three remaining Q -bridges. Claim . At most one of v v and v v is in G . Proof.
Suppose both are in G . We consider a 1-drawing D of G −(cid:104) s (cid:105) . As Q must be crossed in D (it has BOD and s is contained in a planar Q -bridge; applyLemma 5.9), we conclude that r r r r crosses r r r r in D . In particular, s and s cannot be exposed.In order for v v to be not crossed in D , we must have the crossing in r .Likewise, v v implies the crossing is in r . But then neither r r nor r r iscrossed, so Q is not crossed in D , a contradiction. (cid:3) . G EMBEDS WITH ALL SPOKES IN M We note that v v and v v are not symmetric: the embedding Π of G in R P distinguishes these two cases. However, it is easy to add either of these to Ψ sothat the newly added edge is in the closed disc D Ψ bounded by Ψ[ R ] in Ψ.Finally, it remains to show that we may also add M Q to Ψ. Here the argumentdepends slightly on which of v v and v v occurs in G . We will assume, for thesake of definiteness, that it is v v that occurs; the argument in the other case iscompletely analogous. We shall simply import Π[ M Q ] in R P as its embedding inΨ. To this end, let B be any H -bridge contained in M Q so that Π[ B ] ⊆ D . Weshow that either att( B ) ⊆ r r r r [ v , r , a ] or att( B ) ⊆ r r r r [ v , r , a ].We begin by observing that such a B cannot overlap v v (as R -bridges), asboth are are embedded in D by Π. An analogous discussion applies if v v isreplaced by v v .The embedding Π shows B cannot have an attachment in each of (cid:104) r r r (cid:105) and (cid:104) r r r r r (cid:105) . Likewise, B cannot have an attachment in each of (cid:104) r r r (cid:105) and r r r r r . The next claim treats the remaining possibilities. Claim . The H -bridge B does not have an attachment in each of (cid:104) r r r (cid:105) and (cid:104) a , r , v ]. Likewise, B does not have an attachment in each of (cid:104) r r r (cid:105) andeither r or (cid:104) a , r , v ]. Proof.
Suppose by way of contradiction that B has an attachment x in (cid:104) a , r , v ] and an attachment y ∈ (cid:104) r r r (cid:105) . Let P be an H -avoiding xy -pathin B . Since a is an attachment of some element of N + , there is a shortest path S in OD ( Q ) − { v v , v v , M Q } joining some B N in N to a Q -bridge B N + so that B N + has an attachment in [ v , r , x (cid:105) .If B N + ∈ N , then B N + ⊆ D . Lemma 8.8 shows B N + has an attachment in eachof r ∗ and r ∗ +5 ; therefore, B N + is not contained in the closed disc bounded by P anda subpath of r r r r , B N + and P must cross in Π. Therefore, B N + ∈ N + \ N .The neighbour B (cid:48)N + of B N + in S does not have an attachment in [ v , r , x (cid:105) .Since B N + overlaps B (cid:48)N + , it follows that B N + has another attachment in (cid:104) x, r , v ,r , b ]. In particular, the edge e of [ v , r , x ] incident with x is H -green because of B N + .On the other hand, if either x (cid:54) = v or y / ∈ (cid:104) r (cid:105) , then P combines with the xy -subpath of r r r [ v , r , x ] to make another H -green cycle containing e , con-tradicting Theorem 6.7. Therefore, x = v and y ∈ (cid:104) r (cid:105) . But then att( B ) ⊆ Q ,contradicting the fact that B ⊆ M Q .The “likewise” statement has an analogous proof. (cid:3) We now see that Ψ may be extended to include Π[ M Q ], completing the proofwhen Q ∗ = Q .The proof will be completed by now considering the case Q ∗ = Q . The onlydifference in how we proceed is to note that the H -bridges v v and v v , if theyexist, may be transferred to M at the start. To see this, first observe that v v and v v overlap on R and so cannot both be embedded in D . If v v is not containedin M , then we may consider H (cid:48) to be ( H − (cid:104) s (cid:105) ) + v v , relabel H (cid:48) so that v v —the exposed spoke — is s and proceed as above to move v v into M .The following notions will be helpful for the duration of the work. G EMBEDS WITH ALL SPOKES IN M Definition . Let G be a graph, V ∼ = H ⊆ G and let B be an H -bridgein G .(1) If there is an i ∈ { , , , , } so that att( B ) ⊆ Q i , then B is both a local H -bridge and a Q i -local H -bridge .(2) Otherwise, B is a global H -bridge . Corollary . Let G ∈ M and V ∼ = H ⊆ G . Then there is no i so that Q i has BOD and each edge of r i − r i − r i r i +1 is in an H -green cycle consisting ofa global H -bridge and a path in R having at most two H -nodes other than v i . Proof.
By way of contradiction, suppose there is such an i . By Theorem 8.1, G has a representativity 2 embedding in R P so that H ⊆ M . Thus, s i is in a Q i -bridge other than M Q i .By Lemma 6.6 (10), no edge of r i − r i − r i r i +1 can be crossed in any 1-drawing D of G − (cid:104) s i (cid:105) . By hypothesis, Q i has BOD, so Lemma 5.9 implies Q i is crossedin D , which further implies that some edge of r i − r i − r i r i +1 is crossed in D , acontradiction.HAPTER 9 Parallel edges
In this very short chapter, we present some observations on how parallel edgescan occur in 2-crossing-critical graphs. This will be used in later sections, especiallySection 15, where we determine all the 3-connected, 2-crossing-critical graphs thatdo not have a subdivision of V . There are easy generalizations to k -crossing-criticalgraphs. Definition . For an edge e of a graph G , µ ( e ) denotes the number of edgesparallel to e (including e itself). Observation . Let G be a 2-crossing-critical graph and let e and e (cid:48) beparallel edges of G . Then:(1) if G is the underlying simple graph, then G is not planar;(2) the edge e (cid:48) is crossed in any 1-drawing of G − e ;(3) µ ( e ) ≤ ;(4) if e (cid:48) is an edge parallel to e , then G − { e, e (cid:48) } is planar;(5) if cr( G ) > , then G is simple; and(6) if n ≥ and V n ∼ = H ⊆ G , then one of e and e (cid:48) is in the H -rim. Proof.
For (1), a planar embedding of G allows us to introduce all the paralleledges of G with no crossings, showing G is planar, a contradiction.For (2)–(5), let D be a 1-drawing of G − e and suppose e (cid:48) is not crossed in D Then we may add e alongside D [ e (cid:48) ] to obtain a 1-drawing of G , a contradiction.Since D has at most one crossing, it must be of e (cid:48) , which is (2). Adding e alongside D [ e (cid:48) ] yields a 2-drawing of G . Thus we have (4) and (5). Also, (3) follows, sinceany other edge e (cid:48)(cid:48) parallel to e does not cross e (cid:48) in D e . Thus, e (cid:48)(cid:48) is not crossed in D e , which contradicts the second sentence, with e (cid:48)(cid:48) in place of e (cid:48) .Finally, for (6), we may suppose e is not in H . Lemma 3.6 shows that theonly edges that are in every non-planar subgraph of G − e are those in the H -rim.Therefore, e (cid:48) is in the H -rim. (cid:3) HAPTER 10
Tidiness and global H -bridges In this section, we show that, if G ∈ M and V ∼ = H ⊆ G , then there is a V ∼ = H (cid:48) ⊆ G with many useful additional characteristics that we call “tidiness”.The main result is that a tidy subdivision of V has only very particular globalbridges, each of which is an edge. We start with a slightly milder version of tidiness. Definition . Let Π be a representativity 2 embedding of G in R P andlet V ∼ = H ⊆ G . Then H is Π -pretidy if:(1) all H -spokes are embedded in M ; and(2) for every H -quad Q and for every Q -bridge B other than M Q , Q ∪ B hasno non-contractible cycle in Π.The first step in this section is to find an embedding with a pretidy subdivisionof V . Lemma . Let G ∈ M and V ∼ = H ⊆ G . Then G has a representativity 2embedding Π in R P so that H is Π -pretidy. Proof.
By Theorem 8.1, G has a representativity 2 embedding Π in R P so thatall the H -spokes are contained in M and so that, for any Q -bridge B other than M Q , Π[ Q ∪ B ] has no non-contractible cycle. We note that every global H -bridgeis contained in D . We describe a particular representativity 2 embedding Π ∗ of G in R P for which H is Π ∗ -pretidy. Let γ be the non-contractible simple closedcurve that meets Π( G ) at just the two points a and b .The embedding Π ∗ is obtained by adjusting the local H -bridges; we do notadjust those that are Q -local. We start with Π ∗ being the same as Π on H and allthe Q -bridges other than M Q . Let Q be an H -quad other than Q . By Theorem5.23, Q has BOD and all Q -bridges other than M Q are planar. Let ( S, T ) be abipartition of OD ( Q ) labelled so that M Q ∈ T . Let Π Q be a planar embedding of Q and all the Q -bridges other than M Q so that all the Q -bridges in T \ { M Q } areon one side of Π Q [ Q ] and all the Q -bridges in S are on the other side of Π Q [ Q ].Extend Π ∗ to include all the Q -bridges other than M Q by placing the Q -bridgesin S into the H -face in Π ∗ bounded by Π ∗ [ Q ], using Π Q . As every Q -bridge in T \ { M Q } does not overlap M Q , each of these has all its attachments on one of thefour H -branches in Q and these may be embedded in Π ∗ on the other side of Π ∗ [ Q ],and without crossing M Q ∪ γ .The only concern here is that a local H -bridge can be local for distinct H -quads. Such an H -bridge B must have all its attachments on the same H -spoke s i .We claim it is in T for one of Q i − and Q i and in S for the other one of Q i − and Q i . As G is 3-connected, OD ( Q i ) is connected (see [ , Thm. 1], where this isproved for binary matroids). There is a shortest M Q i B -path P = ( B , B , . . . , B n ) H -BRIDGES 63 in OD ( Q i ) (thus, B = M Q i and B n = B ). Let k be least so that B k has anattachment in (cid:104) s i (cid:105) . Claim . For j > k , att( B j ) ⊆ s i , and k ≤ Proof.
If, for some j > k , B j has an attachment not in s i , then j < n . If B j has an attachment in (cid:104) s i (cid:105) , then B j is skew to M Q i and P is not a shortest M Q i B -path, a contradiction. Thus, there is a least j (cid:48) > j so that B j (cid:48) has an attachmentin (cid:104) s j (cid:105) . Since B j (cid:48) overlaps B j (cid:48) − and B j (cid:48) − has no attachment in (cid:104) s i (cid:105) , B j (cid:48) has anattachment not in s i . Again, B j (cid:48) is skew to M Q i , so P is not a shortest M Q i B -path,a contradiction. Thus, for all j > k , att( B j ) ⊆ s i .If k = 0, then obviously k ≤
1, so we may assume k ≥
1. As B k has anattachment in (cid:104) s i (cid:105) and B k − does not, it follows that B k has an attachment not in s i . But then B k is skew to M Q i . Because P is a shortest M Q i B -path, we deducethat k ≤ (cid:3) The claim shows that the Q i -bridges B k +1 , B k +2 , . . . , B n are also Q i − -bridgesand, therefore, ( B k +1 , B k +2 , . . . , B n ) is a path in OD ( Q i − ). Suppose first that k =0. Then M Q i contains a vertex x in (cid:104) s i (cid:105) so that x and v i +1 are skew to B . Thereis a shortest Q i -avoiding path P in M Q i joining x to a vertex in Nuc( M Q i ) ∩ H .Since P is not in the face of Π[ Q i ] contained in M , we deduce that P is containedin the face of Π[ Q i − ] contained in M . But then we conclude that P is containedin a Q i − -local H -bridge B (cid:48) , showing that B (cid:48) is skew to both M Q i − and to B .We deduce that, in OD ( Q i ), M Q i and B are on opposite sides of the bipartition of OD ( Q i ), while M Q i − and B are on the same side of the bipartition of OD ( Q i − ).Since B and B = B n have not changed their relative positions, we see that inone of OD ( Q i ) and OD ( Q i − ), B is on the same side of the bipartition as thecorresponding M¨obius bridge, while in the other B and the other correspondingM¨obius bridge are on opposite sides of the bipartition.The argument works exactly in reverse when k = 1. In this case, B is skew to M Q i and B . Since B ⊆ M Q i − , we conclude that B is skew to M Q i − , and theresult follows analogously to the argument in the preceding paragraph.Finally, suppose B is a global H -bridge. Then, for each H -quad Q , B ⊆ M Q ,so B does not overlap any of the Q -local H -bridges already embedded in D Π ∗ and,since Π[ B ] ⊆ D , B can also be added to Π ∗ .We are now ready to move to tidiness. Definition . Let V ∼ = H ⊆ G and let Π be a representativity 2 embeddingof G . Then H is Π -tidy if:(1) H ⊆ M ;(2) every local H -bridge is contained in M ;(3) for each H -quad Q , no two Q -local H -bridges overlap; and(4) there is no H -avoiding path P in D and an index i ∈ { , , , . . . , } sothat P has both its ends in (cid:104) v i , r i , v i +1 , r i +1 , v i +2 , r i +2 , v i +3 (cid:105) .If V ∼ = H ⊆ G , then H is tidy if there is a representativity 2 embedding Π of G sothat H is Π-tidy.Our aim is the following result. Theorem . Let G ∈ M have a subdivision of V . Then there exists arepresentativity 2 embedding Π in R P of G with a Π -tidy subdivision of V . H -BRIDGES The following concept is central to the proof.
Definition . Let V ∼ = H ⊆ G . Then Loc( H ) denotes the union of H andall the local H -bridges in G . Proof of Theorem 10.4.
For any V ∼ = H ⊆ G , Lemma 10.2 implies there is arepresentativity 2 embedding Π of G in R P so that H is Π-pretidy. Among all H for which Loc( H ) is maximal and all Π so that H is Π-pretidy, we consider the pairs( H, Π) so that G ∩ M Π( H ) is maximal. Among all these pairs ( H, Π), we choose onefor which the number of edges of G in H -spokes in minimized. We claim that this H is Π-tidy. We note that (1) is satisfied by the fact that H is Π-pretidy.If H and Π fail to satisfy either (2) or (4), then either there is an H -quad Q sothat some Q -local H -bridge B is not embedded in M H , or there is an H -avoidingpath P contained in D H and an index i ∈ { , , , . . . , } so that P has both endsin (cid:104) r i r i +1 r i +2 (cid:105) . In the first case, as Q ∪ B has no non-contractible cycles, the onlypossibility is that B has all its attachments in one of the H -rim branches of Q .Thus, the first case is a special case of the second; we now consider the second case.Let P (cid:48) be the subpath of (cid:104) r i r i +1 r i +2 (cid:105) joining the ends u and w of P , with thelabelling chosen so that u is nearer to v i in P (cid:48) than w is. Note that the cycle P ∪ P (cid:48) is an H -green cycle and, therefore, bounds a face of G .We construct a new subdivision H (cid:48) of V in G . The H (cid:48) -rim is obtained from the H -rim by replacing P (cid:48) with P . The spokes s i , s i +3 , and s i +4 of H (cid:48) are also spokesof H (cid:48) . The H -spokes s i +1 and s i +2 might need extension, using the subpaths of r i r i +1 r i +2 joining u and/or w to either v i +1 or v i +2 as necessary, to become spokesof H (cid:48) . Evidently all H (cid:48) -spokes are contained in M H (cid:48) , so H (cid:48) ⊆ G ∩ M H (cid:48) ⊆ Loc( H (cid:48) ).Furthermore, if F is the (closed) face of G bounded by P ∪ P (cid:48) , then M H (cid:48) = M H ∪ F . Claim . Loc( H ) ⊆ Loc( H (cid:48) ). Proof.
Let e be an edge of Loc( H ). If e ∈ M H (cid:48) , then e ∈ Loc( H (cid:48) ), so we mayassume e / ∈ M H (cid:48) . Let B be the local H -bridge containing e . Since e / ∈ M H (cid:48) and M H ⊆ M H (cid:48) , we deduce that B ⊆ D H , and so all attachments of B are in some H -rim branch (recall H is Π-pretidy). Thus, Corollary 5.15 implies B has preciselytwo attachments and therefore is just the edge e . Consequently, B is disjoint from P (it is not in M H (cid:48) ), and so B is an H (cid:48) -bridge, whence e ∈ Loc( H (cid:48) ). (cid:3) If P is not contained in a local H -bridge, then, since P ⊆ Loc( H (cid:48) ), we contradictmaximality of Loc( H ). Therefore, P is contained in, and therefore is, a local H -bridge B . But this implies that H (cid:48) is Π-pretidy and that G has one more edge in M H (cid:48) than it has in M H , contradicting the maximality of G ∩ M H . Therefore, (2)and (4) hold for ( H, Π).It follows that, if H is not Π-tidy, then (3) is violated: there exists an H -quad Q and two Q -bridges B and B (cid:48) in ( M Q ) that overlap. As both B and B (cid:48) arecontained in M , one, say B , is Q -interior in Π, while B (cid:48) is Q -exterior. This impliesthat att( B (cid:48) ) ⊆ s , for some H -spoke s ⊆ Q . Corollary 5.15 implies that B (cid:48) is justan edge uw . We note that B has an attachment x in (cid:104) u, s, w (cid:105) and an attachment y not in [ u, s, w ].Let H (cid:48)(cid:48) be the subdivision of V obtained from H by replacing s with ( s −(cid:104) u, s, w (cid:105) ) ∪ B (cid:48) . We note that H (cid:48)(cid:48) is Π-pretidy, Loc( H (cid:48) ) = Loc( H ), and M H (cid:48)(cid:48) = M H ,so G ∩ M H (cid:48)(cid:48) is maximal. However, the H (cid:48)(cid:48) -spokes have in total at least one feweredge than the H -spokes, contradicting the choice of H .
0. TIDINESS AND GLOBAL H -BRIDGES 65 We now turn our attention to the global H -bridges of a tidy H . Theorem . Let G ∈ M and V ∼ = H ⊆ G . If H is tidy, then any global H -bridge is just an edge, and, in particular, has one of the forms v i v i +2 , v i v i +3 , orhas v i as one end and the other end is in (cid:104) r i − (cid:105) ∪ (cid:104) r i +2 (cid:105) . Proof.
Let Π be a representativity 2 embedding of G for which H is Π-tidy.In particular, all H -spokes and all local H -bridges are in M , and, for each i =0 , , . . . ,
9, no global H -bridge has two attachments in (cid:104) r i r i +1 r i +2 (cid:105) .Let B be a global H -bridge. We note that B ⊆ D . Claim . If there is an i so that att( B ) ⊆ r i r i +1 r i +2 , then either B = v i v i +2 or B = v i +1 v i +3 or B = v i v i +3 or B has v i as one end and the other end is in (cid:104) r i +2 (cid:105) or B has v i +3 as one end and the other end is in (cid:104) r i (cid:105) . Proof.
Because H is tidy, no two attachments of B are in (cid:104) r i r i +1 r i +2 (cid:105) . Thus,at least one of v i and v i +3 is an attachment of B ; for the sake of definiteness, letit be v i . Then tidiness implies no attachment of B can be in (cid:104) r i r i +1 (cid:105) . As tidinessalso implies r i +2 has at most one, and therefore exactly one, attachment of B , theresult follows. (cid:3) Claim . If there is no i so that att( B ) ⊆ r i r i +1 r i +2 , then either att( B ) = { v , v , z } , with z ∈ (cid:104) r (cid:105) ∪ (cid:104) r (cid:105) or att( B ) = { v , v , z } , with z ∈ (cid:104) r (cid:105) ∪ (cid:104) r (cid:105) . Proof.
We may assume that B is embedded in the ( H ∪ γ )-face contained in D and incident with v , v , . . . , v . As H is tidy and B is H -global, there exist i, j ∈ { , , , , , } so that (taking 9 to be equal to − i < j , B has attachments x in r i − v i +1 and y in r j − v j , and j − i ≥
3; choose such i, j so that j − i is assmall as possible. By tidiness, there is no other attachment of B in[ r i − r i r i +1 (cid:105) ∪ (cid:104) r j − r j r j +1 ] . Subclaim . Either i = − j = 4. Proof.
In the alternative, i ≥ j ≤
3. As j − i ≥
3, we conclude that i = 0 and j = 3, so the six H -rim branches r i − , r i , r i +1 , r j − , r j , and r j +1 are alldistinct and cover the entire ab -subpath in the boundary of ( H ∪ γ )-face containing B , with the possible exception of v , in which case both x = v and y = v .Let e be an edge in s and let D be a 1-drawing of G − e . Theorem 5.23 implies Q has BOD; now Lemma 5.9 implies Q is crossed in D . In particular, r r r r crosses r r r r in D .In the case v is an attachment of B , let P and P (cid:48) be H -avoiding v v - and v v -paths in B , respectively. Then the cycles r r [ v , P, v ] and r r [ v , P (cid:48) , v ]are both H -green. Lemma 7.2 (1) implies neither is crossed in D , yielding thecontradiction that r r r r is not crossed in D .Thus, B is the edge xy . Note that B is not a local H -bridge and, therefore,not both v and v are attachments of B . As B is not crossed in D , we deduce thatthe xy -subpath of r r r r is also not crossed in D . Therefore, either r or r iscrossed in D . From this, we conclude that, since Q is crossed in D , r r is crossedin D . Moreover, either s or s is exposed in D . By symmetry, we may assume s is exposed in D .If x (cid:54) = v , then the cycle r r r s r r s r s is clean in D and separates x ∈ (cid:104) r (cid:105) from y ∈ (cid:104) v , r , v ], so B must be crossed in D , a contradiction. If y (cid:54) = v , H -BRIDGES then the cycle r r s r s r r s is clean in D and separates x ∈ [ v , r , v (cid:105) from y ∈ (cid:104) r (cid:105) , and again B is crossed in D , a contradiction. (cid:3) Recall that − Subclaim . (1) If x ∈ [ a, r , v (cid:105) , then there is no attachment in [ v , r , v (cid:105) .(2) If y ∈ (cid:104) v , r , b ], then there is no attachment in (cid:104) v , r , v ].The next two subclaims are rather less trivial. Subclaim . (1) If x ∈ [ a, r , v (cid:105) , then there is no attachment in [ v , r , v (cid:105) .(2) If y ∈ (cid:104) v , r , b ], then there is no attachment in (cid:104) v , r , v ]. Proof.
We prove (1); (2) is symmetric. For (1), suppose there is an at-tachment y (cid:48) in [ v , r , v (cid:105) . By tidiness, there is no attachment other than y (cid:48) in (cid:104) r r r r (cid:105) , and so minimality of j − i implies y (cid:48) = y .The only other possible attachment is in [ v , r , b ]. If there is an attachment z in [ v , r , b ], then either y = v or z = b = v . Thus, either z does not exist and B is the edge xy , or z exists, B has exactly three attachments, namely x , y , and z ,and Lemma 5.19 shows B is a K , . Let P and P (cid:48) be the xy - and yz -paths (thelatter only if z exists) in B .Suppose first that y (cid:54) = v . Then x = v , as otherwise [ y, P, x, r , v ] r r [ v , r , y ]is an H -green cycle with the three H -nodes v , v , v in its interior, contradictingLemma 6.6 (9).Theorem 5.23 (6a) does not apply, as x = v = a implies v (cid:54) = a . If Theorem5.23 (6b) applies, then there is a second H -bridge B (cid:48) attaching at b = v and in r r . But then B and B (cid:48) must cross in Π, a contradiction. Therefore, Theorem5.23 (6) shows Q has BOD.Let e be an edge of s and let D be a 1-drawing of G − e . Lemma 5.9 implies Q is crossed in D . On the other hand, the presence of P and Lemma 7.2 (3a) and(2) imply Q cannot be crossed in D , the desired contradiction.Therefore, y = v . Since x, y ∈ r r r , the hypothesis of the claim implies z must exist. The cycles [ x, P, v ] r r [ v , r , x ] and [ z, P (cid:48) , v ] r r [ v , r , z ] are H -green. Let e be an edge in s and let D be a 1-drawing of G − e . Theorem 5.23implies Q has BOD, so Lemma 5.9 implies Q is crossed in D . However, Lemma7.2 (1) shows that r and r are not crossed. If x (cid:54) = v , then the same result shows r is not crossed and likewise if z (cid:54) = v , then r is not crossed. If, say, x = v , thenLemma 7.2 (3b) implies r can only cross r . However, if z (cid:54) = v , then (2) shows r cannot be crossed.In the remaining case, x = v and z = v . In this case, a = x = v . If Q does not have BOD, then Theorem 5.23 (6) implies b = v and there is a Q -bridge B (cid:48) different from M Q , having attachments at b and in r r , and embedded in D .But then B (cid:48) is an H -bridge different from B that overlaps B on R , while both areembedded in D , a contradiction. (cid:3) Subclaim . (1) If x ∈ [ a, r , v (cid:105) , then there is no attachment in [ v , r , v (cid:105) .(2) If y ∈ (cid:104) v , r , b ], then there is no attachment in (cid:104) v , r , v ]. Proof.
We prove (1); (2) is symmetric. For (1), suppose there is an attach-ment in [ v , r , v (cid:105) . By minimality of j − i , Subclaim 3 and tidiness, this attachmentis y . Also by tidiness, there is no other attachment in (cid:104) r r r r (cid:105) .
0. TIDINESS AND GLOBAL H -BRIDGES 67 Suppose there is also an attachment z in [ v , r , v (cid:105) . The preceding paragraphshows z = v . Tidiness now implies that x is v and, since a ∈ r and x ∈ [ a, r , v (cid:105) , a = v . Let P and P (cid:48) be H -avoiding xz - and yz -paths in B , respectively.Theorem 5.23 (6) implies Q has BOD. If D is any 1-drawing of G − (cid:104) s (cid:105) , thenLemma 5.9 implies Q is crossed in D . But Lemma 7.2 implies (recall z = v )the two H -green cycles [ z, P, x, r , v , r , z ] and [ y, P (cid:48) , z, r , v , r , v , r , y ] are notcrossed in D . Thus, r r r r is not crossed in D (since x = v ), so Q is notcrossed in D , a contradiction.Therefore, there is no attachment in [ v , r , v (cid:105) . Thus, we may assume that theonly attachments in [ a, r , v ] r r r r are x ∈ [ a, r , v (cid:105) and y ∈ [ v , r , v (cid:105) . Tidi-ness further shows there is no attachment in [ v , r , v (cid:105) , so the only other possibleattachment of B is v , in which case y = v .In each of the two cases x (cid:54) = v and x = v , we show that Q has NBOD byshowing that B , M Q , and the Q -bridge B containing s are mutually overlapping.We remark that B and B are in different faces of Π[ H ], so B (cid:54) = B . Obviously, B is skew to M Q . Case 1. x (cid:54) = v .The attachments x and y of B are skew to v and v , so B and B overlap.Also, x and y are skew to v and v , so B and M Q overlap, as required. Case 2. x = v .As x, y ∈ Q and B is not Q -local, there is another attachment z of B . Ourearlier remarks imply z = v and y = v . Now y and z show B and B are skew,while x and y show B and M Q are skew.We now resume our general discussion. Let P xy be the xy -path in B . Since x ∈ [ a, r , v (cid:105) , v (cid:54) = a . Suppose some Q -bridge B (cid:48) has an attachment at b = v and an attachment in r r . Since B is not a Q -bridge and both B and B (cid:48) are H -bridges, B (cid:54) = B (cid:48) . Then P xy and a v [ r r ]-path in B (cid:48) would cross in Π, whichis impossible. Therefore, Theorem 5.23 shows Q has BOD.Let D be a 1-drawing of G − (cid:104) s (cid:105) . Because Q has NBOD, Lemma 5.6 implies D [ Q ] is not clean in D . Since Q has BOD and s is contained in a planar Q -bridge, Lemma 5.9 implies Q is crossed in D . Therefore, s is exposed in D . Thus D [ H − (cid:104) s (cid:105) ] is one of two possible 1-drawings, depending on whether r crosses r r or r crosses r r .If x (cid:54) = v , then P xy cannot be added to D [ H − (cid:104) s (cid:105) ] without introducing asecond crossing, which is impossible. If x = v , then the three attachments of B are not all on the same face of D [ H − (cid:104) s (cid:105) ], so B cannot be added to D [ H − (cid:104) s (cid:105) ]without introducing a second crossing, the final contradiction. (cid:3) We can now complete the proof of Claim 2. Subclaim 1 implies either x ∈ [ a, r , v (cid:105) or y ∈ (cid:104) v , r , b ]. By symmetry, we may assume the former. Subclaims3 and 4 imply y ∈ [ v , r , b ]. If y (cid:54) = v , then Subclaims 2, 3 and 4 (all six state-ments) show that there is no other attachment of B . But then B is Q -local, acontradiction. Therefore, y = v , and, furthermore, there is an attachment z of B in [ v , r , v (cid:105) .If x (cid:54) = v , then both x and z are in (cid:104) r r r (cid:105) , contradicting tidiness. Thus, x = v . H -BRIDGES The claim will be proved once we know z (cid:54) = v . By way of contradiction,suppose z = v . Consider any 1-drawing D of G − (cid:104) s (cid:105) . By Theorem 5.23, Q has BOD. Thus, Lemma 5.9 implies Q is crossed in D . That is, r r r r crosses r r r r in D . In particular, neither s nor s is exposed in D .Since B is global and has attachments at v and v , it must be that D [ B ]is in the face of D [ R ∪ s ∪ s ] incident with s and the crossing. Since v isan attachment of B , v must be in the subpath of r r r r between the crossingand v . But then s is not exposed in D , implying B must cross s in D , acontradiction that shows v is not an attachment of B , completing the proof of theclaim. (cid:3) To complete the proof of the theorem, by way of contradiction assume there isno i so that att( B ) ⊆ r i r i +1 r i +2 . Claim 2 shows either att( B ) = { v , v , z } , with z ∈ (cid:104) r (cid:105) ∪ (cid:104) r (cid:105) or att( B ) = { v , v , z } , with z ∈ (cid:104) r (cid:105) ∪ (cid:104) r (cid:105) . These are all the sameup to the labelling of H , a , and b , so we may assume att( B ) = { v , v , z } , with z ∈ (cid:104) r (cid:105) . Let H (cid:48) be the subdivision of V consisting of H − (cid:104) s (cid:105) , together with the v v -path in B .In order to apply Theorem 7.1, we show that Π is H (cid:48) -friendly. If Π is not H (cid:48) -friendly, then Lemma 6.5 (1) implies (since H and H (cid:48) have the same nodes) v v is an edge and Π[ v v ] is contained in M H (cid:48) , which is the same as M H . But v and v are not incident with the same H -face in M H and, therefore, this is impossible.Thus, Π is H (cid:48) -friendly. However, H (cid:48) violates Theorem 7.1, a contradiction.Therefore, there is an i so that att( B ) ⊆ r i r i +1 r i +2 . Claim 1 implies B hasone of the three desired forms.We can go somewhat further in our analysis of the global H -bridges of a tidy V ∼ = H ⊆ G . Definition . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let B be aglobal H -bridge with attachments x and y .(1) The span of B is the xy -subpath R with the fewest H -nodes.(2) An edge or subpath of R is spanned by B if it is in the span of B .(3) B is: a if, for some i , its attachments are v i and v i +2 ; a if,for some i , its attachments are v i and v i +3 ; or else is a .We remark that Theorem 10.6 implies that, in the case of a 2.5-jump, there isan i so that v i is one attachment and the other attachment is in (cid:104) r i − (cid:105) ∪ (cid:104) r i +2 (cid:105) .Theorem 10.6 further implies a global H -bridge has precisely two attachments andits span has at most four H -nodes. It follows from Definition 6.2 that every global H -bridge combines with its span to form an H -green cycle. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. For each i ∈{ , , , , } , either Q i has BOD or one of v i − v i − and v i +1 v i +4 is a global H -bridge. Proof.
Let Π be an embedding of G in R P so that H is Π-tidy. Suppose neitherof the edges v i − v i − and v i +1 v i +4 occurs in G . The Q i -bridges that are Q i -exteriorconsist of M Q i , those that are contained in M and, therefore, attach along either s i − or s i +1 , and those that are contained in D . Since H is Π-tidy, these lattermust be global. By Theorem 10.6 they are 2-, 2.5-, and 3-jumps.
0. TIDINESS AND GLOBAL H -BRIDGES 69 Consider any global H -bridge. It is embedded in D so that it, together withits spanned path in R , bounds a face of G . In particular, if we are considering a2-jump B that is a Q i -bridge, the 2-jump is either v i − v i +1 or v i +4 v i +6 . In thiscase, Q i ∪ B has no non-contractible cycle in R P and so, by Lemma 5.16, B doesnot overlap any other Q i -exterior Q i -bridge.It is not possible for a 2.5-jump to be a Q i -bridge. The only 3-jumps that canbe a Q i -bridge are v i +1 v i +4 and v i − v i − , and these are assumed not to be in G .We conclude that the Q i -exterior Q i -bridges do not overlap and, therefore, Q i hasBOD. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Then:(1) no two global H -bridges have an H -node in common;(2) at most one global H -bridge is a -jump;(3) there is no i so that v i v i +3 is a 3-jump and some 2.5-jump has an end in (cid:104) v i − , r i − , v i ] ;(4) if B and B are global H -bridges, then, for every i ∈ { , , , , } , thereis some edge of Q i ∩ R that is not spanned by either B or B ; and(5) for each i ∈ { , , , , } , at most one of (cid:104) r i (cid:105) and (cid:104) r i +5 (cid:105) can contain anend of a 2.5-jump. Proof.
We start with (1).
Claim . No two global H -bridges have an H -node in common. Proof. suppose by way of contradiction that the two global H -bridges B and B have the H -node v i in common. For j = 1 ,
2, let P j be the subpath of R spanned by B j . Then each of B j ∪ P j is a green cycle; therefore, Theorem 6.7implies P and P are edge disjoint. We choose the labelling so that r i ∪ r i +1 ⊆ P and r i − ∪ r i − ⊆ P . We treat various cases. Subclaim . At least one of B and B is not a 3-jump. Proof.
Suppose to the contrary that B and B are both 3-jumps, so B = v i v i +3 and B = v i − v i , respectively. Then there is a 1-drawing D i of ( H − s i ) ∪ B ∪ B ; Lemma 10.8 implies Q i has BOD, so Lemma 5.9 implies Q i is crossed in D i . Because of B , Lemma 7.2 (3a) implies r i +1 and r i +2 are not crossed in D i ,while (3b) of the same lemma implies that if r i were crossed, it would cross r i +3 .However, (2) shows r i +3 is not crossed. Therefore, no edge of r i r i +1 r i +2 is crossedin D i . Analogously, no edge of r i − r i − r i − is crossed in D i . These two assertionsshow Q i cannot be crossed in D i , a contradiction. (cid:3) Subclaim . Neither B nor B is a 3-jump. Proof.
By Claim 1, not both B and B are 3-jumps. So suppose for sake ofdefiniteness that B is the 3-jump v i v i +3 and B is a global H -bridge with one endat v i and one end in (cid:104) v i − , r i − , v i − ].The embedding in R P shows that v i +2 v i +5 is not an edge of G (it would cross B ) and Claim 1 shows v i − v i is not an edge of G . Therefore, Lemma 10.8 implies Q i +1 has BOD. Thus, in any 1-drawing D i +1 of G − (cid:104) s i +1 (cid:105) , Lemma 5.9 implies Q i +1 is crossed in D i +1 . H -BRIDGES By Lemma 7.2 (3a) (when B is a 2.5-jump) or (1) (when B is a 2-jump), r i − is not crossed in D i +1 . Likewise, (1) shows that none of r i , r i +1 , and r i +2 is crossedin D . But then Q i +1 is not crossed in D i +1 , a contradiction. (cid:3) By Claim 2, we know that neither B nor B is a 3-jump. By Theorem 6.7,neither v i − v i − nor v i +1 v i +4 can occur in G ; Lemma 10.8 implies Q i has BOD.Let D i be a 1-drawing of G − (cid:104) s i (cid:105) . By Lemma 5.9, Q i is crossed in D i .Lemma 7.2 (1) shows that P and P are both not crossed in D i . This impliesthat r i − r i − r i r i +1 is not crossed in D and, therefore, Q i is not crossed in D i , acontradiction that completes the proof of the claim. (cid:3) We move on to (2).
Claim . There is at most one global H -bridge that is a 3-jump. Proof.
Suppose there are distinct 3-jumps. Claim 1 implies that, up to rela-belling, they are either v i v i +3 and v i +4 v i +7 or v i v i +3 and v i +5 v i +8 . Theorem 6.7and Claim 1 imply that there cannot be a third 3-jump. Thus, Lemma 10.8 implies Q i +1 has BOD.Let C and C be the two H -green cycles containing these 3-jumps. Lemma5.9 implies Q i +1 is crossed in a 1-drawing D i +1 of G − (cid:104) s i +1 (cid:105) . But Lemma 7.2 (1)implies that neither r i r i +1 nor r i +5 r i +6 is crossed in D i +1 , a contradiction provingthe claim. (cid:3) We next turn to (3).
Claim . There is no i so that v i v i +3 is a 3-jump and some 2.5-jump has anend in (cid:104) v i − , r i − , v i ]. Proof.
Suppose to the contrary that there is such an i . From Claim 1, the2.5-jump has an end w ∈ (cid:104) v i − , r i − , v i (cid:105) . Its other end is v i − . Lemma 10.8 andClaim 2 imply that Q i +2 has BOD. Let D i +2 be a 1-drawing of G − (cid:104) s i +2 (cid:105) . Lemma5.9 implies Q i +2 is crossed in D i +2 .By Lemma 7.2 (2), r i +3 is not crossed in D i +2 . The same lemma (1) implies r i r i +1 r i +2 is not crossed in D i +2 . Consequently, Q i +2 is not crossed in D i +2 ,contradicting the preceding paragraph and proving the claim. (cid:3) Now we prove (4).
Claim . If B and B are global H -bridges, then, for every i ∈ { , , , , } ,some edge of Q i ∩ R is not spanned by either B or B . Proof.
Suppose by way of contradiction that the global H -bridge B spansthe side r i ∪ r i +1 of Q i +1 and a second global H -bridge B spans r i +5 ∪ r i +6 . Tosee that Q i +1 has BOD, by Lemma 10.8 it suffices to show that neither of the3-jumps v i v i − and v i +2 v i +5 is in G . For the former, Theorem 6.7 implies v i is anattachment of B , contradicting Claim 1. For the latter, v i +2 is an attachment of B , with the same contradiction. Therefore Q i +1 has BOD.Lemma 5.9 implies that, for any 1-drawing D i +1 of G −(cid:104) s i +1 (cid:105) , Q i +1 is crossed in D i +1 . However, Lemma 7.2 (1) implies that neither r i r i +1 nor r i +5 r i +6 is crossedin D i +1 , showing Q i +1 is not crossed in D i +1 , a contradiction proving the claim. (cid:3)
0. TIDINESS AND GLOBAL H -BRIDGES 71 Finally, we prove (5). Suppose, for j ∈ { i, i + 5 } , (cid:104) r j (cid:105) contains an end of the2.5-jump B j . We may use the symmetry to assume that B i = wv i − . If B i +5 has v i +3 as an end, then we contradict Claim 4. Therefore, B i +5 has v i +8 = v i − as anend, contradicting Claim 1.We conclude this section with two observations about local bridges of a tidysubdivision of V . Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Then no H -bridgehas all its attachments in one H -spoke. Proof.
By way of contradiction, suppose B is an H -bridge and s is an H -spokeso that att( B ) ⊆ s . By Corollary 5.15, B has precisely two attachments, so B isjust an edge uw . Choose B so that no other H -bridge has all its attachments in aproper subpath of [ u, s, w ]. If [ u, s, w ] has no interior vertex, then B and [ u, s, w ]are parallel edges not in the H -rim, contradicting Observation 9.2 (6). Thus, some H -bridge B (cid:48) has an attachment x in (cid:104) u, s, w (cid:105) .Let Π be an embedding of G in R P for which H is Π-tidy. Since H ⊆ M , B (cid:48) is a local H -bridge. Moreover, Corollary 5.15 and the choice of B show that not allattachments of B (cid:48) can be in [ u, s, w ], so B has an attachment y not in [ u, s, w ]. Butthen, for at least one of the two H -quads Q containing s , B and B (cid:48) are overlapping Q -bridges, contradicting the definition of tidiness. Lemma . Let G ∈ M , V ∼ = H ⊆ G , with H tidy. For any H -spoke s , if B is an H -bridge having an attachment in (cid:104) s (cid:105) , then B has no other attachment in [ s ] . Proof.
Suppose B is an H -bridge and s an H -spoke so that B has attachments x, y in s , with x ∈ (cid:104) s (cid:105) . Let Π be an embedding of G in R P for which H is Π-tidy. Then Π shows B is not a global H -bridge. By Lemma 10.10, B has a thirdattachment z not in [ s ]. Let Q be the unique H -quad containing all of x , y , and z .If y is not an H -node, then let r be an H -rim branch of Q not containing z .Then x , y , and z are all contained in Q − [ r ], contradicting Corollary 5.15. Thus, y is an H -node v i . We choose the labelling so that r i ⊆ Q . Corollary 5.15 showsthat z is not in Q − [ r i +5 ] and, therefore, z is in r i +5 . Furthermore, Corollary5.15 now shows that B can have no other attachment, so Theorem 8.2 implies B isisomorphic to K , . Let w be the vertex in Nuc( B ). Claim . The cycles [ y, B, w, B, x, s, y ] and [ z, B, w, B, x, Q − y, z ] bound facesof Π[ G ]. Proof.
For the latter, [ z, B, w, B, x, Q − y, z ] is an H -green cycle, so the resultfollows from Lemma 6.6. The former, call it C , has just one vertex in R , so Lemma5.20 implies it has BOD and every one of its bridges other than the one containing H − (cid:104) s (cid:105) is planar. If it has a second bridge B (cid:48) , then C is clean in any 1-drawing of B (cid:48) , contradicting Lemma 5.9. (cid:3) The chosen labelling shows that Q i − is the other H -quad containing s . Claim . There is no Q i − -local H -bridge that has an attachment in (cid:104) s (cid:105) . H -BRIDGES Proof.
Suppose B (cid:48)(cid:48) is a Q i − -local H -bridge having an attachment x (cid:48) in (cid:104) s (cid:105) .Lemma 10.10 implies B (cid:48)(cid:48) has an attachment z (cid:48) not in [ s ]. If z (cid:48) is in the same H -rimbranch r i − contained in Q i − as y , then [ x (cid:48) , B (cid:48)(cid:48) , z (cid:48) , r, y, B, w, x, s, x (cid:48) ] is an H -greencycle C . As the edge of s incident with y is C -interior, C does not bound a face ofΠ[ G ]. If z (cid:48) is not in r i − , then (cid:2) z (cid:48) , B (cid:48)(cid:48) , x (cid:48) , s, x, B, w, B, z, Q i − y, z (cid:48) (cid:3) is a non-facial H -green cycle. Both conclusions contradict Lemma 6.6 (8). (cid:3) We conclude that s has length 2 and that B is the only H -bridge attaching in (cid:104) s (cid:105) . Let D be a 1-drawing of G − wy . Then D [ s ∪ ( B − wy )] is clean in D and wemay extend D to a 1-drawing of G by adding in wy alongside [ w, B, x, s, y ].HAPTER 11 Every rim edge has a colour
In this section we introduce, for a tidy subdivision H of V in G , H -yellowedges. The main result is that every H -rim edge has a colour: H -green, H -yellow,or red. This is a major step on the route. In the next section, we will analyze rededges, with the main result being that there are red edges. Definition . Let H be a subdivision of V in a graph G .(1) A 3 -rim path is a path contained in the union of three consecutive H -rimbranches.(2) The closure cl( Q ) of an H -quad Q is the union of Q and all Q -local H -bridges.(3) Let H be tidy in G . A cycle C in G is H -yellow if C may be expressedas the composition P P P P of four paths so that:(a) P and P are R -avoiding (recall R is the H -rim) and have length atleast 1;(b) P and P are 3-rim paths and P ∪ P is not contained in a 3-rimpath; and(c) there is an H -green cycle C (cid:48) so that P ⊆ (cid:104) C (cid:48) ∩ R (cid:105) .(4) An H -rim edge e is H -yellow if it is not H -green and is in an H -yellowcycle.We remark that the H -rim edges that are H -yellow are those in P . The nextresult elucidates the nature of an H -yellow cycle. Lemma . Let G ∈ M , V ∼ = H ⊆ G , with H tidy. Let C be an H -yellowcycle, with decomposition P P P P into paths as in Definition 11.1, and let C (cid:48) bethe witnessing H -green cycle. Then:(1) C (cid:48) − (cid:104) C (cid:48) ∩ R (cid:105) is a global H -bridge;(2) for i ∈ { , } , P i is either H -avoiding or decomposes as P i P i , where P i is contained in some H -spoke, including an incident H -node, and P i is H -avoiding;(3) there is only one C -bridge in G ; and(4) there is an i ∈ { , , , , } so that C ⊆ cl( Q i ) . Proof.
Let Π be an embedding of G in R P for which H is Π-tidy; in particular,every H -green cycle bounds a face of Π[ G ].For (1), the alternative is that C (cid:48) is contained in cl( Q ), for some H -quad Q .Lemma 6.6 (8) shows that C (cid:48) bounds a face of G in R P , so P and P are containedin global H -bridges. Each of P and P is in an H -green cycle (as is every global H -bridge) and, since P has an end in (cid:104) C (cid:48) ∩ R (cid:105) , some edge of C (cid:48) ∩ R is in two H -green cycles, contradicting Theorem 6.7.
734 11. EVERY RIM EDGE HAS A COLOUR
For (2), let i ∈ { , } . Since P i has positive length, the end u i of P i in P isdistinct from the end w i of P i in P . Because C (cid:48) bounds a face of G and is containedin D , we see that the edges of P i incident with u i is in M . Since P i is R -avoiding, P i is contained in M , with only its ends in R .Now suppose P i has an edge e not in H . Choose e to be as close to u i in P i aspossible. As w i is in H , there is a first vertex y of P i after e that is in H . If y = w i ,then we are done, so we may assume y (cid:54) = w i . Since P i is R -avoiding, we see that y must be in the interior of some spoke s . Let z be the vertex of P i incident to e sothat e is in [ z, P i , y ].As P i is contained in M , we see that [ u i , P i , y ] is contained in a closed Π[ H ]-facebounded by some H -quad Q . Also, [ z, P i , y ] is H -avoiding and so is contained insome Q -local H -bridge B i . By Lemma 10.11, y is the only attachment of B i in [ s ].Since z (cid:54) = y and both are attachments of B i , we have that z / ∈ [ s ].The path [ u i , P i , z ] is R -avoiding and contained in H . Therefore, either it istrivial or it is contained in some H -spoke s (cid:48) . In the latter case, z (cid:54) = y implies s (cid:48) (cid:54) = s .In the former case, u i = z , so u i / ∈ s . In both cases, [ u i , P i , y ] ∪ Q contains an H -green cycle that contains an H -rim edge incident with u i , contradicting Theorem6.7 and completing the proof of (2).For (3), we start by noting that there exist i and j so that P ⊆ r i r i +1 , . . . , r j and i − ≤ j ≤ i + 2; we assume P has one end in [ v i , r i , v i +1 (cid:105) , one end in (cid:104) v j , r j , v j +1 ], and that j = i − P is just the single H -node v i . Item 2implies P is contained in cl( Q i − ) ∪ cl( Q i ) and that P is contained in cl( Q j ) ∪ cl( Q j +1 ). It follows that P has its ends in r i +4 r i +5 and r j +5 r j +6 . There are atmost ( j + 6) − ( i + 3) ≤ H -rim branches r i +4 r i +5 . . . r j +6 , so P , being a 3-rimpath, must be contained in this path. It follows that C is disjoint from either s i − or s i +2 .Let s be an H -spoke disjoint from C and let M C denote the C -bridge containing s . Set R (cid:48) = ( R − (cid:104) C (cid:48) ∩ R (cid:105) ) ∪ ( C (cid:48) − (cid:104) C (cid:48) ∩ R (cid:105) ). Then R (cid:48) ∪ s contains a non-contractible cycle C (cid:48)(cid:48) disjoint from C . Lemma 5.20 shows C is contractible, hasBOD, and every C -bridge other than M C is planar.Suppose there is a C -bridge B other than M C ; let D be a 1-drawing of B .Lemma 5.9 implies D [ C ] is crossed. Let s , s (cid:48) , and s (cid:48)(cid:48) be the three H -spokes disjointfrom (cid:104) C (cid:48) ∩ R (cid:105) . Then R ∪ s ∪ s (cid:48) ∪ s (cid:48)(cid:48) is a subdivision of V in B that is edge-disjointfrom both P and P ; this shows that some edge of P ∪ P is crossed in D .But now R (cid:48) ∪ s ∪ s (cid:48) ∪ s (cid:48)(cid:48) is another subdivision of V in B . Therefore, thecrossing in D must involve two edges of R (cid:48) ∪ s ∪ s (cid:48) ∪ s (cid:48)(cid:48) . In particular it does notinvolve an edge of C (cid:48) ∩ R , and, since P ⊆ C (cid:48) ∩ R , no edge of P is crossed in D .Likewise, let R (cid:48)(cid:48) be obtained from R (cid:48) by replacing P with P P P . Now R (cid:48)(cid:48) ∪ s ∪ s (cid:48) ∪ s (cid:48)(cid:48) is a third subdivision of V in B that is disjoint from P . Thus,the crossing in D does not involve an edge of P . Thus, none of P , P , P , and P is crossed in D , contradicting the fact that C is crossed in D . We conclude thatthere is no C -bridge other than M C , as claimed.Finally, for (4), suppose first that P is not contained in a single H -rim branch.Then there is an H -node v i in the interior of P . However, P is incident on one sidewith the face bounded by C (cid:48) , so the edge of s i incident with v i is on the other sideof P . Since C is contractible, we conclude that there are at least two C -bridges,contradicting (3). Therefore, there is an i ∈ { , , , , } so that P ⊆ r i .
1. EVERY RIM EDGE HAS A COLOUR 75
If both P and P are contained in cl( Q i ), then so is P , as it is a 3-rim path.Therefore, by symmetry, we may assume that P has some edge not in cl( Q i ). Aswe traverse P from its end in P , we come to a first edge e that is not in cl( Q i ).One end of e is the vertex u that is in either s i or s i +1 ; for the sake of definiteness,we assume the former. Then (2) implies [ v i , s i , u ] ⊆ P and that the remainderof P consists of an H -avoiding uw -path, with w an end of P . It follows that w ∈ r i +4 . Let ˆ e be the edge of s i incident with u and not in P .Switching paths, we know that P has an end x in r i . If x (cid:54) = v i +1 , then (2)implies P ⊆ cl( Q i ). In this case, ˆ e is in a C -bridge other than M C , contradicting(3). Otherwise x = v i +1 , in which case P P [ w, r i +4 , v i +5 , r i +5 , v i +6 , s i +1 , v i +1 ]is an H -yellow cycle (cid:98) C . There is a (cid:98) C -bridge other than M (cid:98) C containing ˆ e , alsocontradicting (3) for (cid:98) C .We now turn our attention to the all-important red edges. We comment that,if n ≥ V n ∼ = H ⊆ G , then any red edge of G is in the H -rim.The remainder of this section is devoted to proving the following. Theorem . Let G ∈ M and let V ∼ = H ⊆ G . If H is tidy, then every H -rim edge is one of H -green, H -yellow, and red. We start with an easy observation.
Lemma . Let G ∈ M and let V ∼ = H ⊆ G . If H is tidy and the H -rimedge e is either H -green or H -yellow, then e is not red. Proof.
Suppose first that e is H -green and let C be the H -green cycle containing e .There are three H -spokes s , s (cid:48) , and s (cid:48)(cid:48) disjoint from (cid:104) C ∩ R (cid:105) . Thus, ( R −(cid:104) C ∩ R (cid:105) ) ∪ ( C − (cid:104) C ∩ R (cid:105) ) together with s , s (cid:48) , and s (cid:48)(cid:48) is a subdivision of V contained in G − e ,showing e is not red.Now suppose e is H -yellow and let C be the H -yellow cycle containing e . Let C (cid:48) be the H -green cycle and P P P P the decomposition of C as in Definition11.1. Then e is in P and there are three H -spokes s , s (cid:48) , and s (cid:48)(cid:48) disjoint from C ∪ (cid:104) C (cid:48) ∩ R (cid:105) . In this case, ( R − ( (cid:104) C (cid:48) ∩ R (cid:105) ∪ (cid:104) P (cid:105) )) ∪ ( C (cid:48) − (cid:104) C (cid:48) ∩ R (cid:105) ) ∪ P P P ,together with s , s (cid:48) , and s (cid:48)(cid:48) is a subdivision of V contained in G − e , showing e isnot red.The following concepts and lemma play a central role in the proof of Theorem11.3. Definition . Let V ∼ = H ⊆ G . Let e and f be two edges of the H -rim R . Then e and f are R -separated in G if G has a subdivision H (cid:48) of V so that the H (cid:48) -rim is R and e and f are in disjoint H (cid:48) -quads.The following two observations are immediate from the definition. Observation . Let V ∼ = H ⊆ G and suppose e and f are two edges of the H -rim R that are R -separated in G .(1) If D is a 1-drawing of G , then e and f do not cross each other in D .(2) If H (cid:48) is a V in G witnessing the R -separation of e and f , then there aretwo H (cid:48) -spokes that have all their ends in the same component of R −{ e, f } . The following is a kind of converse of Observation 11.6 (1).
Lemma . Let G ∈ M be a graph and let V ∼ = H ⊆ G , with H tidy.Suppose G ⊆ G with H ⊆ G . Let e ∈ r i and f ∈ r i +4 ∪ r i +5 ∪ r i +6 be edges thatare both neither H -green nor H -yellow. If e and f are not R -separated in G , thenthere is a 1-drawing of G in which e crosses f . Proof.
Let Π be an embedding of G in R P so that H is Π-tidy.We may write r i = [ v i , . . . , x e , e, y e , . . . , v i +1 ] and, by symmetry, we may assume f is in r i +5 ∪ r i +6 = [ v i +5 , r i +5 , . . . , x f , f, y f , . . . , r i +6 , v i +7 ] . If f ∈ r i +5 , then let J e,f = cl( Q i ) and Q = Q i , while if f ∈ r i +6 , then let J e,f = cl( Q i ) ∪ cl( Q i +1 ) and Q = Q i +1 . The two H -spokes contained in Q are s i and s e,f , which is either s i +1 or s i +2 . Claim . There are not totally disjoint s i s e,f -paths in J e,f − e . Proof.
Because H is Π-tidy, Π[ J e,f ] is contained in the closed disc boundedby Π[ Q ]. Therefore, one of a pair of totally disjoint s i s e,f -paths in J e,f would bedisjoint from r i +5 r i +6 and it, together with a subpath of Q − r i +5 r i +6 yields thecontradiction that e is H -green. (cid:3) Let w e be a cut-vertex in J e,f − e separating s i from s e,f . Then J e,f − e hasa separation ( H e , K e ) with s i ⊆ H e , the other H -spoke s e,f contained in Q iscontained in K e , and H e ∩ K e = (cid:107) w e (cid:107) . Clearly, w e ∈ r i +5 r i +6 .There is also a separation ( H f , K f ) of J e,f − f , so that H f ∩ K f is a singlevertex w f , s i ⊆ H f , and s e,f ⊆ K f . For x ∈ { e, f } , there is a face F x of Π[ J e,f ]incident with both x and w x . If F e = F f , then any vertex of r i r i +1 in the bound-ary cycle C of F e may be selected as w f . Similarly, w e may be any vertex of r i +5 r i +6 that is in C . We choose w e and w f so that they are in different com-ponents of C − { e, f } . Thus, whether F e = F f or not, the cycle Q has the form[ w e , . . . , e, . . . , w f , . . . , f, . . . ]. In particular, e and w e are in the same componentof Q − { w f , f } , while f and w f are in the same component of Q − { w e , e } . Byinterchanging the roles of e and f and exchanging the labels of v j and v j +5 , for j = 0 , , , ,
4, we may assume Q has the form[ w e , . . . , v i +5 , s i , v i . . . , e, . . . , w f , . . . , s e,f , . . . , f, . . . ] . For technical reasons, we choose w e as close as possible to f in r i +5 r i +6 and w f as close as possible to e in r i +5 r i +6 , while respecting the ordering that was justdescribed of these four elements of Q .Set N = K e ∩ H f . Then J e,f − { e, f } = H e ∪ N ∪ K f , H e ∩ N = (cid:107) w e (cid:107) , and K f ∩ N = (cid:107) w f (cid:107) . See Figure 11.1. Claim . N does not have disjoint paths both with ends in the two componentsof N ∩ R . Proof.
Such paths, together with the H -rim and the H -spokes s i − and s i +3 ,would show e and f are R -separated. (cid:3) Let w be a cut-vertex in N separating the two components of N ∩ R , and let( N i , N i +5 ) be a separation of N so that, for j ∈ { i, i + 5 } , N j ∩ R is contained in r j ∪ r j +1 and N i ∩ N i +5 = (cid:107) w (cid:107) . We proceed to describe a new 2-representativeembedding of G in R P that shows that G has a 1-drawing.
1. EVERY RIM EDGE HAS A COLOUR 77 f N K f H e w e w f e Figure 11.1.
The locations of e , f , w e , w f , H e , N , and K f .Let G (cid:48) be the subgraph of G obtained by deleting all the vertices and edges of N that are not in N ∩ R . There is a face of Π[ G (cid:48) ] contained in M and incident withboth e and f . Claim . No global H -bridge has a vertex in (cid:104) N i ∩ R (cid:105)∪(cid:104) N i +5 ∩ R (cid:105) in its span. Proof.
For sake of definiteness, suppose some vertex of (cid:104) N i ∩ R (cid:105) is in the spanof the global H -bridge B . If the H -node v e,f in r i r i +1 incident with s e,f is in theinterior of the span of B , then the cycle bounding F f is H -yellow, contradictingthe fact that f is not H -yellow. Letting z be the vertex of N i nearest e in r i r i +1 ,we conclude that B has an attachment in (cid:104) z, r i r i +1 , v e,f ], and B does not span anyedge of r i +2 .By Theorem 10.6, B is either a 2-, 2.5-, or 3-jump. It follows from the precedingparagraph that e is contained in the span of B , yielding the contradiction that e is H -green. (cid:3) We can now easily complete the proof of the lemma. By Claim 3, we canseparately embed N i and N i +5 in the face outside of M . As no global H -bridge canattach on both paths in R − { e, f } without making at least one of e and f H -green,we can join the two copies of w together to obtain a representativity 2 embeddingΠ (cid:48) of G in R P having a non-contractible simple closed curve meeting Π (cid:48) [ G ] onlyin the interiors of e and f . This implies that G has a 1-drawing, as required.We further investigate the detailed structure of H -rim edges. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. If v i v i +3 is a global H -bridge, then, for j ∈ { i − , i +3 } there is an edge e j ∈ r j that is neither H -yellownor H -green. Proof.
The two sides are symmetric, so it suffices to prove the existence of e i − .Lemmas 10.8 and 10.9 (2) imply that Q i +1 has BOD. Let D be a 1-drawing of G − (cid:104) s i +1 (cid:105) . Lemma 5.9 implies Q i +1 is crossed in D .However, the cycle C consisting of v i v i +3 and the path it spans is H -close, for H = R ∪ s i − ∪ s i ∪ s i +3 . Therefore, Lemmas 5.3 and 5.4 imply that C is notcrossed in D . We conclude from the nature of 1-drawings of V that r i − crosses r i +5 ∪ r i +6 ; let e be the edge in r i − that is crossed in D . Suppose, by way of contradiction, that there is a global H -bridge B spanning e .Theorem 10.6 implies B is either a 2-, 2.5- or 3-jump, while Theorem 6.7 implies B does not span any edge of r i (such an edge is already spanned by v i v i +3 ). Lemma10.9 (1) implies v i is not an attachment of B , so B must be a 2.5-jump with oneend in (cid:104) r i − (cid:105) , contradicting Lemma 10.9 (3). Thus, e is not spanned by a global H -bridge.It follows that, if e is in an H -green cycle C (cid:48) , then C (cid:48) ⊆ cl( Q i − ). But such a C (cid:48) is H -close, for H = R ∪ s i ∪ s i +2 ∪ s i +3 . By Lemmas 5.3 and 5.4, C (cid:48) is notcrossed in any 1-drawing of G − (cid:104) s i +1 (cid:105) . This contradicts the fact that e is crossedin D . We conclude that e is not H -green.So now we suppose e is in the H -yellow cycle C (cid:48) and that C (cid:48)(cid:48) is a witnessing H -green cycle. Then C (cid:48) ⊆ cl( Q i − ) and C (cid:48)(cid:48) contains a global H -bridge B thatspans an edge in r i +4 . This implies B (cid:54) = v i v i +3 , so Lemma 10.9 (2) shows that B is not a 3-jump.Moreover, (3) of the same lemma shows B cannot have an attachment in[ v i +3 , r i +3 , v i +4 (cid:105) , while (4) shows B cannot have v i +7 as an attachment. There-fore, B is a 2- or 2.5-jump v i +4 w , with w ∈ [ v i +6 , r i +6 , v i +7 (cid:105) .The cycle ( R − (cid:104) C (cid:48)(cid:48) ∩ R (cid:105) ) ∪ B , together with the H -spokes s i − , s i +2 , and s i +3 is a subdivision H of V for which C (cid:48) is H -close, showing that e is not crossed inany 1-drawing of G − (cid:104) s i +1 (cid:105) . This contradicts the fact that e is crossed in D and,therefore, e is not H -yellow.The proof of Theorem 11.3 will also depend on the following new concepts. Definition . Let G be a graph and let V ∼ = H ⊆ G , with H tidy. Let Πbe an embedding of G in R P so that H is Π-tidy and has the standard labellingrelative to γ . For i ∈ { , , , . . . , } :(1) ← P i = r i − r i − , P ← i = r i +3 r i +4 , → P i = r i +1 r i +2 , and P → i = r i +6 r i +7 .(2) the spines (cid:61) i and i (cid:60) of Q i consist of the paths ← P i ∪ s i ∪ P ← i and → P i ∪ s i +1 ∪ P → i , respectively (see Figure 11.2);(3) the scope K i of Q i consists of cl( Q i ) ∪ (cid:61) i ∪ i (cid:60) ∪ B i , where B i consistsof all global H -bridges having both attachments either in ← P i ∪ → P i or in P ← i ∪ P → i ; and(4) the complement K (cid:92)i of K i is obtained from M Q i by deleting the edges (butnot their incident vertices) that comprise the H -bridges in B i .(5) The two vertices v i − and v i +3 are the trivial (cid:61) i i (cid:60) -paths in K i . Anyother (cid:61) i i (cid:60) -path in K i is non-trivial .We note that (cid:61) i ∩ i (cid:60) is equal to (cid:107){ v i − , v i +3 }(cid:107) . For our purposes, these arenot “useful” (cid:61) i i (cid:60) -paths.We observe that, for each i ∈ { , , , , } , G = K i ∪ K (cid:92)i .The following lemma plays an important role in the rest of this section. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let e be an edgeof R and let i be such that e ∈ r i . Then G − e has a subdivision of V if and onlyif there are disjoint non-trivial (cid:61) i i (cid:60) -paths in K i − e .
1. EVERY RIM EDGE HAS A COLOUR 79 (cid:61) (cid:60) r r r r r r r r r r ← P → P P ← P → ab ba Figure 11.2.
The paths with small dashes are ← P , P ← , → P , and P → . The spine (cid:61) is the path r r s r r , while (cid:60) is r r s r r .There is some subtlety here; 2-criticality is important. Suppose we have asubdivision H of V embedded in R P with representativity 2 so that all the H -spokes are in M . Give H the usual labelling relative to γ . Now delete (cid:104) r (cid:105) and (cid:104) r (cid:105) , and then add the 2.5-jump av and the 3-jump v v . Then there are disjointnon-trivial (cid:61) (cid:60) -paths in the union H (cid:48) of ( H − (cid:104) r (cid:105) ) − (cid:104) r (cid:105) and the two jumps,but H (cid:48) is planar.We shall need the following. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let e be an edgeof R and let i be such that e ∈ r i . If there are disjoint non-trivial (cid:61) i i (cid:60) -paths in K i − e , then there are two such paths so that at least one of them is contained in cl( Q i ) and the other contains at most one global H -bridge. In the proof, we consider many possibilities for the two disjoint (cid:61) i i (cid:60) -paths.For a given i , some possibilities might not occur because of limitations imposed byΠ. In principle, for i = 2, all of the considered possibilities can occur, while for i = 4, several of the considered possibilities cannot occur. Proof.
Let P and P be the hypothesized disjoint paths. Claim . If there is a (cid:61) i i (cid:60) -path in K i − e disjoint from r i +5 , then thereare disjoint (cid:61) i i (cid:60) -paths so that one of them is contained in cl( Q i ) and the othercontains at most one global H -bridge. Proof.
Suppose that P and r i +5 are disjoint paths. If P contains two (ormore) global H -bridges, then they must be 2.5-jumps having an end in (cid:104) r i (cid:105) . ByTheorem 6.7, they must be of the form v i − w and w v i +3 , with w being no furtherfrom v i in r i than w is. By symmetry, we may assume e is not in [ v i , r i , w ]. Now[ v i , r i , w ]( P − v i − ) and r i +5 are the desired disjoint (cid:61) i i (cid:60) -paths in K i − e . (cid:3) Thus, we may assume both P and P intersect r i +5 . Claim . If either of P and P contains two global H -bridges, then there aredisjoint (cid:61) i i (cid:60) -paths in K i − e so that one of them is contained in cl( Q i ) and theother contains at most one global H -bridge. Proof.
We may assume P contains two global H -bridges B and B . Both B and B are 2.5-jumps. Both have ends in (cid:104) r i (cid:105) ∪ (cid:104) r i +5 (cid:105) . By Lemma 10.9 (5), they both have an end in the same one of (cid:104) r i (cid:105) and (cid:104) r i +5 (cid:105) . We choose the labelling sothat ( B , B ) is either ( v i − w , w v i +3 ) or ( v i +3 w , w v i +8 ). We treat these casesseparately.Suppose ( B , B ) = ( v i − w , w v i +3 ). Assume first that e / ∈ [ w , r , w ]. Then B ∪ [ w , r , w ] ∪ B is disjoint from r i +5 , and we are done by Claim 1. Therefore,we may assume e ∈ [ w , r , w ].In this case, P consists of B , B , and a w w -path P (cid:48) contained in cl( Q i ).We know that P (cid:48) contains a vertex in r i +5 . Lemma 10.9 (5) implies that P consists of a global H -bridge with no vertex in (cid:104) r i +5 (cid:105) . Therefore, we may choose[ v i , r i , w ] ∪ P (cid:48) ∪ [ w , r i , v i +1 ] and P as the desired paths.We conclude the proof of this claim by considering the case ( B , B ) = ( v i +3 w , w v i +8 ).First, by way of contradiction suppose P is not contained in cl( Q i ). Lemma 10.9(5) implies that P consists of a global H -bridge having both ends in ← P i ∪ → P i . Butthen P is disjoint from r i +5 and we are done by Claim 1. Thus, we may assume P ⊆ cl( Q i ).If P is disjoint from either [ v i +5 , r i +5 , w (cid:105) or (cid:104) w , r i +5 , v i +6 ], then we mayreplace either B with the former or B with the latter, and we are done again.Otherwise, there is a [ v i +5 , r i +5 , w (cid:105) (cid:104) w , r i +5 , v i +6 ]-path P (cid:48) contained in P that is r i +5 -avoiding; let its ends be w ∈ [ v i +5 , r i +5 , w (cid:105) and w ∈ (cid:104) w , r i +5 , v i +6 ].If P (cid:48) is r i -avoiding, then P (cid:48) ∪ [ w , r i +5 , w ] is an H -green cycle. Since B together with the subpath of R it spans is also H -green, the edge of [ v i +5 , r i +5 , w ]incident with w is in two H -green cycles, contradicting Theorem 6.7.Therefore, P (cid:48) is not r i -avoiding and so contains two subpaths, one being a w r i -path P (cid:48) and the other being an r i w -path P (cid:48) . For k = 1 ,
2, let u k be the vertexof P (cid:48) k in r i . If e ∈ [ v i , r i , u ], then the paths [ v i +5 , r i +5 , w ] ∪ P (cid:48) ∪ [ u , r i , v i +1 ] and B ∪ [ w , r i +5 , v i +6 ] constitute the required disjoint paths. Otherwise, [ v i , r i , u ] ∪ [ u , P (cid:48) , w , r i +5 , v i +6 ] and [ v i +5 , r i +5 , w ] ∪ B constitute the required disjoint paths. (cid:3) To complete the proof of the lemma, we may now assume that, for each j = 1 , P j contains a unique global H -bridge B j .We first suppose, by way of contradiction, that both B and B have an endin (cid:104) r i (cid:105) ∪ (cid:104) r i +5 (cid:105) . Lemma 10.9 (5) shows that such ends are in the same one of (cid:104) r i (cid:105) and (cid:104) r i +5 (cid:105) ; let i (cid:48) ∈ { i, i + 5 } be such that, for j = 1 , B j has an end w j ∈ (cid:104) r i (cid:48) (cid:105) .We may assume B = v i (cid:48) − w and B = w v i (cid:48) +3 .Theorem 6.7 implies w is closer to v i (cid:48) in r i (cid:48) than w is. The paths P − v i (cid:48) − and P − v i (cid:48) +3 are both in cl( Q i ); the former is a w s i +1 -path, with end x ∈ s i +1 ,and the latter is a w s i -path, with end x ∈ s i .Recall that Π[cl( Q i )] is a planar embedding of cl( Q i ) with Q i bounding a face.The vertices w , w , x , x occur in this cyclic order in Q i , so the disjoint paths P − v i (cid:48) − and P − v i (cid:48) +3 must cross in Π[cl( Q i )], a contradiction. Therefore, atmost one of B and B has an end in (cid:104) r i (cid:105) ∪ (cid:104) r i +5 (cid:105) , while the other is equal to thepath among P and P that contains it.We may choose the labelling so that P consists only of B . Theorem 6.7implies no edge of r i ∪ r i +5 is spanned by both B and B ; since B spans one of r i and r i +5 completely, one of B and B spans edges in r i and the other spans edgesin r i +5 . If either B j spans all of r i , then, as it is disjoint from r i +5 , we are done by
1. EVERY RIM EDGE HAS A COLOUR 81
Claim 1. In particular, B spans r i +5 , edges spanned by B are in r i , and B doesnot span all of r i .Therefore, B is a 2.5-jump with one end w in (cid:104) r i (cid:105) . We may assume the otherend of B is v i +3 . If e / ∈ [ v i , r i , w ], then [ v i , r i , w ] ∪ B is disjoint from r i +5 , andwe are done by Claim 1. If e ∈ [ v i , r i , w ], then ( P − v i +3 ) [ w , r i , v i +1 ] and P arethe desired paths. Proof of Lemma 11.10.
The following claim settles one direction.
Claim . If there are not disjoint non-trivial (cid:61) i i (cid:60) -paths in K i − e , then G − e is planar. Proof.
For this proof, we need to apply Menger’s Theorem; in order to doso, we treat the copies of v i − and v i +3 in (cid:61) i as different from their copies in i (cid:60) .Let u be a cut-vertex of K i − e separating (cid:61) i and i (cid:60) . Let ← K i be the union ofthe (cid:107) u (cid:107) -bridges in K i − e that have an edge in (cid:61) i and let → K i be the union of theremaining (cid:107) u (cid:107) -bridges in K i − e . Then K i − e = ← K i ∪ → K i and ← K i ∩ → K i is just (cid:107) u (cid:107) .Let Π be an embedding of G in R P so that H is Π-tidy. Since r i +5 ⊆ K i − e , u ∈ r i +5 . Because K i − { e, u } is not connected, there is a non-contractible, simpleclosed curve in R P that meets Π[ G − e ] only at u . Thus, there is no non-contractiblecycle in G − { e, u } , showing that G − e is planar. (cid:3) For the converse, Lemma 11.11 shows there are disjoint non-trivial (cid:61) i i (cid:60) -paths P and P in K i − e so that P ⊆ cl( Q i ). In particular, P is an s i s i +1 -path. Itfollows from the embedding Π[ K i ] that P is disjoint from either r i or r i +5 .In every case, we find our V by adding three spokes to the cycle contained in ( R − ( (cid:104) r i (cid:105)∪(cid:104) r i +5 (cid:105) ) ) ∪ P ∪ P ∪ s i ∪ s i +1 and containing ( R − ( (cid:104) r i (cid:105)∪(cid:104) r i +5 (cid:105) ) ) ∪ P ∪ P .If P contains no global H -bridges, then s i +2 , s i +3 , and s i +4 may be chosen asthe spokes.If P contains precisely one global H -bridge B , then B is one of:(1) v i − v i +1 (symmetrically, v i v i +3 );(2) v i − v i +2 ;(3) v i − w and w is in (cid:104) r i (cid:105) (symmetrically, wv i +3 );(4) wv i +1 and w is in (cid:104) r i − (cid:105) (symmetrically, v i w , with w ∈ (cid:104) r i +2 (cid:105) );(5) v i − w and w is in (cid:104) r i +1 (cid:105) (symmetrically, wv i +2 , with w ∈ (cid:104) r i − (cid:105) );(6) v i − v i +1 (symmetrically, v i v i +2 );(7) and the comparable jumps with ends in r i +3 r i +4 r i +5 r i +6 r i +7 .We choose, in all cases, s i − and s i +2 as two of the spokes, with third spoke (takingthe cases in the same order):(1) the P P -subpath of s i +1 (symmetrically, the P P -subpath of s i );(2) s i − ;(3) the P P -subpath of s i +1 (symmetrically, the P P -subpath of s i );(4) the P P -subpath of s i +1 (symmetrically, the P P -subpath of s i );(5) s i − (symmetrically, the same);(6) s i − (symmetrically, the same); and(7) these cases are symmetric to the preceding ones.In every case, we have found a V in G − e , as required. We conclude this section by proving that every rim edge is either red, H -green,or H -yellow. Proof of Theorem 11.3.
Let e be an edge in the H -rim. There is an i so that e ∈ r i . By Lemma 11.10, G is red if and only if there are no disjoint non-trivial (cid:61) i i (cid:60) -paths in K i − e .Now suppose there are disjoint non-trivial (cid:61) i i (cid:60) -paths P and P in K i − e . ByLemma 11.11, we may assume P is contained in cl( Q i ), while P contains at mostone global H -bridge. If P is disjoint from r i +5 , then every maximal r i -avoidingsubpath of P is contained in an H -green cycle. The edge e is in one of these H -green cycles, as required.Thus, we may assume P contains a vertex in r i +5 . If P ⊆ cl( Q i ), thenthe planar embedding of cl( Q i ) shows P is disjoint from r i +5 and the precedingparagraph, with P in place of P , shows e is H -green. Consequently, we mayfurther assume P contains a global H -bridge B . Case 1: B has its ends in ← P i ∪ r i ∪ → P i . In this case, if e is spanned by B , then there is an H -green cycle containing e , namely the cycle consisting of B and the subpath of R that it spans. The onlyother possibility in this case is that B is a 2.5-jump with an end w in (cid:104) r i (cid:105) andthat e is in the one of [ v i , r i , w ] and [ w , r i , v i +1 ] not spanned by B . For the sakeof definiteness, we suppose B = v i − w and that e is in [ w , r i , v i +1 ].Since P ⊆ cl( Q i ), we see that, in this case, P is disjoint from r i +5 and,therefore, we may assume P = r i +5 . We replace P with [ v i , r i , w ] ( P − v i − ) sothat there are disjoint (cid:61) i i (cid:60) -paths contained in cl( Q i ) − e ; a situation resolved inthe paragraph preceding this case. Case 2: B has its ends in P ← i ∪ r i +5 ∪ P → i . In this case, either P is B or, up to symmetry, B is a 2.5-jump wv i +8 , with w ∈ (cid:104) r i +5 (cid:105) , and P is [ v i +5 , r i +5 , w ] ∪ B . On the other hand, P is an s i s i +1 -pathin cl( Q i ) intersecting r i +5 .Let x be the first vertex in r i +5 as we traverse P from s i and let P (cid:48) be the s i x -subpath of P . We note that P prevents x from being in [ v i +5 , r i +5 , w ], so x ∈(cid:104) w, r i +5 , v i +6 ]. Let y be the end of P (cid:48) in s i . The cycle P (cid:48) [ x, r i +5 , v i +6 ] s i +1 r i [ v i , s i , y ]is H -yellow, as witnessed by the H -green cycle containing B . Therefore, e is either H -yellow or H -green (in Definition 11.1, an H -yellow edge is not H -green).HAPTER 12 Existence of a red edge and its structure
In this section, we prove that if G is a 3-connected, 2-crossing-critical graphcontaining a tidy subdivision H of V , then some edge of the H -rim is red. Fur-thermore, we prove that each red edge e has an associated special cycle we call ∆ e .These “deltas” will be the glue that hold successive tiles together and so form avital element of the tile structure.The argument for proving the existence of a red edge depends on whether ornot there is a global H -bridge that is either a 2.5- or 3-jump. Once these cases aredisposed of, matters become simple. However, with the knowledge of the ∆’s, itturns out we can show that there is no 3-jump. This will be our first aim and so,since we need the ∆’s to complete the elimination of 3-jumps, we shall begin bydetermining the structure of the ∆ of a red edge. Theorem . Let G ∈ M , V ∼ = H ⊆ G , with H tidy. Let e = uw be a rededge of G and let i ∈ { , , , . . . , } be such that e ∈ r i . Then there exists a vertex x e ∈ [ r i +5 ] and internally disjoint x e u - and x e w -paths A u and A w , respectively, in cl( Q i ) so that, letting ∆ e = ( A u ∪ A w ) + e :(1) there are at most two ∆ e -bridges in G ;(2) there is a ∆ e -bridge M ∆ e so that H ⊆ M ∆ e ∪ ∆ e , while the other ∆ e -bridge, if it exists, is one of two edges in a digon incident with x e ; and(3) when there are two ∆ e -bridges, let u e and w e be the attachments of theone-edge ∆ e -bridge, labelled so that u e ∈ A u and w e ∈ A w ; otherwise let u e = w e = x e . In both cases, ∆ e − e contains unique uu e - and ww e -paths P u and P w , each containing at most one H -rim edge, which, if it exists,is in the span of a global H -bridge and, therefore, is H -green. Proof.
Let Π be an embedding of G in R P for which H is Π-tidy. We mayassume r i = [ v i , r i , u, e, w, r i , v i +1 ]. Lemma 11.10 implies K i − e has a cut vertex x e separating (cid:61) i and i (cid:60) (again adopting the perspective that v i − and v i +3 aresplit into different copies in (cid:61) i and i (cid:60) ). As r i +5 is a (cid:61) i i (cid:60) -path in K i − e , x e is in r i +5 .Because cl( Q i ) is 2-connected and Π[cl( Q i )] has Q i bounding the exterior face,there is a face F e of G in R P that is in the interior of Q i and incident with both e and x e . As G is 3-connected and non-planar, F e is bounded by a cycle C e and C e − e consists of a ux e -path A u and a wx e -path A w .For (1) and (2), we begin by noticing that C e ⊆ cl( Q i ). Thus, there is a C e -bridge M C e containing the three H -spokes not in Q i . Claim . Each of C e ∩ s i , C e ∩ s i +1 , and C e ∩ r i is connected. Either C e ∩ r i +5 is connected or it has two components that are joined by an edge e (cid:48) of r i +5 and C e has an edge parallel to e (cid:48) . In particular, each of s i , s i +1 , r i , and r i +5 − e (cid:48) iscontained in C e ∪ M C e .
834 12. EXISTENCE OF A RED EDGE AND ITS STRUCTURE
Proof.
Suppose by way of contradiction that C e ∩ s i is not connected. As C e bounds a face of Π, it follows that there is a Q i -local H -bridge having all itsattachments in s i , contradicting Lemma 10.10. Thus, C e ∩ s i is connected.It follows that any part of s i that is not in C e is in the same C e -bridge as either r i − or r i +4 . That is, it is in M C e , and therefore, s i ⊆ M C e ∪ C e .Symmetry shows that this also holds for s i +1 .Now suppose C e ∩ r i is not connected. Then there is a Q i -local H -bridge B having all attachments in r i . Corollary 5.15 implies B has precisely two attachments x and y , and so Lemma 5.19 implies B is just the edge xy . Thus, [ x, r i , y ] B is an H -green cycle C . Lemma 6.6 (8) shows C bounds a face of Π[ G ].By symmetry, we may assume that x and y are both in [ v i , r i , u ]. Suppose that z is any vertex in (cid:104) x, r i , u ].Suppose first that z (cid:54) = u . As G is 3-connected, z has a neighbour z (cid:48) not in[ x, r i , y ]. If zz (cid:48) is in the interior of Q i , it must be parallel to an edge in r i , as anyother edge would go into one of the faces of Π[ G ] bounded by C e and C . Therefore, zz (cid:48) is outside M and, so is an edge of another H -green cycle. But then one of theedges of [ x, r i , y ] incident with z is in two H -green cycles, contradicting Theorem6.7. This same argument, however, also applies if z = u , with the small variationthat, by Lemma 11.4, zz (cid:48) cannot span the red edge e , giving the contradiction thatthe edge of [ x, r i , u ] incident with u is in two H -green cycles. Thus, C e ∩ r i isconnected. As it did for s i , this implies that r i ⊆ M C e ∪ C e .Finally, we consider C e ∩ r i +5 . Proceeding as we did for r i , if C e ∩ r i +5 is notconnected, there is (up to symmetry) a Q i -local H -bridge B having all attachmentsin [ v i +5 , r i +5 , x e ]; B is a single edge and is in an H -green cycle. One end of B is x e , and the H -green cycle containing B consists of two parallel edges.Thus, there are at most two such H -bridges B , each of which is an edge parallelto an edge in r i +5 . If they both exist, then the 3-connection of G implies x e hasanother neighbour, which, as above, is adjacent to x e by an edge not in M , showingone of the edges of r i +5 incident with x e is in two H -green cycles, contradictingTheorem 6.7. (cid:3) We can now define ∆ e . If C e ∩ r i +5 is connected, then ∆ e = C e . Otherwise,∆ e is obtained from C e by replacing the edge of C e incident with x e and not in r i +5 with its parallel mate that is in r i +5 . Notice that the ∆ e - and C e -bridges arethe same, except for these exchanged edges incident with x e . Set M ∆ e to be M C e .The following is evident from what has just preceded. Claim . H ⊆ M ∆ e ∪ ∆ e and ∆ e ∩ r i +5 is connected.Consider again r i ∩ ∆ e . It is connected, so if it is more than just [ u, e, w ],the symmetry shows we may assume it contains an edge xu other than e . The3-connection of G implies that u is adjacent with a vertex y other than x and w .The edge uy is not interior to Q i , as then it would be in the face of G bounded by C e . Thus, uy is not in M , and, as uw is red, Lemma 11.4 implies uy spans xu . Thevertex x is seen to be H -green by the H -green cycle C y containing uy . Since x has at least three neighbours in G , there is a neighbour of x different from the twoneighbours of x in r i − r i . Because C y bounds a face of G (Lemma 6.6 (8)), everyedge incident with x and not in r i − r i is in M . There is a unique neighbour z of
2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 85 x so that z is not in r i − r i and xz is an edge of ∆ e . This shows that x is one endof r i ∩ ∆ e . These observations easily yield the following claim. Claim . Each of A u ∩ r i and A w ∩ r i has at most one edge.We now turn our attention to r i +5 . Claim . (1) No edge of r i +5 ∩ ∆ e is H -yellow.(2) No global H -bridge has x e in the interior of its span. Proof.
For (1), suppose by way of contradiction there were an H -yellow edgein r i +5 ∩ ∆ e . Then Lemma 11.2 (3) shows the witnessing H -yellow cycle must be∆ e . However, the witnessing H -green cycle must have ∆ e ∩ r i in the interior of itsspan, yielding the contradiction that e is H -green.For (2), suppose by way of contradiction that there is a global H -bridge xy with x e in the interior of the span of xy . Then xy ∪ ( r i +5 − x e ) contains a (cid:61) i i (cid:60) -path in K i − { e, x e } , contradicting Lemma 11.10. (cid:3) Claim . (1) If [ v i +5 , r i +5 , x e ] ∩ ∆ e contains three vertices x , y , and x e of r i +5 , then (choosing the labelling of x and y appropriately) [ v i +5 , r i +5 , x e ] ∩ ∆ e = [ x, xy, y, yx e , x e ], y and x e are joined by a digon, and y is incidentwith a global H -bridge that spans x .(2) If [ v i +5 , r i +5 , x e ] ∩ ∆ e does not contain three consecutive vertices of r i +5 ,but has a vertex x other than x e , then either x and x e are joined by adigon, or x e is incident with a global H -bridge that spans x .The symmetric statements also hold for [ x e , r i +5 , v i +6 ] ∩ ∆ e . Proof.
For (1), the fact that ∆ e ∩ r i +5 is connected implies that there are ver-tices x and y so that [ x, xy, y, yx e , x e ] ⊆ [ v i +5 , r i +5 , x e ]. Because G is 3-connected, y is adjacent to a vertex z other than x and x e . The edge yz cannot be in M , asthen it would be in the face of G bounded by C e , a contradiction. Therefore, it isa 2.5-jump. Claim 4 (2) shows yz does not span x e .As G is 3-connected, x has a neighbour x (cid:48) different from the two neighbours of x in R . If the edge xx (cid:48) is in D , then it is in the face bounded by the H -green cyclecontaining yz , a contradiction. Therefore, xx (cid:48) is in M and, in particular, for that x (cid:48) giving the edge nearest to xy in the cyclic rotation about x , xx (cid:48) is in ∆ e and,therefore, no other vertex of [ v i +5 , r i +5 , x e ] is in ∆ e .Since yx e is not R -separated from e in G , Lemma 11.7 implies yx e is either H -yellow or H -green. Claim 4 (1) implies it is not H -yellow; we conclude that yx e is H -green and let C yx e be the witnessing H -green cycle.As pointed out in the first paragraph of the proof, C yx e cannot contain a global H -bridge that spans x e . On the other hand, xy is H -green by the global H -bridge yz . By Theorem 6.7, this is the only H -green cycle containing xy . Thus, the only H -rim edge contained in C yx e is yx e . It follows that C yx e is contained in cl( Q i ).Claim 1 implies C yx e is a digon.For (2), the fact that [ v i +5 , r i +5 , x e ] ∩ ∆ e is connected implies that [ v i +5 , r i +5 , x e ] ∩ ∆ e = [ x, xx e , x e ]. Lemma 11.7 implies that xx e is either H -yellow or H -green, andClaim 4 (1) shows it is not H -yellow. Therefore, it is H -green.Claim 4 (2) shows any global H -bridge spanning xx e has x e as an attachment.Otherwise, the H -green cycle C xx e containing xx e is contained in cl( Q i ). Again,Claim 1 shows C xx e is a digon. (cid:3) There is one more observation to make before we complete the proof of thetheorem. From Claim 5 (1), it seems possible that both [ v i +5 , r i +5 , x e ] ∩ ∆ e and[ v i +5 , r i +5 , x e ] ∩ ∆ e have three vertices. However, this is not possible, as x e musthave a neighbour z different from its neighbours in R . But now x e z cannot be in M , as then it would be in the face bounded by C e , and it cannot be in D , as thenit is a global H -bridge and one of the digons incident with x e is also spanned by x e z , contradicting Theorem 6.7. Therefore, r i +5 ∩ ∆ e has at most three edges, andall such edges are H -green.If there are no edges, then r i +5 ∩ ∆ e is just x e . If no edge of r i +5 ∩ ∆ e is ina digon, then u e and w e are defined in (3) of the statement to be x e . In this case,Claim 5 (1) implies there can be at most one edge of r i +5 ∩ ∆ e on each side of x e ,but any such edge is spanned by a global H -bridge. If there is a digon, then it is u e w e , each of u e and w e is incident with at most one other edge in r i +5 ∩ ∆ e , andany such edge is spanned by a global H -bridge.Finally, By Lemma 10.9 (4), not both u and u e , for example, can be incidentwith such global H -bridges, so P u has at most one H -rim edge. Definition . Let G ∈ M , V ∼ = H ⊆ G , with H tidy, and e a red edgeof G with ends u and w . With u e and w e as in the statement of Theorem 12.1, the peak of ∆ e is the subgraph of G induced by u e and w e . If the peak has just onevertex, then ∆ e is sharp .The following observations are given to summarize important points from The-orem 12.1. Corollary . Let G ∈ M , V ∼ = H ⊆ G , with H tidy, and e a red edgeof G . Then the peak of ∆ e is either a single vertex or a digon and no edge of thepeak is in the interior of the span of a global H -bridge. Proof.
That the peak is either a single vertex or a digon is a rephrasing of Theorem12.1 (2) and (3). In the case the peak is a digon, neither u e nor w e can be in theinterior of the span of a global H -bridge, since then the H -rim edge in the digon isin two H -green cycles, contradicting Theorem 6.7.So suppose the peak is just the vertex u e = w e , let B be a global H -bridgewith u e in the interior of its span, and let i be such that e ∈ r i . If ∆ e ∩ r i +5 hasan edge e (cid:48) , then e (cid:48) is incident with u e and, moreover, is H -green by a global H -bridge B (cid:48) incident with u e . But then B provides a second H -green cycle containing e (cid:48) , contradicting Theorem 6.7. So ∆ e ∩ r i +5 is just u e , in which case B providesa witnessing H -green cycle that shows ∆ e is H -yellow. But then e is H -yellow,contradicting Lemma 11.4.Our next goal is to eliminate 3-jumps. For this the next two lemmas are helpful. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Suppose C is an H -yellow cycle and C (cid:48) is the witnessing H -green cycle. Let e be an edge of G notin C ∪ C (cid:48) ∪ R . Suppose either C (cid:48) does not contain a 3-jump or e is in one of thefour spokes containing an H -node spanned by C (cid:48) . Then no H -yellow edge in C iscrossed in any 1-drawing of G − e . Proof.
There are at least four H -spokes contained in G − e . By hypothesis,at least one of these has no end in C (cid:48) and, therefore, no end in C ∪ C (cid:48) . Therefore,Lemma 7.2 (2) applies. (cid:3)
2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 87 u e u we ∆ e w e Figure 12.1.
One of several examples of a ∆.
Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Suppose C is an H -green cycle in G . Suppose that C does not contain a 3-jump, e is an edge of G not in R ∪ C and D is a 1-drawing of G − e . If an edge e (cid:48) of C is crossed in D ,then C contains a 2.5-jump with an end in (cid:104) r i (cid:105) , for some i , and e (cid:48) is in r i . Proof.
This is a straightforward consequence of Lemma 7.2 (3a and 3b).
Theorem . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Then no global H -bridge is a 3-jump. Proof.
The proof begins by showing that if v i − v i is a global H -bridge that isa 3-jump, then there is a red edge in r i . The next step is to show that the edgeof r i incident with v i is red. The final step is to show that, if e ∗ is the edge of s i incident with v i , then cr( G − e ∗ ) ≥
2, contradicting the criticality of G . Let Π bean embedding of G in R P so that H is Π-tidy. Claim . There is a red edge of G in r i . Proof.
Lemma 10.9 (2) implies neither v i +5 v i − nor v i v i +3 is in G . Thus,Lemma 10.8 implies Q i − has BOD.Let D i − be a 1-drawing of G − (cid:104) s i − (cid:105) . Lemma 5.9 implies Q i − is crossed in D i − . Let H (cid:48) be the subdivision of V consisting of the H -rim R and the threespokes s i , s i − , and s i +1 . Lemma 7.2 implies the cycle r i − r i − r i − [ v i , v i − v i , v i − ]is clean in D i − . In particular, the crossing must be of an edge in r i +3 ∪ r i +4 andan edge e in r i .We prove e is red in G by proving it is neither H -green nor H -yellow. Lemma10.9 (1) and (3) imply that no global H -bridge other than v i − v i has an end in[ v i , r i , v i +1 (cid:105) . Therefore, no H -green cycle containing e can contain a global H -bridge. Thus, any H -green cycle C containing e is contained in cl( Q i ). Lemma12.5 implies C is not crossed in D i − , contradicting the fact that the edge e is in C and is crossed in D i − . We conclude that e is not H -green.So suppose C is an H -yellow cycle containing e and let P P P P be the de-composition of C into paths as in Definition 11.1. By Lemma 11.2, there is a global H -bridge B so that the interior of the span of B contains P . Lemma 10.9 (2) saysthere is at most one 3-jump in G , so B is either a 2- or 2.5-jump.That e is not H -yellow is an immediate consequence of Lemma 12.4. (cid:3) We now aim to show that the edge of r i incident with v i is red. By Claim 1,there is a red edge in r i ; let e be the red edge nearest to v i in r i . Let r (cid:48) i be thecomponent of r i − e containing v i and let u be the end of e in r (cid:48) i . Claim . No edge of r (cid:48) i is H -yellow. Proof.
Suppose some edge e (cid:48) of r (cid:48) i is H -yellow and let C and C (cid:48) be thewitnessing H -yellow and H -green cycles, respectively. Lemma 11.2 (1) implies C (cid:48) contains a global H -bridge B . We note that Lemma 10.9 (1) and (3) imply (because v i − v i is present and v i − = v i +7 ) that B has no vertex in (cid:104) v i +6 , r i +6 , v i +7 ]. Onthe other hand, to make C H -yellow, B must have one end in (cid:104) v i +5 , r i +5 , v i +6 ].Due to the presence of v i − v i , Lemma 10.9 (4) implies v i +3 is not in B .Therefore, Theorem 10.6 implies B has v i +6 as one end and its other end is in (cid:104) v i +3 , r i +3 , v i +4 ]. Theorem 12.1 (3) implies the edge e of ∆ e − e incident with u is not in H ; by Theorem 12.1, it is in cl( Q i ).Let D be a 1-drawing of G − e . By Theorem 5.23, Q i has BOD, so Lemma 5.9implies Q i is crossed in D . Lemma 7.2 implies no edge in r i +4 r i +5 is crossed in D ,so the crossing in D is of r i with r i +6 .Lemmas 12.4 and 12.5 combine with Theorem 11.3 to show that the edge e (cid:48)(cid:48) of r i +6 crossed in D is red in G . Lemma 11.7 implies e (cid:48)(cid:48) and e are R -separated in G and we conclude that they are also R -separated in G − e (cid:48) ; in fact, e (cid:48)(cid:48) is R -separatedfrom r (cid:48) i [ u, e , w ]. It follows that the edge f of r i crossed in D is in [ w, r i , v i +1 ].Lemmas 12.4 and 12.5 combine with Theorem 11.3 to show that f is red in G ;however, e and f are not R -separated in G − e (cid:48) and, therefore, not separated in G ,contradicting Lemma 11.7. It follows that no edge of r (cid:48) i is H -yellow, as required. (cid:3) Claim . u = v i . Proof.
By way of contradiction, suppose that u (cid:54) = v i . By definition of e , noedge of r (cid:48) i is red, and Claim 2 shows no edge of r (cid:48) i is H -yellow. Theorem 11.3 showsthat every edge of r (cid:48) i is H -green. Because of v i − v i , Lemma 10.9 (1) and (3) showsno edge of r (cid:48) i is H -green by a global H -bridge.Let e be the edge of ∆ e − e incident with u ; Theorem 12.1 and the fact that e is not incident with v i imply that e is not in H . Let D be a 1-drawing of G − e .Note that e is in a Q i -local H -bridge. Since Q i has BOD (Theorem 5.23), it iscrossed in D (Lemma 5.9). Every edge of r i − is H -green in G − e ; thus, Lemma6.6 (10) implies the following. Subclaim . No edge in r i − is crossed in D . (cid:50) We next rule out another possibility.
Subclaim . No edge in r i +1 is crossed in D . Proof.
Suppose some edge e Di of r i +1 is crossed in D . Since Q i is crossedin D , the other crossed edge e (cid:48) Di is in r i +5 . By Lemma 12.4, no H -yellow edge in r i +1 ∪ r i +5 can be crossed in D . Since H ⊆ G − e , Lemma 6.6 (10) implies no H -green cycle not containing e can be crossed in D ; in particular, no H -green edgein r i +1 ∪ r i +5 can be crossed in D . Now Theorem 11.3 implies e Di and e (cid:48) Di are bothred in G .Suppose first that e (cid:48) Di is in [ v i +5 , r i +5 , u e ]. (Recall that u e is the vertex in thepeak of ∆ e nearest u in ∆ e − e .) Lemma 11.7 implies e (cid:48) Di and e are R -separated
2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 89 in G ; this implies that ∆ e (cid:48) Di is disjoint from ∆ e . One of the r i r i +5 -paths in ∆ e (cid:48) Di , s i +1 , s i +2 , and s i +3 combine with R to show that e (cid:48) Di is R -separated in G − e fromevery edge in r i +1 , a contradiction.If, on the other hand, e (cid:48) Di is not in [ v i +5 , r i +5 , u e ], then Lemma 11.7 shows e Di and e (cid:48) Di are R -separated in G and there is a subdivision of V that both witnessesthis separation and does not contain e (the spokes are s i +2 , s i +3 , and the “nearer”( r i r i +1 )( r i +5 r i +6 )-paths in ∆ e Di and ∆ e (cid:48) Di ). This shows that e Di and e (cid:48) Di are R -separated in G − e , a contradiction. (cid:3) Since Q i is crossed in D , Subclaims 1 and 2 imply that some edge e Di of r i iscrossed in D . Subclaim . e Di ∈ r (cid:48) i . Proof. If e Di is not in r (cid:48) i , then let e (cid:48) Di be the edge of r i +4 r i +5 r i +6 that iscrossed in D . Then e Di and e (cid:48) Di are not R -separated in G − e . Observe that ∆ e shows no H -green or H -yellow cycle containing e Di can also contain e . Therefore, e Di is red in G and, consequently is R -separated from e (cid:48) Di in G . In particular, e is in every subdivision of V that contains R and witnesses the R -separation of e Di and e (cid:48) Di . This implies that e (cid:48) Di is in [ v i +5 , r i +5 , u e ].As e and e (cid:48) Di are both red in G , by Lemma 11.7 there is a subdivision K of V containing R and witnessing the R -separation of e and e (cid:48) Di . There is an r (cid:48) i r i +5 -path P in K that is disjoint from ∆ e . Moreover, P ⊆ cl( Q i ). But now, P together with the r i r i +5 -path in ∆ e − u , s i +2 , and s i +3 make the four spokes of asubdivision of V containing R and witnessing the R -separation of e Di and e (cid:48) Di in G − e , a contradiction. (cid:3) We now locate the edge e (cid:48) Di . To this end, let ˆ e be the edge of s i − incident with v i − and let (cid:98) D be a 1-drawing of G − ˆ e . By Lemmas 10.8, 10.9 (2), and 5.9, Q i − must be crossed in (cid:98) D . However, Lemma 7.2 shows that none of r i − r i − r i − canbe crossed in (cid:98) D . Since the edges in r (cid:48) i are all H -green and none of the witnessing H -green cycles contains a global H -bridge, Lemma 12.5 implies that no edge of r (cid:48) i is crossed in (cid:98) D . Thus, some edge of r i +3 r i +4 crosses an edge of r i − (cid:104) r (cid:48) i (cid:105) in (cid:98) D . Subclaim . Every edge in r i +4 is H -green in G and no edge in r i +4 is crossedin (cid:98) D . Proof. If e (cid:48) ∈ r i +4 is H -yellow, then v i − v i is in the witnessing H -green cycleand, therefore, the edge of s i − incident with v i +4 is in the interior of an H -yellowcycle containing s i − ; this contradicts Lemma 11.2, so e (cid:48) is not H -yellow.Now we eliminate the possibility that e (cid:48) is red. To do this, it will be helpfulto know that no H -green edge in r i +4 is crossed in (cid:98) D : fortunately, this is justLemma 12.5, combined with Lemma 10.9 (1) and (4) to eliminate the possibility ofa 2.5-jump.Choose e (cid:48) to be the red (in G ) edge in r i +4 that is nearest in r i +4 to v i +5 .Lemma 11.7 implies e (cid:48) is R -separated from e in G ; we may choose the witnessingsubdivision K of V to contain s i − and s i +2 ; in particular, K avoids ˆ e . Therefore, e (cid:48) is R -separated from e in G − ˆ e . Since the edges in r i +4 between e (cid:48) and v i +5 are neither red (choice of e (cid:48) ) nor H -yellow (two paragraphs preceding), they are H -green (Theorem 11.3), we know they are not crossed in (cid:98) D (preceding paragraph).The subgraph K shows that none of the rest of r i +3 r i +4 can be crossed in (cid:98) D , whichis a contradiction. Therefore, no edge of r i +4 is red in G ; since none is H -yellowby the preceding paragraph, Theorem 11.3 shows they are all H -green. (cid:3) It follows that an edge of r i +3 is crossed in (cid:98) D and it must cross some edge in[ u, r i , v i +1 ]. This further implies that the uu e -subpath P u of ∆ e − e intersects s i as otherwise each edge of [ u, r i , v i +1 ] is R -separated from r i +3 in G − ˆ e .We now return to consideration of D . No edge in r i +4 is red in G and, because P u intersects s i , every edge (if there are any) of [ v i +5 , r i +5 , u e ] is H -green. Thiscombines with Lemma 10.9 (1) and (4) to show that no edge in r i +4 [ v i +5 , r i +5 , u e ]is in the span of a global H -bridge; therefore, Lemmas 12.4 and 12.5 imply that noedge of r i +4 [ v i +5 , r i +5 , u e ] is crossed in D . Thus, the edge e (cid:48) Di that crosses e Di in D is in [ u e , r i +5 , v i +6 ] r i +6 .Because of v i − v i , no edge in [ u e , r i +5 , v i +6 ] r i +6 is in the span of a global H -bridge. Therefore, Lemmas 12.4 and 12.5 imply e (cid:48) Di is red in G . But now Lemma11.7 implies e (cid:48) Di is R -separated in G from e ; there is a witnessing subdivision K of V that contains s i − , s i , and the nearer ( r i r i +1 )( r i +5 r i +6 )-paths in ∆ e and∆ e (cid:48) Di . Note that the path taken from ∆ e does not contain e . Therefore, K is alsocontained in G − e ; Observation 11.6 (1) shows that these edges cannot be crossedin D , the final contradiction that proves the claim. (cid:3) We now move into the final phase of the proof that there is no 3-jump. Let e ∗ be the edge of s i incident with v i and let D ∗ be a 1-drawing of G − e ∗ . Lemma10.9 (2) implies v i − v i is the only 3-jump of G , so Lemma 10.8 implies Q i hasBOD. Lemma 5.9 implies Q i is crossed in D ∗ . In particular, there is an edge e in r i +3 r i +4 r i +5 r i +6 that is crossed in D ∗ . Lemma 7.2 shows that r i +3 is not crossedin D ∗ . Claim . e is red in G . Proof. If e is H -yellow in G , then Lemma 12.4 shows that e is not crossed in D ∗ . Thus, e is not H -yellow.Suppose e is H -green in G , and let C be the witnessing H -green cycle. Lemma10.9 (2) implies C does not contain a 3-jump and Lemma 7.2 implies both that itdoes not contain a 2-jump and is not contained in the union of some Q j togetherwith a Q j -local H -bridge. Therefore, C contains a 2.5-jump b and Lemma 7.2implies e is in the H -rim branch that contains the end x of b that is not an H -node.The edge e has already been shown to be in r i +4 r i +5 r i +6 . Suppose e is in r i +4 . If b = v i +2 x , then we contradict Lemma 10.9 (4) — v i − v i and b span theopposite sides of Q i − , a contradiction. The other alternative is that b = xv i − ,which violates Lemma 10.9 (1). Thus, e / ∈ r i +4 .If e ∈ r i +5 , then either b = xv i − or b = xv i +3 . The former does not occur, asotherwise the edges of r i − are all in two H -green cycles, contradicting Theorem6.7. If the latter occurs, then we contradict Lemma 10.9 (4) — v i − v i and b spanthe opposite sides of Q i − . Thus, e / ∈ r i +5 .So e ∈ r i +6 . In this case b is either xv i − or xv i +4 . For the former, the edges of r i − r i − are all in two H -green cycles, contradicting Theorem 6.7. For the latter,the edge e of r i incident with v i is red by Claim 3. The existence of b shows Q i is
2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 91 H -yellow, contradicting the fact that e is red. This is the final contradiction thatshows e is red. (cid:3) Recall that the edge e is in r i +3 r i +4 r i +5 r i +6 , since it is involved in a crossingwith Q i . We have already observed that e is not in r i +3 .Suppose first that e ∈ r i +4 . Lemma 11.7 implies e and e are R -separated in G ; in particular, v i is not in ∆ e . But then v i − v i shows ∆ e ⊆ cl( Q i − ) − v i to bean H -yellow cycle, contradicting the fact that e is red.Therefore, e ∈ r i +5 r i +6 . Let ˆ e be the edge crossed by e in D ∗ . Since Lemma7.2 implies r i − is not crossed in D ∗ , ˆ e / ∈ r i − . Since e and e are both red in G ,Lemma 11.7 implies they are R -separated in G ; there is a witnessing subdivision K of V that contains s i − and s i − . This K does not contain e ∗ , and so is containedin G − e ∗ . Therefore, K separates e from r i − in G − e ∗ , and so, in D ∗ , e does notcross r i − . Thus, ˆ e is not in r i − .Therefore, ˆ e ∈ r i r i +1 . Lemma 10.9 (4) implies there is no 2.5-jump xv i +4 —it and v i − v i would span the opposite sides of Q i − . Also, Lemma 10.9 (3) impliesthere is no 2.5-jump xv i +3 with x ∈ (cid:104) r i (cid:105) .It follows from Lemmas 12.4 and 12.5 (the preceding pararaph is used here)that the edge ˆ e crossed by e in D ∗ is red in G . This implies that e and ˆ e are R -separated in G and this in turn implies that e and ˆ e are R -separated in G − e ∗ ,the final contradiction. Corollary . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Then every H -hyperquad has BOD. Proof.
By Theorem 12.6, no global H -bridge is a 3-jump. By Lemma 10.8, every H -hyperquad has BOD.We are now prepared for the main result of this section. Theorem . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Then there is ared edge in the H -rim. Proof.
We prove this by first considering the case there is a global H -bridge. ByTheorem 12.6, there is no 3-jump. By Theorem 10.6, a global H -bridge is either a2.5- or a 2-jump. Claim . If G has a 2.5-jump, then G has a red edge. Proof.
By symmetry, we may assume wv i +2 is a 2.5-jump with w ∈ (cid:104) r i − (cid:105) .By way of contradiction, we assume that G has no red edge. We first treat twospecial cases. Case 1: there is a 2.5-jump v i − w (cid:48) , with w (cid:48) ∈ (cid:104) r i − (cid:105) . In this case, let D be a 1-drawing of G − (cid:104) s i +2 (cid:105) . Corollary 12.7 and Lemma 5.9show that Q i +2 is crossed in D . Lemma 7.2 implies each of the cycles consisting ofone of these two 2.5-jumps and the subpath of R it spans is clean in D . The samelemma implies that neither r i +3 nor r i +5 is crossed in D . The combination of factsimply that some edge e in r i +2 crosses some edge e in r i +6 .Since G has no red edge, Theorem 11.3 implies each of e and e is either H -yellow or H -green in G . There is complete symmetry between them (relativeto the two 2.5-jumps), so we treat e . If e is H -yellow in G , then it is in some witnessing H -yellow cycle C for which there is a witnessing H -green cycle C (cid:48) . Theonly possibility is that C (cid:48) contains wv i +2 .We have that C ⊆ cl( Q i +1 ) − v i +2 . Let C = P P P P be the composition ofpaths showing C is H -yellow, as in Definition 11.1. Since P ⊆ (cid:104) C (cid:48) ∩ R (cid:105) , we have P ⊆ r i +1 − v i +2 . Choose the labelling of P and P so the r i +1 -end of P is nearer v i +2 in r i +1 than the r i +1 -end of P is.If P is not disjoint from (cid:104) s i +2 (cid:105) , then the edge of r i +1 incident with v i +2 is intwo H -green cycles, contradicting Theorem 6.7. Therefore, C ∪ C (cid:48) is disjoint from (cid:104) s i +2 (cid:105) . But then Lemma 12.4 implies e is not crossed in D and, therefore, e isnot H -yellow. Likewise, e is not H -yellow.Therefore, e is H -green, so Lemma 12.5 implies e is spanned by some 2.5-jump J and, moreover, is not in either H -rim branch fully contained in the span of J . ByTheorem 6.7, no H -rim edge is in two H -green cycles. Thus, the only possibility forthe 2.5-jump J spanning e is v i +4 w , with w ∈ (cid:104) r i +6 (cid:105) . An analogous argumentapplies to e , so e is spanned by the 2.5-jump J w v i +5 , with w ∈ (cid:104) r i +2 (cid:105) . Butnow we have that every edge of r i +4 is in the distinct H -green cycles containing J and J , contradicting Theorem 6.7, completing the proof in Case 1. Case 2:
There is a 2.5-jump v i − w (cid:48) , with w (cid:48) ∈ r i − . Let D be a 1-drawing of G − (cid:104) s i +1 (cid:105) . Corollary 12.7 and Lemma 5.9 imply Q i +1 is crossed in D . Lemma 7.2 (1) shows none of [ w, r i − v i ], r i , r i +1 , and r i +6 is crossed in D , while (2) of the same lemma shows r i +2 is not crossed. It followsthat some edge e ∈ r i +5 crosses an edge e ∈ [ v i +9 , r i +9 , w ].Since e is not red, Theorem 11.3 shows it is either H -yellow or H -green. If e is H -yellow as witnessed by the H -yellow cycle C and the H -green cycle C (cid:48) ,then the global H -bridge J in C (cid:48) is a 2- or 2.5-jump (Theorems 10.6 and 12.6) and C ⊆ cl( Q − ) (Lemma 11.2 (4)). Lemma 12.4 implies that e is not crossed in D , acontradiction.Likewise, if e is H -green, the Lemma 12.5 shows it is not crossed in D , thefinal contradiction completing the proof in Case 2. Case 3:
All the remaining cases.
Let e i be the edge of s i incident with v i and let D i be a 1-drawing of G − e i .Corollary 12.7 and Lemma 5.9 imply Q i is crossed in D i .Since G (in particular, r i − ) has no red edge, Lemma 12.4 shows any H -yellowedge in r i − is not crossed in D i , while Lemma 12.5 implies that, as we are not inCase 2, no H -green edge of r i − is crossed in D i . Lemma 7.2 (1) implies no edgeof [ w, r i − , v i ] r i r i +1 is crossed in D i . Therefore, it must be that some edge e i − of[ v i − , r i − , w ] is crossed in D i .As e i − is not red in G , Theorem 11.3 implies e i − is either H -green or H -yellow. If it is H -green, then, because we are not in Case 1, Lemma 12.5 implies e i − is in an H -green cycle C contained in cl( Q i − ) and e i ∈ C . But then everyedge in [ w, r i − , v i ] is in two H -green cycles, contradicting Theorem 6.7.We conclude that e i − is H -yellow. Let C and C (cid:48) be the witnessing H -yellowand H -green cycles, respectively, and let B be the global H -bridge contained in C (cid:48) .Lemma 12.4 implies e i ∈ C . Moreover, v i +5 is in the span of B , as otherwise B attaches at v i +2 , contradicting Lemma 10.9 (1). By Lemma 10.9 (4), v i +7 is not inthe span of B .
2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 93 If B has an end in (cid:104) r i +2 (cid:105) , then the other end of B is v i +5 . The R -avoidingpath (one of P and P in the decomposition of the H -yellow cycle as in Definition11.1) in C containing e i contains a positive-length H -avoiding subpath joining avertex of (cid:104) s i (cid:105) to a vertex of [ v i +4 , r i +5 , v i +5 (cid:105) . This yields the contradiction thatthe edge of r i +4 incident with v i +5 is in two H -green cycles. Therefore, B has oneattachment in [ r i +5 r i +6 (cid:105) and one attachment in r i +3 .Let D i +1 be a 1-drawing of G − (cid:104) s i +1 (cid:105) . Lemma 12.4 implies no H -yellow edgein either r i − or r i +2 is crossed in D i +1 . An H -green edge of r i +2 is not spannedby a global H -bridge (there is no room for such a jump between B and wv i +2 ), soLemma 12.5 implies no H -green edge of r i +2 is crossed in D i +1 . Because we are notin Case 1 and there is no 3-jump, Lemma 12.5 implies no H -green edge of either r i − or r i +2 is crossed in D i +1 .Lemma 7.2 (1) implies no edge of r i r i +1 is crossed in D i +1 . Thus, none of r i − r i r i +1 r i +2 is crossed in D i +1 , and therefore Q i +1 cannot be crossed in D i +1 .However, Corollary 12.7 and Lemma 5.9 imply that Q i +1 is crossed in D i +1 . Thiscontradiction completes the proof that G has a red edge when there is a 2.5-jump. (cid:3) At this point, we may assume G has no 2.5-jump and no 3-jump. Claim . If G has a 2-jump v i v i +2 , then either r i − or r i +2 has a red edge. Proof.
In this case, let D i +1 be a 1-drawing of G − (cid:104) s i +1 (cid:105) . Corollary 12.7and Lemma 5.9 imply that Q i +1 is crossed in D i +1 . Lemma 7.2 (1) shows noedge of r i r i +1 is crossed in D i +1 . Therefore, some edge of r i − ∪ r i +2 must becrossed in D i +1 . Lemmas 12.4 and 12.5 imply that no H -yellow or H -green edgein r i − ∪ r i +2 is crossed in D i +1 . Therefore, Theorem 11.3 shows some edge in r i − ∪ r i +2 is red. (cid:3) In the final case, there are no global H -bridges. Therefore, there are no H -yellow cycles and every H -green cycle is contained in cl( Q i ), for some i . For j ∈{ , , , , } , let e j be the edge in s j incident with v j and let D j be a 1-drawing of G − e j . Corollary 12.7 and Lemma 5.9 imply that Q j is crossed in D j , so some edgein r j +3 r j +4 r j +5 r j +6 is crossed in D j . Since e j cannot be in any H -green cyclecontaining an edge in r j +3 r j +4 r j +5 r j +6 , Lemma 12.5 implies no H -green edge in r j +3 r j +4 r j +5 r j +6 can be crossed in D j . Therefore the edge in r j +3 r j +4 r j +5 r j +6 crossed in D j is red in G .We conclude this section with the technical lemma (12.14) below that will beused in the next section. We start with four lemmas leading to a more refined un-derstanding of R -separation in cases of interest for us. The first three are primarilyused in the proof of the fourth. (Recall that an RR -path is an R -avoiding pathwith both ends in R .) Lemma . Let G ∈ M and let V ∼ = H ⊆ G , with H tidy, witnessed by theembedding Π . Let P be an RR -path in G . If B is a global H -bridge so that oneend of P is in the interior of the span of B , then there is an H -quad Q so that P ⊆ cl( Q ) and the two cycles in R ∪ P containing P are non-contractible in R P . Proof. As P is R -avoiding, Theorem 6.7 implies P is not contained in D . If P isjust an H -spoke, then both conclusions are obvious. Otherwise, as we traverse P from an end u in the interior of the span of B , there is a first edge e that is not in H . Since P ⊆ M , there is an H -quad Q so that e ∈ cl( Q ). Let P (cid:48) be the H -bridgein H ∪ P containing e . Then P (cid:48) is an H -avoiding path with both ends in H , so P (cid:48) ⊆ cl( Q ).Since P is R -avoiding, if both ends of P (cid:48) are in R , then P (cid:48) = P and P ⊆ cl( Q ),as claimed. Otherwise, one end w of P (cid:48) is in the interior of some H -spoke s i . Ourtwo claims eliminate many possibilities for the other end x of P (cid:48) . We choose thelabelling so that u ∈ r i − r i . Claim . x is not in (cid:104) s i − r i − r i s i +1 (cid:105) . Proof.
Suppose first that u is an end of P (cid:48) . The choice of e implies P (cid:48) =[ u, P, w ] is just the edge e . If u is an end of s i , then e is an H -bridge having all itsattachments in s i , contradicting Lemma 10.10. If u is not an end of s i , then thereis an H -green cycle that contains an edge f of R incident with u . But then f is intwo H -green cycles, contradicting Theorem 6.7. Thus, u is not an end of P (cid:48) .If x ∈ (cid:104) s i − r i − r i s i +1 (cid:105) − u , then u = v i and P (cid:48) = [ w, P, x ] is contained ineither cl( Q i − ) − r i +4 or cl( Q i ) − r i +5 . In this case, we again have the contradictionthat some edge of R incident with u is in two H -green cycles. (cid:3) Claim 1 implies u = v i and [ u, P, w ] ⊆ s i . Moreover, x is in Q i − ∪ Q i andeither P (cid:48) ⊆ cl( Q i − ) or P (cid:48) ⊆ cl( Q i ). The next claim eliminates another possibilityfor x . Claim . x / ∈ (cid:104) s i (cid:105) . Proof.
Suppose by way of contradiction that x ∈ (cid:104) s i (cid:105) . Let B (cid:48) be the H -bridgecontaining P (cid:48) . Observe that B (cid:48) is H -local and that w and x are both attachmentsof B (cid:48) in (cid:104) s i (cid:105) . Corollary 5.15 implies that these are the only attachments of B (cid:48) ,contradicting Lemma 10.10. (cid:3) We conclude from Claims 1 and 2 that x is in r i +4 r i +5 . Evidently, P is incl( Q i − ) or cl( Q i ), respectively, as required for the first conclusion. Furthermore,both cycles in Π[ R ∪ P ] that contain P are non-contractible in R P . Lemma . Let G ∈ M and let V ∼ = H ⊆ G , with H tidy, witnessed bythe embedding Π . For i ∈ { , , , , } and j ∈ { i + 3 , i + 4 , i + 5 } , let e ∈ r i and f ∈ r j be edges that are not H -green. Suppose P is an RR -path in M having bothends in the component R (cid:48) of R − { e, f } containing r i +6 r i +7 r i +8 r i +9 and so thatthe cycle in Π[ R (cid:48) ∪ P ] is non-contractible. Then P ⊆ (cid:18) cl( Q j ) − [ v j , s j , v j − (cid:105) (cid:19) ∪ (cid:91) j − Choose the labelling u and w of the ends of P so that u is nearer in R (cid:48) tothe end incident with f than w is.Let γ be a non-contractible curve meeting Π[ G ] in just the two points a and b ;we note that u and w are on different ab -subpaths of R (allowing a or b to be anend of P ). We may choose the labelling of a and b so that a ∈ R (cid:48) , and if both a and b are in R (cid:48) , then a is closer to the end of R (cid:48) incident with f than b is. Claim . (1) If v j and w are on the same ab -subpath of R , then P ∩ (cid:104) s j (cid:105) is empty. 2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 95 (2) If v i +1 and u are in the same ab -subpath of R , then P ∩ (cid:104) s i +1 (cid:105) is empty. Proof. The statements are symmetric, so it suffices to prove the first. Supposeto the contrary that P ∩ (cid:104) s j (cid:105) is not empty. As we traverse P from w (which is notincident with s j ), let x be the first vertex in (cid:104) s j (cid:105) and let P (cid:48) denote the wx -subpathof P . Evidently, P (cid:48) is contained in one component M (cid:48) of M \ ( γ ∪ s j ). On the otherhand, f is between v j and w , and so f is in M (cid:48) . If w ∈ r j , then P (cid:48) and f are in an H -green cycle, a contradiction.Otherwise, v j +1 ∈ M (cid:48) and P (cid:48) intersects (cid:104) s j +1 (cid:105) . In this case, some (cid:104) s j +1 (cid:105) (cid:104) s j (cid:105) -subpath of P (cid:48) is in an H -green cycle with f , also a contradiction. (cid:3) If both v i +1 and w are in the same ab -subpath of R and both v i +4 and u are inthe same ab -subpath of R , then Claim 1 implies P is trapped between s j and s i +1 ,as required. By symmetry, we may assume that v i +1 is not in the same ab -subpathof R as w . Let R w denote the ab -subpath of R containing w and let R i +1 denotethe other ab -subpath of R , so v i +1 ∈ R i +1 .This implies that v i +1 , v i +2 , . . . , v j are all in R i +1 . We noted above that u / ∈ R w , so u is also in R i +1 . From Claim 1 (1), we conclude that P is disjoint from (cid:104) s j (cid:105) . Thus, P is contained in the component of M − s j disjoint from v i +2 .It follows from the fact that all the H -spokes are in M that v j − is on the same ab -subpath as w . This combines with the fact that v i +1 is not in that ab -subpathand the fact that P meets γ at most in a to tell us that P ⊆ (cid:18) cl( Q j ) − (cid:104) s i − (cid:105) (cid:19) ∪ (cid:91) j − Let R (cid:48) be the component of R −{ e, f } containing r i +1 r i +2 and let C be thecycle in R ∪ P that contains P and is contained in R (cid:48) ∪ P . Since R is contractible,the other cycle in R ∪ P containing P is homotopic to C ; thus, it suffices to show C is contractible.Let u e be the end of R (cid:48) incident with e . Suppose there is a ([ u e , r i , v i +1 ] s i +1 ) s i -path P (cid:48) in P contained in cl( Q i ). Since C is disjoint from r i +1 , P (cid:48) is contained inan H -green cycle containing e , a contradiction.Thus, there is no ([ u e , r i , v i +1 ] s i +1 ) s i -path in P contained in cl( Q i ). Since C is disjoint from r i +5 , there is an arc in the disc bounded by Π[ Q i ] joining a pointof [ v i , r i , u e (cid:105) to r i +5 that is disjoint from C ; this shows that C is contractible, asrequired.Our next lemma takes us one step closer to the useful description of R -separation. Lemma . Let G ∈ M and let V ∼ = H ⊆ G , with H tidy. Suppose e ∈ r i and f ∈ r i +3 r i +4 are R -separated as witnessed by the subdivision H (cid:48) of V . If e is not H -green, then the component of R − { e, f } containing both ends of some H (cid:48) -spoke is the one containing r i +5 r i +6 r i +7 r i +8 r i +9 . Proof. Let Π be an embedding of G in R P so that H is Π-tidy. Recall that R is also the H (cid:48) -rim. Observation 11.6 (2) shows that two of the four H (cid:48) spokes haveall their ends in the same component of R − { e, f } . Of the four H (cid:48) -spokes, at mostone can be in D . Thus, of the two that have both ends in the same component R (cid:48) of R − { e, f } , there is at least one, call it s , that is in M .In particular, the two cycles in R ∪ s containing s are non-contractible. NowLemma 12.11 shows the two ends of the RR -path s are not in the componentof R − { e, f } containing r i +1 r i +2 and so must be in the component containing r i +5 r i +6 . . . r i +9 , as claimed.Our next lemma in the series gives a quite refined description of R -separation. Lemma . Let G ∈ M and let V ∼ = H ⊆ G , with H tidy. Let e ∈ r i and f ∈ r i +4 r i +5 be edges that are both not H -green. If e and f are R -separated in G ,then there is a witnessing subdivision H (cid:48) of V having s i +2 and s i +3 as H (cid:48) -spokesand the other two H (cid:48) -spokes are in cl( Q i − ) ∪ cl( Q i ) . Proof. Let Π be an embedding of G in R P for which H is Π-tidy. Let H bea subdivision of V witnessing the R -separation of e and f . Let s be an H spokehaving both ends in the same component R (cid:48) of R − { e, f } . Claim . The cycles in Π[ R ∪ s ] containing s are non-contractible. Proof. Suppose first by way of contradiction that Π[ s ] in not contained in M .Since H is Π-tidy, s is a global H -bridge. Theorems 10.6 and 12.6 show s is eithera 2- or a 2.5-jump. By hypothesis, it is not possible for both e and f to be in thespan of s and, therefore, neither is. On the other hand, each of the other three H -spokes has precisely 1 end in the span of s , and is contained in M . Let thesespokes appear in the order t , t , t in the span of s .We claim that the t i imply the existence of an H -yellow cycle that does notbound a face of Π[ G ], contradicting Lemma 11.2 (3). Let P be the span of s and,for i = 1 , , 3, let u i be the end of t i that is not in P . Because Π[ s ∪ P ] bounds aclosed disc, both cycles in Π[ R ∪ t i ] containing t i are non-contractible. Thus, t i hasan end in each of the ab -subpaths of R .Lemma 12.9 implies that each t i is contained in an H -quad. Thus t ∪ t ∪ t is contained the the union of the closures of the H -quads that have an edge in P . In particular, u , u , and u occur in a 3-rim path P having u and u asends. Letting P be the minimal subpath of P containing the ends of the t i , we seethat P t P t is an H -yellow cycle C . However, Π[ C ] bounds a face of Π[ G ]; thecontradiction is that t and s are on different sides of Π[ C ].Thus, s is contained in M . Since s is one of four H -spokes, the two cycles inΠ[ R ∪ s ] that contain s are non-contractible. (cid:3) In particular, s has an end in each of the two ab -subpaths of R determined bythe standard labelling of Π[ G ].In the case f ∈ r i +5 , we may, if necessary, use the reflective symmetry j ↔ − j (for 0 ≤ j ≤ s f of s is, in Π[ R (cid:48) ], between the end u f 2. EXISTENCE OF A RED EDGE AND ITS STRUCTURE 97 of f in R (cid:48) and a , say, while the other end s e of s is between a and the end u e of e . In particular, v i +1 , v i +2 , v i +3 , and v i +4 are not in R (cid:48) . Lemma 12.12 shows thisalways holds when f ∈ r i +4 .Let s (cid:48) be the other H (cid:48) -spoke having both ends in R (cid:48) . The arguments above for s apply equally well to s (cid:48) . Lemma 12.10 shows that ( s ∪ s (cid:48) ) ⊆ cl( Q i − ) ∪ cl( Q i ). Inparticular, s and s (cid:48) are disjoint from s i +2 and s i +3 , so these H -spokes may replacethe two H -spokes having ends in both components of R − { e, f } , as required.The final technical lemma of this section will be used in the next. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. If e and e (cid:48) are rededges in the same H -rim branch, then ∆ e and ∆ e (cid:48) are disjoint. Proof. We may choose the labelling of e and e (cid:48) so that e = uw and e (cid:48) = xy aresuch that r i = [ v i , r i , u, w, r i , x, y, r i , v i +1 ]. As we follow ∆ e − e from w , there is afirst edge f that is not in R . In fact, Theorem 12.1 (3) implies f is incident with w , as there can be no global H -bridge spanning e (cid:48) .Observe that f is not in H , so H ⊆ G − f . Moreover, if f is in an H -yellowcycle, then either e or e (cid:48) is H -yellow, a contradiction. Thus, Lemmas 12.4 and 12.5imply the colours of an edge of R are the same in G and G − f , unless the edge isin an H -green cycle in G that contains f . Such an edge is necessarily in [ w, r i , x ].Let D be a 1-drawing of G − f and let e and e be the edges of G − f crossed in D . Since f is incident with w ∈ (cid:104) r i (cid:105) , Theorem 5.23 and Lemma 5.9 imply that Q i iscrossed in D , so we may assume e ∈ r i − r i r i +1 and e ∈ r i +4 r i +5 r i +6 . Moreover,no H -green cycle containing e contains f , so e is red in G . In particular, Lemma11.7 implies e is R -separated from both e and e (cid:48) .Let u e and w e be the first vertices in r i +5 as we traverse ∆ e − e from u and w ,respectively. Likewise, we have x e (cid:48) and y e (cid:48) in r i +5 ∩ (∆ e (cid:48) − e (cid:48) ). Claim . e ∈ (cid:104) u e , r i +5 , y e (cid:48) (cid:105) . Proof. Suppose by way of contradiction that e ∈ r i +4 [ v i +5 , r i +5 , u e ]; a sim-ilar argument will treat the case e ∈ (cid:104) y e (cid:48) , r i +5 , v i +6 (cid:105) r i +6 .If e ∈ r i − [ v i , r i , u ], then e is red in G , so e and e are R -separated in G .Note that either e ∈ r i or e ∈ r i +5 . Lemma 12.13 implies there is a witnessingsubdivision H (cid:48) of V that contains s i +2 and s i +3 , while the other two spokes arein cl( Q i − ) ∪ cl( Q i ). Furthermore, ∆ e shows that f / ∈ H (cid:48) ; therefore, H (cid:48) ⊆ G − f shows that e and e are R -separated in G − f , and therefore cannot cross in D , acontradiction.The other possibility is that e ∈ [ u, r i , v i +1 ] r i +1 . Since e and e are both redin G , Lemma 11.7 implies e is R -separated from e in G − f . As in the precedingparagraph, we may choose the witnessing subdivision H (cid:48) of V to contain s i +2 and s i +3 , while the other two spokes are in cl( Q i − ) ∪ (cl( Q i ) − f ). Again H (cid:48) witnessesthe R -separation of e and e in G − f , a contradiction. (cid:3) Theorem 12.1 (2) shows that any edge in either ∆ e ∩ r i +5 or ∆ e (cid:48) ∩ r i +5 is in adigon in G and so is not e . Thus, e is further restricted to be in (cid:104) w e , r i +5 , x e (cid:48) (cid:105) .Lemma 11.7 implies ∆ e and ∆ e are disjoint, as are ∆ e and ∆ e (cid:48) , which furtherimplies that ∆ e and ∆ e (cid:48) are disjoint, as required.HAPTER 13 The next red edge and the tile structure We now know that there are red edges and every red edge comes equippedwith a ∆. The tiles are determined by what is between “consecutive” red edges. Inthis section, we explain what “consecutive” means, show that consecutive red edgesdetermine one of the tiles, and complete the proof of our main result, Theorem 2.14,by demonstrating that every red edge has a consecutive red edge on each side. Definition . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let e = uw be a red edge in r i , labelled so that r i = [ v i , r i , u, e, w, r i , v i +1 ]. A red edge e w is w -consecutive for e if:(1) e w ∈ [ w e , r i +5 , v i +6 ] r i +6 r i +7 (recall that w e is the vertex in the peak of∆ e nearest w in ∆ e − e );(2) there is no red edge in [ w e , r i +5 , v i +6 ] r i +6 r i +7 between w e and e w ;(3) there is no red edge in [ w, r i , v i +1 ] r i +1 r i +2 between w and the peak of∆ e w ;(4) if e w is the edge of P w nearest w that is not in R , then there is a 1-drawing D of G − e w in which e crosses e w .(5) There is an analogous definition for u -consecutive .Our first main goal is, therefore, the following. Theorem . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let e = uw bered in G . Then there is a w -consecutive red edge and a u -consecutive red edge for e . The next lemma will be helpful in the proof. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. Let e = uw and ˆ e bered edges in G , with e ∈ r i and the labelling chosen so that r i = [ v i , r i , u, e, w, r i , v i +1 ] and ˆ e ∈ [ w e , r i +5 , v i +6 ] r i +6 r i +7 . If e w is the w -nearest edge of P w that is not in R and e and ˆ e are not R -separated in G − e w , then e has a w -consecutive red edge. Proof. Suppose there is a red edge e (cid:48) in r i r i +1 r i +2 between w and the peak of∆ ˆ e . Then e (cid:48) is R -separated from ˆ e in both G and G − e w , showing that e and ˆ e are R -separated in G − e w , a contradiction. Thus, no such red edge exists.Let ˆ e (cid:48) be the w e -nearest red edge in [ w e , r i +5 , v i +6 ] r i +6 r i +7 . Lemma 11.7implies ˆ e (cid:48) is R -separated from e in G ; if ˆ e (cid:48) were also R -separated from e in G − e w ,then so would ˆ e , which contradicts the hypothesis. But now Lemma 11.7 impliesthere is a 1-drawing of G − e w in which e crosses ˆ e (cid:48) , as required.And now the final major proof needed to prove Theorem 2.14. Proof of Theorem 13.2. It obviously suffices to prove the existence of a w -consecutive red edge for e . Let r i be the H -rimbranch containing e . Let e w be theedge of P w nearest w and not in R . There are two principal cases. Case 1: e w is incident with w . We note that e w is contained in a Q i +1 -bridge that is not M Q i +1 . Let D be a1-drawing of G − e w . Corollary 12.7 and Lemma 5.9 show that Q i +1 is crossed in D . Let • f be the edge of r i +4 r i +5 r i +6 r i +7 that is crossed in D and • f (cid:48) be the other edge crossed in D ; thus, f (cid:48) ∈ r i − r i r i +1 r i +2 . Claim . If f is not red in G , then there is a w -consecutive red edge for e . Proof. Because we are in Case 1, no global H -bridge has w in the interiorof its span and, therefore, e w is not in any H -yellow cycle that could witness the H -yellowness of any edge in r i +4 r i +5 r i +6 r i +7 , (in particular, the H -yellowness of f ). Therefore, Lemma 12.4 shows f is not H -yellow. Since f is not red, Theorem11.3 implies f is H -green. Lemma 12.5 implies there is a 2.5-jump J that spans f and so that f is in the H -rim branch whose interior contains an end of J . We notethat if v i +6 is in the span of J , then Lemma 7.2 (1) shows no edge in the span of J is crossed in D . Therefore, v i +6 is not in the span of J . Furthermore, if e w is not in s i +1 , then H ⊆ G − e w and, therefore Lemma 6.6 (10) implies f is not crossed in D ,a contradiction. This implies w = v i +1 . We summarize these remarks as follows. Subclaim . • w = v i +1 and • there is a 2.5-jump J so that: – f is spanned by J ; – f is in the H -rim branch whose interior contains an end of J ; and – v i +6 is not in the span of J . (cid:50) Subclaim . Let j ∈ { i + 4 , i + 5 , i + 6 , i + 7 } so that f is in the H -rim branch r j . Then no edge of r j is H -yellow. Proof. Suppose some edge e (cid:48) of r j is H -yellow. This implies e (cid:48) is not H -greenand, therefore, is not spanned by J . Let C and C (cid:48) be the witnessing H -yellow and H -green cycles, respectively.Suppose first that j ∈ { i +4 , i +5 } . Then r j = [ v j , r j , f, r j , e (cid:48) , r j , v j +1 ]. Because e ∈ r i is not H -green, v j +5 ∈ { v i − , v i } is in the interior of C (cid:48) ∩ R . This impliesthere is an H -yellow cycle containing s j and the portion of r j from v j to e (cid:48) . ByLemma 11.2 (3), this H -yellow cycle must be C and, therefore, f ∈ C . Now the factthat f is crossed in D contradicts Lemma 12.4. A completely analogous argumentholds for j ∈ { i + 6 , i + 7 } . (cid:3) Let (cid:98) w be the vertex in r i +5 that is nearest w in P w . Observe that (cid:98) w is notnecessarily in the peak of ∆ e . (See Figure 12.1, where (cid:98) w is the vertex of ∆ e at thetop right hand corner of ∆ e .) The following claim will be helpful in completing theproof of Case 1. Subclaim . If (cid:98) w (cid:54) = v i +6 , then [ (cid:98) w, r i +5 , v i +6 ] is in an H -green cycle containedin cl( Q i ). 00 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Proof. Let P (cid:48) w be the (cid:98) ws i +1 -subpath of P w . Since e w ∈ s i +1 , P (cid:48) w ⊆ P w − w .Let (cid:98) w e be the end of P (cid:48) w in s i +1 . Since (cid:98) w / ∈ s i +1 and (cid:98) w e ∈ s i +1 , (cid:98) w (cid:54) = (cid:98) w e . Bydefinition of (cid:98) w , P (cid:48) w − (cid:98) w is disjoint from r i +1 . Therefore, P (cid:48) w [ (cid:98) w e , s i +1 , v i +6 , r i +5 , (cid:98) w ]is an H -green cycle containing [ (cid:98) w, r i +5 , v i +6 ], as required. (cid:3) The proof of Claim 1 is completed now by treating separately each of the fourpossibilities for f : f ∈ r i +4 , f ∈ r i +5 , f ∈ r i +6 , and f ∈ r i +7 . Subcase 1: f ∈ r i +4 .In this case, J has an end x (cid:48) ∈ (cid:104) r i +4 (cid:105) and the other end of J is v i +2 . Lemma7.2 (3b) implies f (cid:48) ∈ r i r i +1 . We claim that if f (cid:48) ∈ r i +1 , then there is another1-drawing of G − e w in which f crosses e .Since f ∈ r i +4 and f (cid:48) ∈ r i +1 , we see that s i is exposed in the 1-drawing D of G − e w . Note that D [ Q i − ] consists of a simple closed curve crossed by D [ f (cid:48) ], with D [ r i ] on one side (the inside of D [ Q i − ]) and most of D [ H ] on the other side (thisis the outside of D [ Q i − ]).We claim that we may reroute f inside D [ Q i − ] so that it crosses e insteadof f (cid:48) . If this fails, then there is an ( H − (cid:104) s i +1 (cid:105) )-avoiding path P having one endin the component of r i +1 − f (cid:48) that contains v i +1 , and having its other end in Q i − ∪ [ v i , r i , u ].We note that D [ s i +1 − v i +1 ] (which is possibly just v i +6 ) is completely outside D [ Q i − ]. Therefore, P is H -avoiding. In R P , we conclude that P cannot startinside Q i +1 . Thus, P is contained in a global H -bridge. Therefore, P is a global H -bridge; we note that P has one end in the component of r i +1 − f containing v i +1 .No edge of r i +2 can be spanned by P , as such an edge is already spanned by J andtherefore would contradict Theorem 6.7. In the other direction, P cannot span e ,as e is red and not H -green. This contradiction shows that f may be redrawn asclaimed. Consequently, we may assume f (cid:48) ∈ r i .Observe that no global H -bridge can have an end y in (cid:104) r i (cid:105) , since yv i +3 shows e is H -green, a contradiction, and yv i − shows f is H -yellow and, therefore, byLemma 12.4 cannot be crossed in D . It follows from this, using Lemmas 12.4 and12.5 and Theorem 11.3, that f (cid:48) is red in G .Suppose first that some edge e (cid:48) of [ x (cid:48) , r i +4 , v i +5 ] is red in G . Then ∆ e and∆ e (cid:48) are R -separated in G as witnessed by a subdivision H (cid:48) of V consisting of R , s i − , s i − , and two RR -paths P and P , contained in ∆ e and ∆ e (cid:48) , respectively.The paths P and P are disjoint from s i +1 except that, possibly P contains v i +6 .Thus, H (cid:48) and Lemma 7.2 show that f cannot be crossed in D , a contradiction.Therefore, there is no red edge in [ x (cid:48) , r i +4 , v i +5 ].Furthermore, no global H -bridge other than J has an end in [ x (cid:48) , r i +4 , v i +5 (cid:105) , asotherwise either e is H -yellow, or f is in two H -green cycles, both contradictions,the latter of Theorem 6.7. We conclude that each edge of [ x (cid:48) , r i +4 , v i +5 ] is either H -yellow or contained in an H -green cycle in cl( Q i − ). Subclaim 2 shows thefollowing. Subcase 1 Observation: Each edge of [ x (cid:48) , r i +4 , v i +5 ] is in an H -green cyclecontained in cl( Q i − ) . Suppose there is a red edge e (cid:48) in r i +5 . By Lemma 11.7, e (cid:48) is R -separated from e in G . Therefore, P u is disjoint from s i and now we see that G − e w containsthe subdivision H (cid:48) of V consisting of ( H − (cid:104) s i +1 (cid:105) ) ∪ P u . But J is in an H (cid:48) -green 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 101 cycle C and so, by Lemma 6.6 (10), C , and in particular, f , is not crossed in D , acontradiction.Thus, no edge of r i +5 is red in G . We consider next a 1-drawing D i − of G −(cid:104) s i − (cid:105) . By Corollary 12.7 and Lemma 5.9, Q i − is crossed in D i − . From Lemmas12.4, 12.5, and 7.2 (1), no edge in r i +2 r i +3 r i +4 is crossed in D i − . Therefore, itis some edge f (cid:48)(cid:48) in r i +5 that is crossed in D i − . Since no edge of r i +5 is red in G ,Lemmas 12.4 and 12.5 imply that f (cid:48)(cid:48) is spanned by a 2.5-jump J (cid:48)(cid:48) = x (cid:48)(cid:48) v i − , with x (cid:48)(cid:48) ∈ (cid:104) r i +5 (cid:105) .Now consider a 1-drawing D i +3 of G − (cid:104) s i +3 (cid:105) . As for Q i − in the precedingparagraph, Q i +3 is crossed in D i +3 . In this case, r i +1 is contained in the H -yellowcycle Q i +1 (with witnessing H -green cycle containing J (cid:48)(cid:48) ). Therefore, r i +1 is notcrossed in D i +3 . Lemma 7.2 (1) implies no edge in the span of J is crossed in D i +3 . Subcase 1 Observation combines with Lemma 12.5 to show that no edge in[ x (cid:48) , r i +4 , v i +5 ] is crossed in D i +3 . But now we see that Q i +3 cannot be crossed in D i +3 , a contradiction that shows Subcase 1 cannot occur. Subcase 2: f ∈ r i +5 .In this case, J has an end x (cid:48) ∈ (cid:104) r i +5 (cid:105) . Subclaim 1 implies that v i +6 is notspanned by J , so the other end of J is v i +3 . Lemma 7.2 implies the edge f (cid:48) (crossed by f in D ) is in r i +2 .We first show that there is no global H -bridge spanning any edge in r i r i +1 r i +2 .For if J (cid:48) is a global H -bridge that spans such an edge, then J (cid:48) does not span e ,while Lemma 10.9 (1) shows it cannot be the 2-jump v i +1 v i +3 . Theorem 6.7 shows J (cid:48) cannot span any edge in r i +3 , so no edge of r i +1 r i +2 is spanned by a global H -bridge. On the other side, J (cid:48) would have to span r i − r i − . In that case, J and J (cid:48) contradict Lemma 10.9 (4).We also conclude that no edge of r i +5 r i +6 r i +7 is H -yellow.Our next principal aim is to show that each edge of [ x (cid:48) , r i +5 , v i +6 ] is H -green,witnessed by a cycle in cl( Q i ). We have already seen that none of the edges in[ x (cid:48) , r i +5 , v i +6 ] is H -yellow; to see they are H -green, it suffices by Theorem 11.3 toshow none is red.If e (cid:48) is one of these edges that is red, then Lemma 11.7 implies it is R -separatedfrom e . We note that ∆ e and ∆ e (cid:48) are disjoint, both are in cl( Q i ), and w = v i +1 .Therefore, e (cid:48) is in r i +5 , between x (cid:48) and the peak of ∆ e . However, this shows e (cid:48) and e are R -separated in G − e w and, therefore, f and r i +2 are R -separated in G − e (cid:48) ,showing that f cannot cross anything in D , a contradiction. Therefore, no edge of[ x (cid:48) , r i +5 , v i +6 ] is red, and so they are all H -green.We next show they are not spanned by a global H -bridge. Recall that (cid:98) w is thevertex in r i +5 that is nearest w in P w .If (cid:98) w (cid:54) = v i +6 , then ( P w − e w ) ∪ ( s i +1 − e w ) ∪ [ (cid:98) w, r i +5 , v i +6 ] contains an H -greencycle that contains [ (cid:98) w, r i +5 , v i +6 ] and is contained in cl( Q i ). Theorem 6.7 shows noedge of [ (cid:98) w, r i +5 , v i +6 ] is spanned by a global H -bridge, so no edge of [ x (cid:48) , r i +5 , v i +6 ] is H -green by a global H -bridge. In this case, every edge of [ x (cid:48) , r i +5 , v i +6 ] is H -greenby a local cycle.So suppose (cid:98) w = v i +6 . By way of contradiction, we suppose there is a global H -bridge J (cid:48)(cid:48) spanning the edge of r i +5 incident with v i +6 . Then J (cid:48)(cid:48) must be x (cid:48)(cid:48) v i +8 ,for some x (cid:48)(cid:48) ∈ [ x (cid:48) , r i +5 , v i +6 ]. All edges in [ x (cid:48) , r i +5 , x (cid:48)(cid:48) ] are H -green by local cycles.For j ∈ { i + 3 , i + 8 } , let e j be the edge of s i +3 incident with v j and let D j be a 02 13. THE NEXT RED EDGE AND THE TILE STRUCTURE G − e j . Corollary 12.7 implies Q i +3 has BOD and Lemma 5.9 implies Q i +3 is crossed in D j . Lemma 7.2 (3a) implies neither r i +6 r i +7 nor r i +3 r i +4 iscrossed in D j , while (2) of the same lemma implies neither r i +9 nor r i +1 is crossedin D j . Therefore, r i +8 crosses r i +2 .If the edge e (cid:48) i +8 of r i +8 that is crossed in D i +3 is H -green because of some2.5-jump, then Lemma 7.2 implies e (cid:48) i +8 can cross only r i +1 in D i +3 . Therefore,Theorem 11.3 and Lemmas 12.4 and (because no H -green cycle containing e (cid:48) i +8 can contain e i +3 ) 12.5 imply e (cid:48) i +8 is red in G . Likewise the edge e (cid:48) i +2 of r i +2 thatis crossed in D i +8 is red in G .By Lemma 11.7, e (cid:48) i +2 and e (cid:48) i +8 are R -separated in G . Moreover, the nearerof the ( r i +7 r i +8 )( r i +2 r i +3 )-paths P in ∆ e (cid:48) i +2 and P in ∆ e (cid:48) i +8 , along with s i and s i +1 witness their R -separation. We now show that P is contained in cl( Q i +8 ) andmust be disjoint from s i +4 .If P intersects s i +4 at a vertex other than v i +4 , then P ∪ s i +4 ∪ r i +8 containsan H -green cycle that includes e (cid:48) i +8 . Otherwise, P and s i +4 intersect just at v i +4 ,in which case P ∪ s i +4 ∪ r i +8 contains a cycle C that includes e (cid:48) i +8 . The H -greencycle containing J shows C is H -yellow. Both possibilities contradict the fact that e (cid:48) i +8 is red.Symmetrically, we use J (cid:48)(cid:48) to show that P is disjoint from s i +2 . Thus, G contains a subdivison of V consisting of R , P , P , s i − , s i , s i +1 and s i +2 . Butthen G − e w contains a subdivision of V , yielding the contradiction that f cannotbe crossed in D . Therefore, there is no global H -bridge J (cid:48)(cid:48) spanning the edge of r i +5 incident with v i +6 .We conclude that every edge of [ x (cid:48) , r i +5 , v i +6 ] is in an H -green cycle containedin cl( Q i ).We are now in a position to show that r i +6 has a red edge. By way of contra-diction, we suppose r i +6 has no red edge. If there were a global H -bridge havingan end in (cid:104) r i +6 (cid:105) , then r i +2 is H -yellow; Lemma 12.4 shows r i +2 is not crossed in D , a contradiction. Thus, no global H -bridge has an end in (cid:104) r i +6 (cid:105) .Let D i be a 1-drawing of G − (cid:104) s i (cid:105) . Then Corollary 12.7 and Lemma 5.9 imply Q i is crossed in D i . However, Lemma 7.2 shows none of r i +3 r i +4 r i +5 r i +6 can becrossed in D i , a contradiction.Thus, r i +6 has a red edge e (cid:48) . Then e is R -separated from e (cid:48) in G . If e is R -separated from e (cid:48) in G − e w , then f is R -separated from r i +2 in G − e w and so f cannot be crossed in D , a contradiction. Therefore, e is not R -separated from e (cid:48) in G − e w , so Lemma 13.3 implies there is w -consecutive red edge for e , completingthe proof in Subcase 2. Subcase 3: f ∈ r i +6 .In this case, J has an end x (cid:48) ∈ (cid:104) r i +6 (cid:105) and the other end is v i +9 . Also, Lemma7.2 implies f (cid:48) (crossed by f in D ) is in r i +9 .Suppose by way of contradiction that no edge of r i +6 is red in G . We showthat no edge of r i +6 is H -yellow. As every edge in [ x (cid:48) , r i +6 , v i +7 ] is H -green (be-cause of J ), we assume by way of contradiction that there is an H -yellow edgein [ v i +6 , r i +6 , x (cid:48) ]. Let C and C (cid:48) be the witnessing H -yellow and H -green cycles,respectively. Lemma 11.2 (1) implies there is a global H -bridge B contained in C (cid:48) ,while (4) shows C ⊆ cl( Q i +1 ). 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 103 The edges of the span P B of B are all H -green, so P B does not contain thered edge e . One end of B is in [ w, r i , v i +1 , r i +1 , v i +2 ] and the other end is in r i +3 .Furthermore, Lemma 10.9 (4) and the presence of J shows v i +4 is not the otherend of B .Write C = P P P P as in Definition 11.1 ( H -yellow). Because C bounds aface Π[ G ], C ⊆ cl( Q i +1 ), so that P = r i +1 ∩ C . In particular, e w / ∈ C .Choose the labelling of P and P so that the end of P in r i +6 is nearer to v i +6 than is the corresponding end of P . Since there is an H -yellow cycle containing P and s i +2 , Lemma 11.2 (3) shows this must be C . It follows that P = s i +2 .Consider the subdivision H (cid:48) of V whose rim consists of ( R −(cid:104) P B (cid:105) ) −(cid:104) x (cid:48) , r i +6 , v i +6 (cid:105) , B , C − (cid:104) r i +6 ∩ C (cid:105) , and whose spokes are s i − , s i , and s i +3 , [ v i +3 , r i +3 , z ]. Then H (cid:48) does not contain e w and so must contain the unique crossing in D . Since f is notin H (cid:48) , this is a contradiction, showing that no edge of [ v i +6 , r i +6 , x (cid:48) ] is H -yellow.Because of J , a global H -bridge spanning an edge in [ v i +6 , r i +6 , x (cid:48) ] would haveto be a 2.5-jump having v i +4 as an end. But then e is in an H -yellow cycle, whichis impossible. Thus, for each edge ¯ e of [ v i +6 , r i +6 , x (cid:48) ], ¯ e is in an H -green cycle C ¯ e contained in cl( Q i +1 ). Theorem 6.7 implies C ¯ e is disjoint from s i +2 .Let D i +2 be a 1-drawing of G − (cid:104) s i +2 (cid:105) . We know that Q i +2 is crossed in D i +2 (Corollary 12.7 and Lemma 5.9). Lemma 12.5 shows no edge in [ v i +6 , r i +6 , x (cid:48) ] iscrossed in D i +2 , while J and Lemma 7.2 show no edge in [ x (cid:48) , r i +6 , v i +7 ] r i +7 r i +8 is crossed in D i +2 . Therefore, the crossing in D i +2 must be of an edge f (cid:48)(cid:48) in r i +5 crossing r i +1 r i +2 .If f (cid:48)(cid:48) is red in G , then Lemma 11.7 implies f (cid:48)(cid:48) and e are R -separated in G .Since e w ∈ s i +1 , f (cid:48)(cid:48) is between (in r i +5 ) v i +5 and the peak of ∆ e . Thus, f (cid:48)(cid:48) and e are R -separated in G − (cid:104) s i +2 (cid:105) (using s i +3 and s i +4 as two of the four spokes). Inturn, this implies f (cid:48)(cid:48) cannot cross r i +1 r i +2 in D i +2 , a contradiction that shows f (cid:48)(cid:48) is not red. Therefore, Lemmas 12.4 and 12.5 imply f (cid:48)(cid:48) is spanned by a 2.5-jump v i +3 x (cid:48)(cid:48) , with x (cid:48)(cid:48) ∈ (cid:104) r i +5 (cid:105) .Now let D i +3 be a 1-drawing of G − (cid:104) s i +3 (cid:105) . We know that Q i +3 is crossed in D i +3 . However: • Lemma 12.5 implies [ v i +6 , r i +6 , x (cid:48) ] is not crossed in D i +3 ; • Lemma 7.2 (1) implies [ x (cid:48) , r i +6 , v i +7 ] r i +7 r i +8 is not crossed in D i +3 ; and • Lemma 12.4 implies r i +9 is not crossed in D i +3 .These three observations imply the contradiction that Q i +3 cannot be crossed in D i +3 , showing that some edge e (cid:48) in r i +6 is red in G .Obviously, e (cid:48) ∈ [ v i +6 , r i +6 , x (cid:48) ]. By way of contradiction, suppose e and e (cid:48) are R -separated in G − e w . Because e ∈ r i and e (cid:48) ∈ r i +6 , Lemmas 12.12 and 12.13imply that there is a a witnessing subdivision H (cid:48) of V with two H (cid:48) -spokes incl( Q i ) ∪ cl( Q i +1 ) and the other two H (cid:48) -spokes are s i +3 and s i +4 . Furthermore, sixof the eight ends of the H (cid:48) -spokes are in the component R (cid:48) of R − { e, e (cid:48) } containing r i +1 r i +2 r i +3 r i +4 .Let y be the end of e (cid:48) in R (cid:48) . Because w = v i +1 and x (cid:48) ∈ [ v i +6 , r i +6 , x (cid:105) , R (cid:48) iscontained in r i +1 r i +2 r i +3 r i +4 r i +5 [ v i +6 , r i +6 , x (cid:105) . In particular, J is not an H (cid:48) -spoke and at most two of the H (cid:48) -spokes have ends inthe span of J . Lemma 7.2 (1) implies the contradiction that the span of J , whichincludes f , cannot be crossed in D . We conclude that e and e (cid:48) are not R -separatedin G − e w . Lemma 13.3 implies that e has a w -consecutive edge, as required. 04 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Subcase 4: f ∈ r i +7 .In this case, J has an end x (cid:48) ∈ (cid:104) r i +7 (cid:105) . If the other end of J is v i +5 , then Lemma7.2 (3b) implies f (cid:48) is in r i +3 . The contradiction is that Q i +1 is not crossed in D .Therefore, the other end of J is v i . Lemma 7.2 (3b) implies f (cid:48) is in r i r i +1 .Suppose there is no red edge in r i +6 r i +7 . Let e i +8 be the edge of s i +3 incidentwith v i +8 and let D i +8 be a 1-drawing of G − e i +8 . Corollary 12.7 and Lemma5.9 imply Q i +3 is crossed in D i +8 . No edge in r i +6 is spanned by a 2.5-jumphaving an end in (cid:104) r i +6 (cid:105) , as otherwise e is H -yellow. Therefore, Lemmas 12.4 and12.5 imply no edge of r i +6 is crossed in D i +8 . Lemma 7.2 (1) shows that no edgeof [ x (cid:48) , r i +7 , v i +8 ] r i +8 r i +9 is crossed in D i +8 . We conclude that some edge ˆ f of[ v i +7 , r i +7 , x (cid:48) ] is crossed in D i +8 .Lemmas 12.4 and 12.5 imply that there is a 2.5-jump v i +5 x (cid:48)(cid:48) , with x (cid:48)(cid:48) ∈(cid:104) v i +7 , r i +7 , x (cid:48) ], and, furthermore, that ˆ f ∈ [ v i +7 , r i +7 , x (cid:48)(cid:48) ]. Lemma 7.2 (3b) im-plies ˆ f crosses an edge e (cid:48) in r i +4 . Lemmas 12.4 and 12.5 imply e (cid:48) is red in G .Let y be the end of e (cid:48) nearest v i +5 in r i +4 . The r i r i +5 -path P contained in the uu e -subpath of ∆ e − e must have v i +5 as an end, since otherwise e is either H -greenor H -yellow. Symmetrically, the r i +4 r i +9 -path P contained in the yy e (cid:48) -subpath of∆ e (cid:48) − e (cid:48) has v i as an end.Lemma 11.7 implies e (cid:48) is R -separated from e in G . Therefore, P and P aredisjoint. This implies that R ∪ P ∪ P ∪ s i +2 ∪ s i +3 ∪ s i +4 is a subdivision V in G − e w , showing that f cannot be crossed in D , a contradiction that proves thereis a red edge e (cid:48)(cid:48) in r i +6 r i +7 .Suppose e and e (cid:48)(cid:48) are R -separated in G − e w . Lemma 12.12 implies thata witnessing subdivision H (cid:48) of V is such that the component R (cid:48) of R − { e, f } containing six of the eight ends of H (cid:48) -spokes contains r i +1 r i +2 r i +3 r i +4 r i +5 .However, J spans [ x (cid:48) , r i +7 , v i +8 ] r i +8 r i +9 , so at most two H (cid:48) -spokes have endsthat are in the span of J . Lemma 7.2 (1) combines with H (cid:48) to yield the contradictionthat the span of J , including f , cannot be crossed in D . It follows that e and e (cid:48)(cid:48) are not R -separated in G − e w , and now Lemma 13.3 implies e has a w -consecutivered edge, completing the proof of Claim 1. (cid:3) With Claim 1 in hand, we may assume f is red. Recall that f and f (cid:48) are theedges crossed in D , with f ∈ r i +4 r i +5 r i +6 r i +7 and f (cid:48) ∈ r i − r i r i +1 r i +2 . Theproof in Case 1 is completed by finding a w -consecutive red edge for e . We proceedin four cases, basically depending on which side of ∆ e each of f and f (cid:48) is on. Subcase 1: f is in r i +4 [ v i +5 , r i +5 , u e ] and f (cid:48) is in r i − [ v i , r i , u ] . Since f and f (cid:48) are not R -separated in G − e w and, therefore, not R -separatedin G , f (cid:48) cannot be red (Lemma 11.7). If f (cid:48) is H -yellow in G , then Lemma 12.4shows it is not crossed in D . Therefore, Theorem 11.3 implies f (cid:48) is H -green in G . Lemma 12.5 says there is a 2.5-jump J spanning f (cid:48) so that f (cid:48) is in the partial H -rim branch spanned by J . As J cannot span e ( e is not H -green), Lemma 7.2(3b) and our current context ( f in r i +4 r i +5 and f (cid:48) in r i − r i ) implies this is possibleonly if f (cid:48) ∈ r i − and f ∈ r i +5 . However, the red edges f and e are R -separatedin G , implying that G − e w still has five spokes (we may replace s i +1 with the r i r i +5 subpath of P u ). Thus, f (cid:48) is H (cid:48) -green in G − e w , for some H (cid:48) ∼ = V . This isimpossible, as f (cid:48) is crossed in D (Lemma 6.6 (10)). Subcase 2: f ∈ r i +4 [ v i +5 , r i +5 , u e ] and f (cid:48) ∈ [ u, r i , v i +1 ] r i +1 r i +2 . 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 105 In this subcase, f is R -separated in G from e . The witnessing subdivision H (cid:48) of V can be chosen to contain the “nearer” ( r i − r i )( r i +4 r i +5 )-paths, one from eachof ∆ f and ∆ e , together with the H -spokes s i +2 and s i +3 to construct H (cid:48) .We claim that this H (cid:48) also shows that f is R -separated from f (cid:48) in G − e w . If f ∈ r i +4 , then, since Q i +1 is crossed in D , f (cid:48) ∈ r i r i +1 . In this case, H (cid:48) containsthe spokes s i +2 and s i +3 , so indeed f and f (cid:48) are in disjoint H (cid:48) -quads, as required.If f ∈ r i +5 , then f (cid:48) ∈ r i +2 by Lemma 7.2 (3b), and again f and f (cid:48) are in disjoint H (cid:48) -quads, showing f and f (cid:48) are R -separated in G − e w . Observation 11.6 (1) yieldsthe contradiction that f and f (cid:48) do not cross each other in D . Subcase 3: f ∈ [ w e , r i +5 , v i +6 ] r i +6 r i +7 and f (cid:48) ∈ r i − [ v i , r i , u ] . If f is R -separated from e in G − e w , then it cannot cross f (cid:48) in D , a contradiction.Otherwise, Lemma 13.3 implies there is a w -consecutive red edge for e . Subcase 4: f ∈ [ w e , r i +5 , v i +6 ] r i +6 r i +7 and f (cid:48) ∈ [ u, r i , v i +1 ] r i +1 r i +2 .If f (cid:48) = e , then we are done: Lemma 13.3 implies e has a w -consecutive edge.So we assume f (cid:48) (cid:54) = e . If f (cid:48) is red in G , then Lemma 11.7 implies it is R -separatedfrom f in G . Therefore, f (cid:48) is R -separated from f in G − e w , a contradiction; so f (cid:48) is not red in G .Suppose by way of contradiction that f (cid:48) is H -yellow, with witnessing H -yellowand H -green cycles C and C (cid:48) , respectively. If e w is not in C , then Lemma 12.4yields the contradiction that f (cid:48) is not crossed in D .If e w is in C , then let P be the RR -subpath of C containing e w , let P (cid:48) be the RR -subpath of ∆ e − e that contains e w , and let J be the global H -bridge containedin C (cid:48) . The end of P (cid:48) in r i +5 cannot be in the interior of the span of J , as then eitherthe peak of ∆ e is a vertex, in which case we have that ∆ e is H -yellow, yielding thecontradiction that e is H -yellow, or the peak of ∆ e consists of parallel edges, bothin the span of J , contradicting Theorem 6.7.It follows that P (cid:48) has its end in r i +5 , but not in the interior of the span of J .On the other hand, P has, by Definition 11.1, one end in the interior of the span of J . But now ( P ∪ P (cid:48) ) − e w contains an R -avoiding subpath that intersects at mostthe one spoke s i +1 . Therefore, this subpath is in an H -green cycle and contains anedge spanned by J , contradicting Theorem 6.7. It follows that f (cid:48) is H -green.Theorem 12.6 implies that H has no 3-jumps. If f (cid:48) is H -green by a 2.5-jump J , then, because J cannot span e , Lemma 7.2 (3b) implies f ∈ [ w e , r i +5 , v i +6 ] r i +6 and f (cid:48) ∈ r i +2 . Let x be the end of f closest to w e in r i +5 r i +6 . Let H (cid:48) be thesubdivision of V obtained from H − (cid:104) s i +1 (cid:105) by replacing s i +2 with P x (recall thisis defined in Theorem 12.1 (3)). Now f and f (cid:48) violate Lemma 7.2 (3b) relative to H (cid:48) . Therefore, f (cid:48) is not H -green by a 2.5-jump.Lemma 7.2 implies f (cid:48) is not H -green by a 2-jump, as then it is not crossed in D . Thus, f (cid:48) is H -green by a local H -green cycle C . Lemma 12.5 implies e w is in C .Since f cannot be R -separated from f (cid:48) in G − e w , we see that f is not R -separatedfrom e in G − e w . Now Lemma 13.3 implies there is a w -consecutive red edge for e , concluding the proof for Case 1. Case 2: e w not incident with w . By Theorem 12.1 (3), w is incident with a global H -bridge J w . Since w is notincident with e w , w (cid:54) = v i +1 , and therefore J w is the 2.5-jump wv i +3 . 06 13. THE NEXT RED EDGE AND THE TILE STRUCTURE We observe that, since e w is not incident with w , its incident vertex in r i isin the interior of the span of J w . Moreover, e w is the first edge of an R -avoiding r i r i +5 -path P in ∆ e − e , which, together with a subpath of r i r i +1 , s i +2 , and asubpath of r i +5 r i +6 makes an H -yellow cycle C . By Lemma 11.2 (3), there is onlyone C -bridge in G and, therefore, P = s i +1 . In particular, e w ∈ s i +1 . Claim . No edge in r i +7 r i +8 is H -yellow. Proof. Suppose some edge e (cid:48) in r i +7 r i +8 is H -yellow. Let C and C (cid:48) be thewitnessing H -yellow and H -green cycles, respectively. By Lemma 11.2 (1), C (cid:48) contains a global H -bridge J (cid:48) .In the case e (cid:48) is in r i +7 , the span of J (cid:48) contains a vertex of r i +2 in its interior.Theorem 6.7 implies J (cid:48) = J w . But now C ∪ Q i +1 contains an H -yellow cycle C (cid:48)(cid:48) for which there is a C (cid:48)(cid:48) -interior C (cid:48)(cid:48) -bridge containing an edge of s i +2 , contradictingLemma 11.2 (3). Therefore, no edge in r i +7 is H -yellow.Now we suppose e (cid:48) is in r i +8 . Lemma 10.9 (1) shows J (cid:48) does not have v i +3 asan end, so J (cid:48) has one end x ∈ (cid:104) r i +3 (cid:105) and its other end is v i +6 . But now C ∪ Q i +4 contains an H -yellow cycle C (cid:48)(cid:48) having a C (cid:48)(cid:48) -interior C (cid:48)(cid:48) -bridge containing an edgeof s i +4 , contradicting Lemma 11.2 (3). (cid:3) Claim . Some edge of r i +7 is red. Proof. Suppose no edge of r i +7 is red. By Theorem 11.3 and Claim 2, everyedge in r i +7 is H -green. Subclaim . If there is a red edge in either r i +3 r i +4 or r i +8 r i +9 , then thereis a red edge in r i +8 r i +9 . Furthermore, among all such red edges, the one e (cid:48)(cid:48) withan end x (cid:48)(cid:48) nearest v i +8 in r i +8 r i +9 is such that ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) is not incident with x (cid:48)(cid:48) (thatis, Case 1 does not apply to e (cid:48)(cid:48) and x (cid:48)(cid:48) ). Proof. We first suppose no edge of r i +3 r i +4 is red. Then there is a red edgein r i +8 r i +9 . For any such red edge e (cid:48)(cid:48) , if the end x (cid:48)(cid:48) of e (cid:48)(cid:48) nearest to v i +8 isincident with ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) , then Case 1 shows there is an x (cid:48)(cid:48) -consecutive red edge ˆ e for e (cid:48)(cid:48) . By Definition 13.1 (1), ˆ e ∈ r i +1 r i +2 r i +3 r i +4 . Since the edges in r i +1 r i +2 are H -green, ˆ e / ∈ r i +1 r i +2 . But then ˆ e is a red edge in r i +3 r i +4 , a contradiction.Therefore, x (cid:48)(cid:48) is not incident with ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) , as required.The alternative is that there is a red edge in r i +3 r i +4 . Among all such edges,let e (cid:48) be the one having an incident vertex x (cid:48) nearest v i +3 in r i +3 r i +4 . Because ofTheorem 6.7 and J w , x (cid:48) is not incident with a 2.5-jump x (cid:48) v i +1 or x (cid:48) v i +2 . Therefore, x (cid:48) is incident with ( e (cid:48) ) x (cid:48) , and we conclude from Case 1 that there is an x (cid:48) -consecutivered edge e (cid:48)(cid:48) for e (cid:48) . Because of J w , every edge in r i +6 is either H -yellow or H -greenand so, in particular, is not red. By assumption, no edge of r i +7 is red. By Definition13.1 (1), e (cid:48)(cid:48) ∈ r i +8 r i +9 . Also, ∆ e (cid:48)(cid:48) separates s i +3 from ∆ e (cid:48) in cl( Q i +3 ) ∪ cl( Q i +4 ).Let x (cid:48)(cid:48) be the end of e (cid:48)(cid:48) nearest v i +8 in r i +8 r i +9 . By way of contradiction,suppose x (cid:48)(cid:48) is incident with ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) . Then Case 1 shows there is an x (cid:48)(cid:48) -consecutivered edge ˆ e for e (cid:48)(cid:48) . But ˆ e is not in r i +1 r i +2 because J w makes every one of thoseedges H -green. Therefore, ˆ e is in r i +3 r i +4 . Since ∆ ˆ e separates s i +3 from ∆ e (cid:48)(cid:48) incl( Q i +3 ) ∪ cl( Q i +4 ), we see that ˆ e is nearer to v i +3 than e (cid:48) is, contradicting thechoice of e (cid:48) . Therefore x (cid:48)(cid:48) is not incident with ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) , as required. (cid:3) Subclaim . No edge in either r i +3 r i +4 or r i +8 r i +9 is red. 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 107 Proof. Suppose by way of contradiction that there is a red edge in either r i +3 r i +4 or r i +8 r i +9 . By Subclaim 1, there is a red edge e (cid:48)(cid:48) in r i +8 r i +9 so thatthe end x (cid:48)(cid:48) of e (cid:48)(cid:48) nearest v i +8 in r i +8 r i +9 is not incident with ( e (cid:48)(cid:48) ) x (cid:48)(cid:48) . Therefore,Theorem 12.1 (3) implies x (cid:48)(cid:48) is incident with a 2.5-jump that is either x (cid:48)(cid:48) v i +6 or x (cid:48)(cid:48) v i +7 . It cannot be the former, as the 2.5-jumps x (cid:48)(cid:48) v i +6 and J w contradict Lemma10.9 (4). Therefore, x (cid:48)(cid:48) is in the interior of r i +9 and the 2.5-jump is x (cid:48)(cid:48) v i +7 . Thecontradiction is obtained by showing that cr( G ) ≤ D be a 1-drawing of G − (cid:104) r i +7 (cid:105) . There is still a subdivision H (cid:48) of V in G − (cid:104) r i +7 (cid:105) consisting of the rim ( R − (cid:104) r i +7 (cid:105) ) ∪ x (cid:48)(cid:48) v i +7 and the four spokes s i , s i +1 , s i +2 and s i +3 r i +8 [ v i +9 , r i +9 , x (cid:48)(cid:48) ]. We note that x (cid:48)(cid:48) v i +7 is an H (cid:48) -rim branch,contained in an H (cid:48) -quad Q consisting of s i +2 , r i +2 , s i +3 r i +8 [ v i +9 , r i +9 , x (cid:48)(cid:48) ], and x (cid:48)(cid:48) v i +7 .We aim to show D [ Q ] is clean, so by way of contradiction, we assume D [ Q ] isnot clean. The H (cid:48) -rim branches of Q are r i +2 and x (cid:48)(cid:48) v i +7 . Since r i +1 r i +2 is notcrossed in D (Lemma 7.2 (3a)), we deduce that x (cid:48)(cid:48) v i +7 is crossed in D . Furthermore,the cycle r i +3 s i +4 r i +8 s i +3 (which is Q i +3 in G ) is H (cid:48) -close and, therefore Lemmas5.3 and 5.4 imply Q i +3 is not crossed in D . It follows that x (cid:48)(cid:48) v i +7 crosses r i +4 in D , so s i +3 r i +8 [ v i +9 , r i +9 , x (cid:48)(cid:48) ] is exposed in D , from which D [ H (cid:48) ] is completelydetermined. (See Figure 13.1.) v i +7 v i +6 v i +1 v i +5 x (cid:48)(cid:48) ev i v i +2 e (cid:48)(cid:48) v i +4 v i +3 v i +8 v i +9 y (cid:48)(cid:48) Figure 13.1. D [ H (cid:48) ]Our contradiction is obtained from a detailed consideration of ∆ e (cid:48)(cid:48) . We firstshow that v i +4 is in the peak of ∆ e (cid:48)(cid:48) . To see this, we note that the r i +9 r i +4 -subpathof ∆ e (cid:48)(cid:48) − e (cid:48)(cid:48) that starts nearest x (cid:48)(cid:48) is simply s i +4 , as otherwise there is an H -yellowcycle C with more than one C -bridge. Theorem 12.1 (3) implies the subpath of∆ e (cid:48)(cid:48) − e (cid:48)(cid:48) from x (cid:48)(cid:48) to the peak of ∆ e (cid:48)(cid:48) has at most one edge in R ; therefore, there isno edge of r i +4 between v i +4 and the peak of ∆ e (cid:48)(cid:48) . That is, v i +4 is in the peak of∆ e (cid:48)(cid:48) .Let y (cid:48)(cid:48) be the end of e (cid:48)(cid:48) different from x (cid:48)(cid:48) . Because y (cid:48)(cid:48) is too close to J w , it isnot incident with a global H -bridge. Thus, the edge of ∆ e (cid:48)(cid:48) − e (cid:48)(cid:48) incident with y (cid:48)(cid:48) is not in R and, therefore, is the first edge of an r i +9 r i +4 -subpath P of ∆ e (cid:48)(cid:48) − e (cid:48)(cid:48) .Let z (cid:48)(cid:48) be the other end of P . 08 13. THE NEXT RED EDGE AND THE TILE STRUCTURE We note that z (cid:48)(cid:48) (cid:54) = v i +4 , as D [ P ] cannot cross D [ H (cid:48) ]. Therefore, z (cid:48)(cid:48) ∈ (cid:104) v i +4 , r i +4 ,v i +5 ]. If z (cid:48)(cid:48) is in the peak of ∆ e (cid:48)(cid:48) , then z (cid:48)(cid:48) and v i +4 are joined by parallel edges,one of which is not in H (cid:48) . That one must cross D [ H (cid:48) ], which is a contradiction.Therefore, z (cid:48)(cid:48) is not in the peak of ∆ e (cid:48)(cid:48) . But now Theorem 12.1 (3) implies z (cid:48)(cid:48) isin the interior of the span of a global H -bridge J (cid:48)(cid:48) that has an end in the peak of∆ e (cid:48)(cid:48) ; therefore, this end of J (cid:48)(cid:48) is in r i +4 .The end of J (cid:48)(cid:48) in r i +4 must be v i +4 , as otherwise J (cid:48)(cid:48) is a 2.5-jump with one endbeing v i +7 , which, together with x (cid:48)(cid:48) v i +7 , contradicts Lemma 10.9 (1). Therefore, J (cid:48)(cid:48) is either v i +4 v i +6 or v i +4 u (cid:48)(cid:48) , with u (cid:48)(cid:48) ∈ (cid:104) r i +6 (cid:105) . However, Lemma 7.2 (1) or (3a)and J (cid:48)(cid:48) show that r i +4 cannot be crossed in D , a contradiction that finally shows D [ Q ] is clean.We can now obtain the claimed 1-drawing of G . Observe that x (cid:48)(cid:48) v i +7 is in an H -green cycle that, by Lemma 6.6 (8), has only one bridge. Also, if there is a Q i +2 -bridge other than M Q i +2 , then cl( Q i +2 ) has an edge f not in Q i +2 . But Theorem5.23 and Lemma 5.9 imply Q i +2 would be crossed in any 1-drawing of G − f ;however, both r i +2 and r i +7 are H -green courtesy of J w and x (cid:48)(cid:48) v i +7 . Therefore, Q i +2 has only one bridge. It follows that there are only two Q -bridges in G , one ofwhich is r i +7 . Since D [ Q ] is clean, it bounds a face of D [ G − (cid:104) r i +7 (cid:105) ] and it is easyto put r i +7 into this face so as to obtain a 1-drawing of G . That is, cr( G ) ≤ 1, acontradiction completing the proof of the subclaim. (cid:3) We are now in a position to finish the proof of Claim 3. Let e be the edgeof s i +3 incident with v i +3 and let D be a 1-drawing of G − e . Corollary 12.7and Lemma 5.9 imply Q is crossed in D . It follows that there is an edge ˆ e in r i +6 r i +7 r i +8 r i +9 that is crossed in D .The H -yellow cycle Q i +1 contains r i +6 , so Lemma 12.4 implies r i +6 is notcrossed in D . By assumption for r i +7 and by Subclaim 2 for r i +8 r i +9 , no edgeof r i +7 r i +8 r i +9 is red. Lemmas 12.4 and 12.5 imply that ˆ e is spanned by some2.5-jump J (cid:48) , and, moreover, ˆ e is in the H -rim branch whose interior contains theend x (cid:48) of J (cid:48) .If ˆ e ∈ r i +7 , then J (cid:48) is either x (cid:48) v i +5 or x (cid:48) v i . Suppose first that J (cid:48) = x (cid:48) v i +5 .Lemma 7.2 (3b) implies ˆ e crosses an edge in r i +4 . But Theorem 6.7 shows r i +4 cannot be in the span of a 2.5-jump, so Lemmas 12.4 and 12.5 imply no edge of r i +4 is crossed in D . Thus, J (cid:48) (cid:54) = x (cid:48) v i +5 .Now we suppose J (cid:48) = x (cid:48) v i . In this case, Lemma 7.2 (3b) implies ˆ e crosses anedge in r i +1 , while (1) of the same lemma implies no edge in the span of J , whichincludes r i +1 , is crossed in D . We conclude that ˆ e / ∈ r i +7 .If ˆ e ∈ r i +8 , then J (cid:48) is either x (cid:48) v i +6 or x (cid:48) v i +1 . Theorem 6.7 shows the latterdoes not happen. Lemma 10.9 (4) shows the former does not happen. Therefore,ˆ e / ∈ r i +8 .The last possibility is that ˆ e ∈ r i +9 . In this instance, J (cid:48) is either x (cid:48) v i +7 or x (cid:48) v i +2 . Theorem 6.7 precludes the latter possibility, so we assume J (cid:48) = x (cid:48) v i +7 .However, in this case, Lemma 7.2 (3b) implies ˆ e crosses an edge ˜ e in r i +5 , in whichcase neither ˆ e nor ˜ e is in Q i +3 , contradicting the fact that Q i +3 is crossed in D . (cid:3) We now finish the proof of Case 2 and, therefore, Theorem 13.2. By Claim 3,we may let e (cid:48) = xy be the red edge in r i +7 that is nearest v i +7 in r i +7 , labelledso that x is nearer v i +7 in r i +7 than y is. We look for the x -consecutive red edge 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 109 for e (cid:48) . As the edges spanned by J w are H -green, e is the only possibility for the x -consecutive red edge for e (cid:48) .Suppose first that e (cid:48) and x satisfy the condition for Case 1. We have provedthere is an x -consecutive red edge for e (cid:48) and, as just mentioned, this can only be e . This implies that x = v i +7 . To see that e (cid:48) is the w -consecutive red edge for e , it remains to show that e and e (cid:48) can be crossed in G − e w . (This is the onlyasymmetric condition in the definition of consecutive.)The H -quad Q i +1 is also an H -yellow cycle and so (Lemma 11.2 (3)) boundsa face of G . It follows that e and e (cid:48) are not R -separated in G − e w and, thereforeLemma 11.7 implies there is a 1-drawing of G − e w in which e and e (cid:48) are crossed,as required.The alternative is that e (cid:48) and x do not satisfy the condition for Case 1. Then,just as for w above, there is a 2.5-jump J x = xv i +5 incident with x . Also, the edge e x of ∆ e (cid:48) − e (cid:48) that is nearest x and not in R is in s i +7 . Since Q i +1 bounds a faceof G , e and e (cid:48) are not R -separated in G − e w and, therefore, Lemma 11.7 impliesthere is a 1-drawing of G − e w in which they are crossed.The following is a consequence of Definition 13.1 and Theorem 13.2. Lemma . Let G ∈ M and V ∼ = H ⊆ G , with H tidy. With the la-belling of e = uw and e w as in Definition 13.1, if x is the end of e w nearest w e in [ w e , r i +5 , v i +6 ] r i +6 r i +7 , then e is x -consecutive for e w . Proof. By Theorem 13.2, there is an x -consecutive red edge e (cid:48)(cid:48) for e w . Conditions(2) and (3) of Definition 13.1 applied to e w being w -consecutive for e and the sameconditions applied to e (cid:48)(cid:48) being x -consecutive for e w imply that e = e (cid:48)(cid:48) .The main goal of this work is to prove Theorem 2.14. The following lemmawill be very helpful. Lemma . Let G ∈ M , V ∼ = H ⊆ G , and let Π be an embedding of G in R P so that H is Π -tidy. Let C be a contractible cycle contained in M so that C is the union of a 3-rim path C ∩ R (recall Definition 11.1 (1)) and an R -avoidingpath P . Then, for every edge e of C ∩ R , there is an H -green cycle containing e and contained in H ∪ P . Proof. The graph H ∪ P is 2-connected and not planar, so every face of Π[ H ∪ P ]is bounded by a cycle. There is a face F of H ∪ P contained in M and incidentwith e ; by the preceding remark, F is bounded by a cycle C (cid:48) .Let j be the index so that e ∈ r j ; thus, F is Q j -interior. Since F is also C -interior, C (cid:48) ∩ H ⊆ (cid:104) s j r j s j +1 (cid:105) . In particular, there is at least one edge of C (cid:48) thatis in P but not in H .Observe that (cid:104) s j r j s j +1 (cid:105)− e has two components K and K . Since C (cid:48) containsa vertex in each of K and K (namely the ends of e ), C (cid:48) contains an (cid:104) s j r j s j +1 (cid:105) -avoiding K K -path P (cid:48) . Thus, P (cid:48) ⊆ P .Let C (cid:48)(cid:48) be the cycle in (cid:104) s j r j s j +1 (cid:105) ∪ P (cid:48) . Then C (cid:48)(cid:48) is evidently an H -green cyclecontaining e , as required.Now for the main result. Theorem 2.14 If G is a 3-connected, 2-crossing-critical graph containing a subdi-vision of V , then G ∈ T ( S ) . 10 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Proof. By Theorem 10.4, G contains a tidy subdivision H of V ; let Π bean embedding of G in R P so that H is Π-tidy. The strategy is to show that,between every red edge e = uw and its w -consecutive red edge e w , there is one ofthe thirteen pictures (as defined just before Lemma 2.11). This is accomplishedby showing that e produces “one side” of the picture and e w produces the other.Let i ∈ { , , , . . . , } be such that e ∈ r i ; we choose the labelling so that r i =[ v i , r i , u, e, w, r i , v i +1 ]. Thus, e w ∈ r i +5 r i +6 r i +7 .Let x be the end of e w so that e is the x -consecutive red edge for e w . Let P be the w e x -subpath of R that is a 3-rim path (Definition 11.1 (1)); likewise P isthe x e w w -subpath of R that is a 3-rim path. Claim . Let B be a global H -bridge spanning an edge of P . Then:(a) B has ends w e and x ;(b) w e = v i +5 ; and(c) e w ∈ s i +1 and ( e w ) x ∈ s i +2 .The analogous claims holds for P . Proof. We remark that the span of B does not include in its interior a peakvertex of ∆ e , and does not include e w . Therefore, B has both its attachments in P . Consequently, the attachments of B are contained in r i +5 r i +6 [ v i +7 , r i +7 , v i +8 (cid:105) .Theorem 10.6 implies one end of B is v i +5 and the other end is in [ v i +7 , r i +7 , v i +8 (cid:105) .It follows that w e = v i +5 . At the other end, we claim x is in B . We note that e w is in r i +7 , so that H − (cid:104) s i +4 (cid:105) shows that e and e w are R -separated. Let x (cid:48) bethe end of B in r i +7 .If ( e w ) x is not in s i +2 , then let e i +7 be the edge of s i +2 incident with v i +7 andlet D be a 1-drawing of G − e i +7 . Corollary 12.7 and Lemma 5.9 imply Q i +2 iscrossed in D . The presence of J and B combine with Lemma 7.2 (1) to show thatneither [ w, r i , v i +1 ] r i +1 r i +2 nor r i +6 [ v i +7 , r i +7 , x (cid:48) ], respectively, is crossed in D .It follows that some edge e (cid:48) in [ x (cid:48) , r i +7 , v i +8 ] is crossed in D . Let P w be thepath in ∆ e described in Theorem 12.1 (3). Since P w does not have v i +1 as one end,and its other end is v i +5 , its only intersection with s i +1 can be in (cid:104) s i +1 (cid:105) . Such anintersection produces an H -green cycle that shows the edge of r i incident with v i +1 is in two H -green cycles, contradicting Theorem 6.7. Therefore, P w is disjoint from s i +1 .Using P w , s i +1 , s i +3 and s i +4 as spokes and R as the rim yields a V that shows r i +7 is R -separated in G − e i +7 from [ v i , r i , w ]; thus, Observation 11.6 (1) shows e (cid:48) crosses an edge e (cid:48)(cid:48) of r i +3 in D . Lemmas 12.4 and 12.5 shows e (cid:48) and e (cid:48)(cid:48) are red in G .Lemma 11.7 shows that e (cid:48) and e (cid:48)(cid:48) are R -separated in G . Lemma 12.13 shows thata witnessing V can be chosen to avoid e i +7 . But now D contradicts Observation11.6 (1). Therefore, ( e w ) x is in s i +2 and incident with v i +7 .If B has an end in (cid:104) r i +7 (cid:105) , then Theorem 12.1 (3) implies P x ∩ r i +7 has just oneedge, namely xv i +7 and, consequently, x is in B .If, on the other hand, v i +7 is an end of B , then Theorem 12.1 (3) implies x must be incident with e x and, therefore x = v i +7 . Again, we see that x is in B .Observe that J w and B are now seen to be completely symmetric with respectto ( e, w ) and ( x, e w ); in particular, we conclude that e w ∈ s i +1 . (cid:3) 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 111 If there is a global H -bridge B spanning an edge of P , then we let P (cid:48) = B .Otherwise, we let P (cid:48) = P . A completely analogous discussion holds for P to yieldthe wx w e -path P (cid:48) .Our next claim identifies the cycle that is the boundary of our picture. Claim . The closed walk P w P (cid:48) P x P (cid:48) is a cycle. Proof. If the edge of P w incident with w is in R , then Theorem 12.1 (3) showsthat w is incident with a global H -bridge B . Claim 1 implies B = P (cid:48) . Thus, w hasdegree 2 in the closed walk P w P (cid:48) P x P (cid:48) . Otherwise, P (cid:48) = P , w is incident with e w ,and again w has degree 2 in P w P (cid:48) P x P (cid:48) .The other “corners” w e , x , and x e w are treated similarly. (cid:3) Definition . Let e and e (cid:48) be red edges and let w and x be the ends of e and e (cid:48) , respectively, so that e (cid:48) is the w -consecutive red edge for e and e is the x -consecutive red edge for e (cid:48) . Let P be the x ∆ e -path in R that is a 3-rim pathand let P be the w ∆ e (cid:48) -path in R that is a 3-rim path. Let P w be the ww e -pathin ∆ e − e and let P x be the xx e (cid:48) -path in ∆ e (cid:48) − e (cid:48) . For i = 1 , 2, let P (cid:48) i be P i unlessthere is a global H -bridge B i spanning an edge of P i , in which case P (cid:48) i = B i .The cycle C e is the composition P w P (cid:48) P x P (cid:48) .We will see that C e is the outer boundary of the one of the thirteen pictures thatoccurs. We observe that C e is in the boundary of the closed disc in R P consistingof the union of the closed discs bounded by r i r i +1 r i +2 s i +3 r i +7 r i +6 r i +5 s i , P (cid:48) P (if P (cid:48) (cid:54) = P ), and P (cid:48) P (if P (cid:48) (cid:54) = P ). Therefore, C e is the boundary of a closed disc D e in R P .We now prove three claims that will be useful for finding the various parts ofthe picture. Claim . Let C be a cycle contained in D e . If either C ∩ P (cid:48) or C ∩ P (cid:48) is empty,then C bounds a face of Π[ G ]. Proof. By symmetry, we may suppose C ∩ P (cid:48) is empty. Let M be the C -bridgecontaining s i +4 . Subclaim . If B is a C -bridge different from M , then Π[ C ∪ B ] is contractiblein R P . Proof. We start by noting that Π[ B ] ⊆ M , since P (cid:48) is either just an edgethat is a global H -bridge (and so in D and forcing B to be in M ) or P (cid:48) = P and there is no global H -bridge having an attachment in (cid:104) P (cid:105) . In the latter case,any global H -bridge having an attachment at an end of P (say w ), has its otherattachment in the H -rim R −(cid:104) P (cid:105) . Such an attachment is in Nuc( M ), contradictingthe assumption that B (cid:54) = M .It follows that Π[ C ∪ B ] is contained in M and totally disjoint from s i +4 .Therefore, Π[ C ∪ B ] is contractible, as claimed. (cid:3) Let H (cid:48) be the subgraph of H ∪ P (cid:48) ∪ P (cid:48) consisting of ( R − ( (cid:104) P (cid:105)∪(cid:104) P (cid:105) )) ∪ ( P (cid:48) ∪ P (cid:48) )and the three H -spokes s i +3 , s i +4 , and s i . The following claim shows that H (cid:48) is asubdivision of V . (The notation (cid:107) y (cid:107) is in Definition 4.1 (1).) Subclaim . C e ∩ s i ⊆ (cid:107) v i +5 (cid:107) and C e ∩ s i +3 ⊆ (cid:107) v i +3 (cid:107) . 12 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Proof. Recall that P w is contained in ∆ e . Theorem 12.1 (the existence of A u and A w , together with (3)) implies P w is internally disjoint from P u and, therefore,cannot intersect s i , except possibly at their common end point v i +5 . The analogousargument using ∆ e w applies for s i +3 . (cid:3) If C does not bound a face of Π[ G ], then let e (cid:48) be any edge of any C -interior C -bridge and let D be a 1-drawing of G − e . Subclaim 2 implies that C is H (cid:48) -close(Definition 5.2). Lemmas 5.3 and 5.4 imply C is clean in D . Therefore, D containsa 1-drawing of C ∪ M in which C is clean and Lemma 5.6 implies C has BOD. Itnow follows from Corollary 4.7 that cr( G ) ≤ 1, the final contradiction. (cid:3) We find structures in the C e -interior that lead to the pictures. Our discussionwill be w -centric; there is a completely analogous discussion for x .A useful observation is the following. Recall that P w is the ww e -path in ∆ e − e (Theorem 12.1 (3)) and P x is the analogous xx e w -path in ∆ e w . Claim . (1) No C e -interior C e -bridge has an attachment in each of thecomponents of ( C e − P x ) − e w .(2) No C e -interior C e -bridge has an attachment in each of the components of( C e − P w ) − e x . Proof. Let H (cid:48) be a subdivision of V witnessing the R -separation of e and e w . As e and e w are R -separated in neither G − e w nor G − e x , e w and e x are bothin H (cid:48) . Since e and e w are in disjoint H (cid:48) -quads, e w and e x are in disjoint H (cid:48) -spokes,which we denote as P w and P x , respectively; P w and P x are contained in the closeddisc bounded by Π[ C e ]. Subclaim . There is such an H (cid:48) so that P x = P x . Proof. As a first case, suppose C e ∩ s i = ∅ . Then we may choose H (cid:48) to be R , s i , s i +4 , P w , and P x , and we are done. In the second case, C e ∩ s i +3 = ∅ ; replace s i with s i +3 .In the final case, C e ∩ s i and C e ∩ s i +3 are not empty. In this instance, e w ∈ r i +7 .We may choose H (cid:48) to consist of R , s i +4 , s i , s i +1 , and P x , the latter being containedin cl( Q i +2 ). (cid:3) By symmetry, it suffices to prove (1). Suppose by way of contradiction thatthere is a C e -interior C e -bridge B having an attachment in each component of( C e − P x ) − e w . Subclaim 1 implies there is a subdivision H (cid:48) witnessing the R -separation of e and e w so that P x ⊆ H (cid:48) . Let P w be the other H (cid:48) -spoke containedin the interior of C e .Let C (cid:48) be the cycle bounding the C e -interior face of C e ∪ P w that is incidentwith e w . The C e -bridge B contains a subpath P (cid:48) joining the two components of( C (cid:48) − P x ) − e w . Now (( C (cid:48) − P x ) − e w ) ∪ P (cid:48) contains an R -avoiding path P (cid:48)(cid:48) that canreplace P w in H (cid:48) to get another subdivision of V that witnesses the R -separationof e and e w in G − e w . However, this contradicts the fact that e and e w are not R -separated in G − e w . (cid:3) Here is our final preliminary claim. Claim . Let B be a C e -interior C e -bridge. Then B is just an edge and itsends. 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 113 Proof. Suppose to the contrary that B is a C e -interior C e -bridge with at leastthree attachments. Subclaim . B has at most two attachments in each of C e − P (cid:48) and C e − P (cid:48) . Proof. By symmetry, it suffices to prove the first of these. Suppose B has atleast two attachments in C e − P (cid:48) . Let y and z be the ones nearest the two ends of C e − P (cid:48) . There is a cycle in B ∪ C e consisting of a C e -avoiding yz -path in B andthe yz -subpath of C e − P (cid:48) . Claim 3 implies this cycle bounds a face of Π[ G ] and,therefore, B can have no other attachment in C e − P (cid:48) . (cid:3) Subclaim . att( B ) ∩ P (cid:48) ⊆ { x, w e } and att( B ) ∩ P (cid:48) ⊆ { w, x e w } . Proof. By symmetry, it suffices to prove the first of these. By way of con-tradiction, suppose B has an attachment y in (cid:104) P (cid:48) (cid:105) . Because B has at least threeattachments, Subclaim 1 implies B has an attachment z in P (cid:48) . Any C e -avoiding yz -path in B contradicts Claim 4. (cid:3) From these two subclaims, we easily deduce that: • B has at most four attachments; • one of w and x e w is an attachment of B ; and • one of x and w e is an attachment of B .Observe that Claim 4 (1) implies that not both w and w e are attachments of B , while (2) implies that not both x and x e w are attachments of B . Therefore,att( B ) ∩ ( P (cid:48) ∪ P (cid:48) ) is either { w, x } or { w e , x e w } . Subclaim . att( B ) ∩ ( P (cid:48) ∪ P (cid:48) ) = { w e , x e w } . Proof. Suppose by way of contradiction that att( B ) ∩ ( P (cid:48) ∪ P (cid:48) ) = { w, x } .As B has at least three attachments, there is an attachment y in (cid:104) P w (cid:105) ∪ (cid:104) P x (cid:105) . Bysymmetry, we may assume y ∈ (cid:104) P w (cid:105) . Let P yw be a C e -avoiding yw -path in B .Then the union of P yw and the yw -subpath of P w is a cycle C yw in D e .Since y and w are in P w − w e , C yw is disjoint from P (cid:48) . Claim 3 implies C yw bounds a face of Π[ G ]. On the other hand, P w is contained in the boundary of theface bounded by ∆ e and, therefore, C yw ∩ P w is in the boundary of two faces ofΠ[ G ]. We deduce that C yw ∩ P w is just the edge wy .Furthermore, Claim 4 implies w and y are in the same component of P w − e w .Therefore, the definition of e w implies wy is in R , and consequently P (cid:48) is a global H -bridge spanning wy . However, any edge of B incident with w — and there is atleast one such — must be in the interior of the face of Π[ G ] bounded by the H -greencycle containing P (cid:48) (Lemma 6.6 (8)). This contradiction proves the subclaim. (cid:3) We are now ready to complete the proof of the claim. Any vertex in att( B ) \{ w e , x e w } is in (cid:104) P w (cid:105) ∪ (cid:104) P x (cid:105) . Subclaim 1 implies there is at most one of these. Since B has at least three attachments, there is at least one of these. We conclude thereis exactly one such attachment y . We may choose the labelling so that y ∈ (cid:104) P w (cid:105) .Lemma 5.19 implies B is isomorphic to K , .The vertex y is in the interior of P w . Thus, both edges of P w incident with y are in the boundary of the face bounded by Π[∆ e ]. Consequently, any edge of G incident with y is in D e .Let c be the vertex of degree 3 in B . Claim 3 implies that the cycles [ y, c, w e , y ]and [ y, c, x e w , P (cid:48) , w, P w , y ] both bound faces in D e . Therefore, y has degree 3 in G . 14 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Let e (cid:48) be the edge cw e of B and let D (cid:48) be a 1-drawing of G − e (cid:48) . Consider thesubdivision H (cid:48) of V consisting of ( R − ( (cid:104) P (cid:105) ∪ (cid:104) P (cid:105) ) ∪ ( P (cid:48) ∪ P (cid:48) ), P x , s i , and s i +4 .Then H (cid:48) shows that P w ∪ ( B − e (cid:48) ) is not crossed in D (cid:48) .The path P (cid:48) = [ c, cy, y, P w , w e ] is not crossed in D (cid:48) . Since y has degree 3 in D (cid:48) ,we may add the edge w e c to D (cid:48) alongside P (cid:48) without crossing to obtain a 1-drawingof G . This is the final contradiction that shows B has only two attachments. Lemma5.19 shows B is just an edge and its ends. (cid:3) We now have our preliminary lemmas in hand and proceed to complete theproof of Theorem 2.14. Definition . Let C e be decomposed as P w P (cid:48) P x P (cid:48) as in Definition 13.6.(1) If f is an edge not in C e with ends w and x e w and P (cid:48) has length 1, then f is a w -chord .(2) If f is an edge not in C e joining w to a vertex y ∈ (cid:104) P x (cid:105) and the yx e w -subpath of P x has length 1, then f is a w -slope .(3) If f and f (cid:48) are edges not in C e , with f joining w with z ∈ (cid:104) P (cid:48) (cid:105) and f (cid:48) joining z to z (cid:48) ∈ (cid:104) P x (cid:105) , and if P (cid:48) has length 2, while the z (cid:48) x e w -subpath of P x has length 1, then { f, f (cid:48) } is a w -chord+ w -slope .(4) If f is an edge not in C e joining x e w to a vertex y in (cid:104) P w (cid:105) , and both P (cid:48) and the yw -subpath of P w have length 1, then f is a w -backslope .(5) If f is an edge not in C e joining y ∈ (cid:104) P w (cid:105) and z ∈ (cid:104) P x (cid:105) , and the paths P w and P x have length 2, while P (cid:48) and P (cid:48) have length 1, then f is a crossbar .The five situations in Definition 13.7 are illustrated in Figure 13.2. w x e w w e P (cid:48) P (cid:48) P w P x f (cid:48) fff f fw x e w w e P (cid:48) P (cid:48) P x P w w x e w w e P (cid:48) P (cid:48) w x e w w e P (cid:48) P (cid:48) P x zz (cid:48) yP w w x e w w e P (cid:48) P (cid:48) P w P x P w P x xx x x x Figure 13.2. Definition 13.7. Claim . If e w is in neither an H -yellow nor an H -green cycle, then everyedge of P is H -green. If C is the set of H -green cycles containing edges of P , then C e ∪ ( (cid:83) C ∈C ) C contains either:(a) C e plus a w -chord;(b) C e plus a w -slope; or(c) C e plus a w -chord+ w -slope. Proof. Because e w is not in an H -yellow cycle, Theorem 12.1 (3) implies w is incident with e w . Case 1: some edge of P is spanned by a global H -bridge. Let B be a global H -bridge spanning an edge of P . Claim 1 implies B hasends x e w and w , x e w = v i +3 , e w ∈ s i +1 , and e x ∈ s i +2 . Since w is incident with e w ,we have w = v i +1 . 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 115 We show (b) occurs by proving that r i +1 is a w -slope. We show that r i +1 and r i +2 are both paths of length 1, starting with the latter.We note that P x is equal to s i +2 r i +2 . Moreover, r i +2 has the face of Π[ G ]bounded by the H -green cycle C g containing B on one side and the face boundedby ∆ e w on the other. Thus, r i +2 is just a single edge.Claim 5 shows that the C e -bridge B (cid:48) containing r i +1 is just an edge and itsends. Thus, r i +1 is B (cid:48) and so has length 1, as required, completing the proof inCase 1. Case 2: no edge of P is spanned by a global H -bridge. In this case, P (cid:48) = P . We start by showing that every edge of P is H -green.Because e w is w -consecutive for e , Definition 13.1 implies no edge of P isred. By Theorem 11.3, we need only show that none is H -yellow. Suppose to thecontrary that there is an H -yellow edge f in P , as witnessed by the H -yellow cycle C y and the H -green cycle C g . Lemma 11.2 implies there is a global H -bridge B contained in C g .The face of Π[ G ] bounded by C y (Lemma 11.2 (3)) is in M . Now the faces ofΠ[ G ] bounded by ∆ e and ∆ e w separate M into two parts, one of which contains f , and therefore P . It follows that P is also in this part and C y has at least avertex in P . We conclude that B spans an edge of P . Claim 1 implies B = P (cid:48) , w e = v i +5 , e w ∈ s i +1 , and e x ∈ s i +2 . Because P (cid:48) = P , and e w ∈ s i +1 , we deducethat w = v i +1 .Since e w is not in an H -yellow cycle, we conclude that Q i +1 is not an H -yellowcycle. The other attachment of B , namely x , which is in [ v i +7 , r i +7 , v i +8 (cid:105) , musttherefore be v i +7 .If the H -yellow edge f is in r i +1 , then C y ∩ P is contained in the interior ofthe span of B . This implies that s i +1 is in an H -yellow cycle and, therefore, e w isin an H -yellow cycle, contrary to the hypothesis.We have noted that x = v i +7 is an end of B . Consequently, no edge of[ v i +2 , r i +2 , x e w ] can be H -yellow. That is, every edge of P is H -green.We now complete the proof in Case 2. Let C be the H -green cycle containingthe edge of P that is incident with w . Because e w is not in any H -green cycle, w isincident with an edge e (cid:48) in C that is not in C e . Let B be the C e -bridge containing e (cid:48) . Claim 5 implies B is just an edge with the two ends w and a second vertex z .The path C ∩ P is in the boundary of the face of Π[ G ] bounded by C (Lemma 6.6(8)). Also, there is no global H -bridge spanning an edge of P (we are in Case 2).These two facts imply C ∩ P is just an edge.Suppose first that z ∈ P x − x e w . Because C is H -green, it is disjoint from P .Thus, Claim 3 implies that the cycle C (cid:48) that is the union of the wz -subpath of P P x and B bounds a face of Π[ G ]. This face is contained in M , as is the face boundedby C . Both are incident with the edge of P incident with w and so they are thesame face. We conclude that C = C (cid:48) .Now C ∩ P x is in the boundary of a face inside the disc bounded by ∆ e (cid:48) onone side and the face bounded by C on the other. Because G is 3-connected, thissubpath has length 1. In this case, we have (b).The other possibility is that z is in P . We have already shown that w and z are the ends of a digon. If z = x e w , then we have (a). Therefore, we may suppose z (cid:54) = x e w . 16 13. THE NEXT RED EDGE AND THE TILE STRUCTURE Since G is 3-connected, z has a neighbour y distinct from its neighbours in P .Let B (cid:48) be the C e -bridge containing zy . Claim 5 implies B (cid:48) is just an edge joining z and y .The choice of y shows y (cid:54) = w . Claim 4 (1) and (2) imply, respectively, that y / ∈ (cid:104) P w P (cid:48) (cid:105) and that y (cid:54) = x . If y ∈ P , then (just as for w and z ) z and y arethe ends of a digon, so y is a neighbour of z in P , contradicting the choice of y .Therefore, y ∈ (cid:104) P x (cid:105) .Let C (cid:48) be the cycle consisting of zy and the zy -subpath of P P x . Claim 3implies C (cid:48) bounds a face of Π[ G ].To see that (c) holds, notice that C (cid:48) ∩ P x is in the boundary of the faces boundedby C (cid:48) and ∆ e w . Again, the 3-connection of G shows C (cid:48) ∩ P x is a path of length 1.Likewise, C (cid:48) ∩ P is in the boundary of the face bounded by C (cid:48) . There is no global H -bridge spanning any edge of P , so C (cid:48) ∩ P is also a path of length 1, completingthe proof that (c) occurs and the proof of Claim 6. (cid:3) It remains to consider the possibilities that e w is in either an H -yellow or an H -green cycle. We do the latter first. Claim . If e w is in an H -green cycle C , then either(d) C e ∪ C contains C e plus a back-slope or(e) C e ∪ C is C e plus a crossbar. Proof. Let F be the face bounded by C (Lemma 6.6 (8)). Obviously F is notinside the face bounded by ∆ e , and, since F is contained in M , F is C e -interior.Let y be the end of e w nearer w in P ; then y ∈ r i . From the definition of H -greencycle (Definition 6.2), the edge of the yx e w -subpath of P incident with y is in C .If w is an attachment of a global H -bridge, then every edge of C ∩ R is in two H -green cycles, which is impossible by Theorem 6.7. Therefore, P = P (cid:48) , y = w ,and C is the union of the wz -path C ∩ P (this defines z ) and an R -avoiding wz -path P . The path P contains a subpath P (cid:48) joining a vertex of the zx -subpath of P P x to a vertex of the component of P w − e w containing w e ; we may assume P (cid:48) is C e -avoiding. Claim 3 implies that the cycle contained in P (cid:48) ∪ P w P P x bounds a faceof Π[ G ]. As z is in this cycle, it must be that z is an end of P (cid:48) and, moreover, thiscycle is C . In particular, P is just P (cid:48) plus a subpath of P w . We know that C ∩ P is just an edge. Since the path C ∩ P w is in the boundary of the faces bounded byboth C and ∆ e , it is also just the edge e w .If z (cid:54) = x e w , then P (cid:48) = P and the zw e -path contained in P ∪ P w contradictsClaim 4 (1). Therefore, z = x e w .Let B be the C e -bridge containing P (cid:48) . Claim 5 implies B has precisely twoattachments w (cid:48) ∈ P w and x (cid:48) ∈ P x : therefore, B is just the edge w (cid:48) x (cid:48) (this is also P (cid:48) ). If x (cid:48) is x e w , then B is a w -backslope.Finally, suppose x (cid:48) is in P x − x e w . Then C bounds a face incident with C ∩ P x .Since C ∩ P x is also in the boundary of the face bounded by ∆ e w , it has length 1.On the P (cid:48) side, B together with the w (cid:48) x (cid:48) -subpath of (cid:104) P w P (cid:48) P x (cid:105) is a cycle C (cid:48) disjoint from P (cid:48) . By Claim 3, C (cid:48) bounds a face of Π[ G ]. As above, each of C (cid:48) ∩ P w , C (cid:48) ∩ P x , and P (cid:48) all have length 1. Therefore, B is a crossbar. (cid:3) Our final case is that e w is in an H -yellow cycle. Claim . If e w is in an H -yellow cycle C , then either 3. THE NEXT RED EDGE AND THE TILE STRUCTURE 117 (d) C e ∪ C contains C e plus a back-slope or(e) C e ∪ C is C e plus a crossbar. Proof. Let C (cid:48) be the H -green cycle witnessing that the cycle C containing e w is H -yellow. Then C (cid:48) contains an H -jump J and either both ends of J are in P or both ends of J are in P . In either case, Claim 1 implies the span of J is allof P or P . We treat these two possibilities separately. Subclaim . If both ends of J are in P , then (e) occurs. Proof. In this case, Claim 1 implies J has ends w and x e w , x e w = v i +3 , e w ∈ s i +1 , and e x ∈ s i +2 .Because e w is both incident with v i +1 and in an H -yellow cycle as witnessedby the H -green cycle C (cid:48) containing J , v i +1 is in the interior of the span of J ;consequently, w ∈ (cid:104) r i (cid:105) . Therefore, the edge of r i incident with v i +1 is H -green.We observe that J witnesses that Q i +1 is an H -yellow cycle. It follows fromLemma 11.2 (3) that C = Q i +1 . The same part of the same lemma combines withthe fact that e is not H -green to show that P w consists of [ w, r i , v i +1 , s i +1 , v i +6 ]and that P w has length precisely two. Symmetrically, P x consists of s i +2 r i +2 andhas length 2. Therefore, we have (e), as required. (cid:3) It remains to consider the possibility that both ends of J are in P . Claim 1implies J = w e x , w e = v i +5 , e w ∈ s i +1 , and e x ∈ s i +2 . Also, r i +5 is in the H -greencycle C (cid:48) containing J , and so P w contains r i +5 s i +1 . Since Theorem 12.1 (3) implies P w has at most one H -rim edge, we conclude that w = v i +1 . Recall that P w isin the boundary of the face of Π[ G ] bounded by ∆ e . The path r i +5 is also in theboundary of the face bounded by C (cid:48) and so is just an edge. The path s i +1 is alsoin the boundary of the face bounded by C , so it too is just an edge.If J is not incident with v i +7 , then the situation is precisely that Subclaim1 with the roles of ( e, w ) and ( e w , x ) interchanged. Therefore, C e ∪ C is (e), asrequired.Therefore, we may assume J is incident with v i +7 . At this point, we knowthat s i +1 r i +5 , J and at least the edge e x of s i +2 are contained in C e . There is a C e -bridge containing r i +6 ; Claim 5 implies this C e -bridge is precisely r i +6 and thisis just an edge.The cycle C has a second edge e (cid:48) incident with v i +6 . There is a C e -bridge B containing e (cid:48) . Claim 5 implies B has precisely two attachments, namely v i +6 andsome other vertex y .If y ∈ P x − x e w , then B together with the yv i +6 -subpath of C e − P (cid:48) containsa cycle disjoint from P (cid:48) and yet does not bound a face (it contains r i +6 ). Weknow that r i +5 r i +6 J bounds a face of Π[ G ], so y is not in J r i +5 . Claim 4 implies y / ∈ P (cid:48) − x e w . Thus, y = x e w .To finish the proof that (d) occurs, note first that s i +1 and B are both edges;thus, it suffices to prove that P (cid:48) is just an edge. In fact, Claim 3 implies P (cid:48) B s i +1 bounds a face of Π[ G ]. In particular, P is not inside this face; therefore, P (cid:48) = P .Consequently, P (cid:48) = P is just an edge. (cid:3) In order to determine the 13 pictures, we remark that, from the perspective ofboth e and e w , any of (1)-(5) in Definition 13.7 can occur. However, if (5) occursfor either, then Claim 3 implies C e and this crossbar is all that is in D e . In thecases (2)(4) and (3)(4), there are two possibilities, as the slope and the back-slope 18 13. THE NEXT RED EDGE AND THE TILE STRUCTURE can have either distinct or common ends in the spoke; the latter is denoted by a + in the listing below. There is no third possibility, since the slope and back-slopedo not cross in D e . Thus, there are the 13 pictures (1)(1), (1)(2), (1)(3), (1)(4),(2)(2), (2)(3), (2)(4), (2)(4) + , (3)(3), (3)(4), (3)(4) + , (4)(4), and (5)(5).Label the red edges in G as e , e , . . . , e k − so that, for i = 0 , , . . . , k − e i has ends u i and v i and so that, reading indices modulo k , e i +1 is the v i -consecutivered edge for e i . This implies that e i is the u i +1 -consecutive red edge for e i +1 .Since there are no red edges between e i − and e i +1 on the “peak of ∆ e i ” portionof R , defining adjacency to mean “consecutive” shows the set of red edges makea cycle. Furthermore, v i and u i +1 are both in the cycle C e i that determines thepicture P i between e i and e i +1 . Taking any v i u i +1 -path P i in P i , we see that P i together with either of the v i u i +1 -subpaths of R makes a non-contractible cycle in R P . In this sense, e i and e i +1 are on opposite sides of R .If we think of e as being on “top” and e on the “bottom”, then e , e , . . . areall on top and e , e . . . , are on the bottom. When we get back to e from e k − ,we have gone once around the M¨obius strip, so e is now on the bottom. It followsthat e k − is on top and, therefore, k − k is odd.It follows that G contains a subgraph H that is in T ( S ). (There may beedges in the interior of C e “between” the structures we identified “near” P (cid:48) and P (cid:48) .) However, Theorem 5.5 implies H ∈ M , so we conclude G = H . That is, G ∈ T ( S ).HAPTER 14 Graphs that are not 3-connected The rest of this work is devoted to: describing all the 2-crossing-critical graphsthat are not 3-connected, discussed in this section; finding all 3-connected 2-cros-sing-critical graphs that do not contain a subdivision of V , treated in Section 15;and showing that the number of 3-connected 2-crossing-critical graphs that do notcontain a subdivision of V n is finite, which is Section 16. These last two combinewith the preceding work to show that there are only finitely many 3-connected 2-crossing-critical graphs to be determined, namely those that have a subdivision of V but no subdivision of V .In this section we show that every 2-crossing-critical graph that is not 3-connected is either one of a few known examples or is obtained from a graph in M by replacing 2-parallel edges with a “digonal” path (that is, a path in whichevery edge is duplicated). We remark that we continue assuming that the minimumdegree is at least 3, as subdividing edges does not affect crossing number. We firstdetermine all the 2-crossing-critical graphs that are not 2-connected. Since the crossing number is additive over components, any 2-crossing-criticalgraph can have at most two components, each of them equal to either K , or K .Thus, there are only three different such graphs: two disjoint copies of K , twodisjoint copies of K , , and disjoint copies of each.Similarly, the crossing number is easily seen to be additive over blocks. Thus,the blocks of a connected, but not 2-connected, 2-crossing-critical graph must be1-critical graphs, and therefore all such graphs can be obtained from the afore-mentioned disconnected 2-crossing-critical graphs by identifying two vertices fromdistinct components. The identified vertex may be a new vertex that subdividessome edge. For example, there are three possibilities in which both blocks are K :the identified vertex is a node in both, or only in one, or in neither. Likewise for K , . There are four 2-crossing-critical graphs in which one block is a subdivisionof K and the other is a subdivision of K , . Proposition . The thirteen graphs in Figure 14.1 are precisely those 2-crossing-critical graphs that are not 2-connected. In this subsection, we treat 2-crossing-critical graphs that are 2-connected, butnot 3-connected. With 36 exceptions, these all arise from 3-connected 2-crossing-critical graphs that have digons (i.e., two edges with the same two ends). Thedigons may be replaced with arbitrarily long “digonal paths” — these are simplypaths in which every edge is converted into a digon. Figure 14.1. The 2-crossing-critical graphs that are not 2-connected.Tutte [ 34, 35 ] developed a decomposition theory of a 2-connected graph intoits cleavage units , which are either 3-connected graphs, cycles of length at least4, or for k ≥ k -bonds (a k -bond is a graph with k edges, all having the sametwo ends). We provide here a brief review of this theory. A 2 -separation of a 2-connected graph G is a pair ( H, K ) of edge-disjoint subgraphs of G , each having atleast two edges, so that H ∪ K = G and H ∩ K = (cid:107){ u, v }(cid:107) (recall (cid:107){ u, v }(cid:107) is thegraph with just the vertices u and v and no edges.). Notice that a 3-cycle and a3-bond have no 2-separations and, therefore, are to be understood in this contextto be 3-connected graphs.The 2-separation ( H, K ) with H ∩ K = (cid:107){ u, v }(cid:107) is a hinge-separation if atleast one of H and K is a (cid:107){ u, v }(cid:107) -bridge and at least one of them is 2-connected.Another way to say the same thing, but in terms of H ∩ K : (cid:107){ u, v }(cid:107) is a hinge if either there are at least three (cid:107){ u, v }(cid:107) -bridges, not all just edges, or there areexactly two (cid:107){ u, v }(cid:107) -bridges, at least one of which is 2-connected.The theory of cleavage units develops as follows. Let G be a 2-connected graph.(1) If (cid:107){ u, v }(cid:107) is a hinge and ( H, K ) is a hinge-separation (possibly of anotherhinge), then there is some (cid:107){ u, v }(cid:107) -bridge containing either H or K .(2) G has no hinge if and only if G is 3-connected, a cycle of length at least4, or a k -bond, for some k ≥ 4. (Recall that a 3-cycle and a 3-bond are3-connected.) In each of these cases, G is its own cleavage unit.(3) If ( H, K ) is a hinge-separation and H ∩ K = (cid:107){ u, v }(cid:107) , then the cleavageunits of G are the cleavage units of the two graphs H + uv and K + uv obtained from H and K by adding a virtual edge between u and v ,respectively. This inductively determines the cleavage units.(4) There is a decomposition tree T whose vertices are the cleavage units of G and whose edges are the virtual edges. A virtual edge joins in T thetwo cleavage units of G containing it.(5) G contains a subdivision of each of its cleavage units.(6) If G contains a subdivision of some 3-connected graph H , then somecleavage unit of G contains a subdivision of H . In attempting to reconstruct G from its decomposition tree and its cleavageunits, each time we combine two graphs along a virtual edge, there are two possi-bilities for how to identify the vertices of the corresponding hinge. This ambiguitywill play a small role in constructing the 2-crossing-critical graphs that are 2- butnot 3-connected.It is easy to see that G is planar if and only if every cleavage unit is planar. (Wecould apply Kuratowski’s Theorem and Item 6 or prove it more directly.) Since weare interested in non-planar graphs, there are two relevant possibilities: one or morethan one of the cleavage units of G is not planar. We start by treating the lattercase. We remark that the following discussion makes clear that the crossing numberis not additive over cleavage units. Related discussions can be found in ˇSir´aˇn [ ],Chimani, Gutwenger, and Mutzel [ ] (but see [ ] for significant comments aboutthe latter), Beaudou and Bokal [ ], and Lea˜nos and Salazar [ ]. Lemma . Let G be a 2-connected graph. If two cleavage units of G are notplanar, then cr( G ) ≥ . It is an important consequence that, if G is 2-crossing-critical, 2-connected, andhas 2 non-planar cleavage units, then G is simple, i.e., has no digons. Proof of Lemma 14.2. Among all 2-separations ( H, K ) of G , we choose theone that has K minimal so that both H + uv and K + uv are not planar, where H ∩ K = (cid:107){ u, v }(cid:107) . If the crossing number of G is not at least 2, then cr( G ) ≤ D is a 1-drawing of G .Let P K and P H be uv -paths in K and H respectively. Since G contains thesubdivision H ∪ P K of H + uv , G is not planar. Therefore, D has a crossing.Evidently, D ( H ∪ P K ) and D ( K ∪ P H ) both contain the crossing. We concludethat the crossing in D is of an edge of P H with an edge of P K . It follows thatthere are not edge-disjoint uv -paths in either H or K and that the crossed edgesare cut-edges in their respective subgraphs.Let w and x be the ends of the edge in K that is crossed, labelled so that w isnearer to u in P K than x is. Let K u and K v be the two components of K − wx , withthe former containing u . Since K + uv is not planar, either K u + uw or K v + vx isnot planar. We may assume it is the former. Notice that ( H ∪ K v ) + xu containsa subdivision of H + uv and, therefore, is not planar. But then (( H ∪ K v ) , K u ) isa 2-separation contradicting the minimality of K .We are now in a position to determine the 2-connected, 2-crossing-criticalgraphs having two non-planar cleavage units. Theorem . Let G be a 2-connected, 2-crossing-critical graph having twonon-planar cleavage units. Then G is one of the 36 graphs in Figures 14.2 and14.3. Proof. Let C and C be non-planar cleavage units of G . Claim . G has at most three cleavage units: C , C and possibly a 3- or4-cycle; if there are three, then the 3- or 4-cycle is the internal vertex in the de-composition tree. Proof. For i = 1 , 2, let { u i , v i } be the hinge of G contained in C i such that C and C are contained in different (cid:107){ u i , v i }(cid:107) -bridges. For any other virtual edge xy 22 14. GRAPHS THAT ARE NOT 3-CONNECTED Figure 14.2. Figure 14.3. C i , there is a path P xy in G that is C ∪ C -avoiding. Let (cid:101) C i be C i ∩ G (i.e., C i with none of its virtual edges) together with all these P xy . Let H be the subgraph of G consisting of (cid:101) C ∪ (cid:101) C ∪ Q , where Q consists of two disjoint { u , v }{ u , v } -pathsin G . Evidently, H is 2-connected and C and C are cleavage units of H .Lemma 14.2 implies cr( H ) ≥ 2. Since H ⊆ G and G is 2-crossing-critical, H = G . Since G has no vertices of degree 2, G consists of either two or threecleavage units, namely C , C , and possibly a 3- or 4-cycle between them. (cid:3) We next determine the possibilities for C and C . Claim . For each i = 1 , 2, one of the following occurs:(1) C i is K ;(2) C i is K , ;(3) C i − u i v i is a subdivision of K , . Proof. Hall proved that every 3-connected non-planar graph is either K orcontains a subdivision of K , [ ]. Since G is simple and C i is 3-connected, wededuce that C i is either K or contains a subdivision of K , . So suppose C i contains a subdivision K of K , .Suppose C i − u i v i has an edge e for which C i − e is not planar. Since C i − e is 2-connected, G − e is 2-connected and has at least two non-planar cleavage units ( C − i and another contained in C i − e ). By Lemma 14.2, cr( G − e ) ≥ 2, contradicting2-criticality of G . So C i − u i v i ⊆ K . Thus, either C i = K or C i − u i v i = K , asclaimed. (cid:3) Claim . There are five possibilities for C i , namely:(1) C i is K ;(2) C i is K , ;(3) C i − u i v i is K , and u i v i joins two non-adjacent nodes of K , ;(4) C i − u i v i is K , with one edge subdivided once and u i v i joins the degree2 vertex to a node of K , that is not incident with the subdivided edge;and(5) C i − u i v i is K , with two non-adjacent edges both subdivided once and u i v i joins the two degree 2 vertices. Proof. If C i is neither K nor K , , then it must be a subdivision K of K , with the additional edge u i v i . Clearly K has at most two vertices of degree 2. If K has no vertices of degree 2, then, since C i is simple, we have (3). Likewise, if K hasonly one vertex of degree 2, that vertex (one of u i and v i ) cannot be in a branchincident with the other one of u i and v i , which is (4). Finally, suppose u i and v i are both of degree 2 in K . Then their containing branches cannot be incident witha common vertex w , as otherwise, we could delete the edge u i w and still have twonon-planar cleavage units, contradicting 2-criticality. This proves (5). (cid:3) Note that in all five cases of Claim 3, there is only one possibility for C i , up toisomorphism. Only (4) has non-isomorphic labellings of u i and v i . Claim . If G has just two cleavage units, then G is one of the 16 graphs inFigure 14.2. Proof. If neither C nor C is (4) from Claim 3, then, with repetition allowed,there are 10 possible unordered pairs for C and C . Each of the pairs uniquelyproduces the graph G . There are four graphs having C but not C satisfying Claim3 (4), and there are two graphs having both C and C satisfying Claim 3 (4). (cid:3) Claim . If G has three cleavage units, then at least one of C and C is either K or K , . Proof. Let e be an edge of G in the third cleavage unit of G ; recall that thiscleavage unit is either a 3- or a 4-cycle. The blocks of G − e include C − u v and C − u v ; if these were both non-planar, then cr( G − e ) ≥ 2, contradicting 24 14. GRAPHS THAT ARE NOT 3-CONNECTED G . Hence, at least one of C − u v and C − u v is planar. ByClaim 3, such a one must be either K or K , . (cid:3) Claim . If G has three cleavage units, then G is one of the 20 graphs in Figure14.3. Proof. There are three pairs in which both C and C are one of K and K , and two possibilities for the third cleavage unit, yielding six graphs. Now suppose C is one of K and K , and C is not. There are three possibilities for C andtwo possibilities for the third bridge. However, when the third bridge is a 3-cycle,there are two ways to attach C when it is of Type (4) from Claim 3. Thus, thereare 6 graphs with the third cleavage unit a 4-cycle and 8 when it is a 3-cycle. (cid:3) From the claims, we see that the 36 graphs shown in Figures 14.2 and 14.3 areall the cases in which G is 2-connected, but not 3-connected, and has two non-planarcleavage units.In the remaining cases of 2-connected, but not 3-connected, 2-crossing-criticalgraphs, there is only one non-planar cleavage unit C . The graph C is simple. Thefollowing result shows how to obtain G from a 3-connected 2-crossing-critical graph.It requires the following definition. Definition . A digonal path is a graph obtained from a path P by adding,for every edge e of P , an edge parallel to e . Theorem . Let G be a 2-crossing-critical graph with minimum degree atleast 3. Suppose that G is 2-connected but not 3-connected and has only exactly onenon-planar cleavage unit, C . The graph (cid:101) C obtained from C by replacing each of itsvirtual edges with a digon is 2-crossing-critical and 3-connected. The graph G isrecovered from (cid:101) C by replacing these virtual edge pairs by digonal paths. Proof. That (cid:101) C is 3-connected is a trivial consequence of the fact that C is 3-connected.As for the rest, let uv be a virtual edge in C . Then (cid:107){ u, v }(cid:107) is a hinge of G .We consider the (cid:107){ u, v }(cid:107) -bridges in G ; let B uv be the one that contains C ∩ G , andlet B uv be the union of the remaining (cid:107){ u, v }(cid:107) -bridges. We have two objectives: toshow that (cid:101) C is 2-crossing-critical and that, for each uv , B uv is a digonal uv -path.For the former, we first show cr( (cid:101) C ) ≥ 2. Otherwise (cid:101) C has a 1-drawing D .Obviously no edge in a digon of (cid:101) C is crossed in D . For each virtual edge uv of C , B uv + uv is planar, so it may be inserted into D in place of the uv -digon in D toobtain a 1-drawing of G , which is a contradiction. Therefore, cr( (cid:101) C ) ≥ B uv consists of digonal uv -paths. Assume first that B uv has a cut-edge e separating u and v . Since G has no vertices of degree 2and B uv is not just a single edge, B uv contains some edge e (cid:48) so that B uv − e (cid:48) stillcontains a uv -path.If no edge of B uv is crossed in a 1-drawing D e (cid:48) of G − e (cid:48) , then, since B uv − e (cid:48) contains a uv -path, B uv may be substituted for B uv − e (cid:48) in D e (cid:48) to obtain a 1-drawingof G , which is impossible. So some edge of B uv is crossed in D e (cid:48) . Deleting edgesfrom B uv − e (cid:48) to leave only a uv path shows that D e (cid:48) restricts to a 1-drawing of B uv + uv in which there is at most one crossing; if there is a crossing, then uv iscrossed. Since every planar embedding of B uv + uv has uv and e on the same face, the 1-drawing of B uv + uv and a planar embedding of B uv + uv may be merged toproduce a 1-drawing of G in which e is crossed. This contradiction that shows B uv contains edge-disjoint uv -paths.Let e be an edge of B uv . Then a 1-drawing D e of G − e must have a crossingof some edge e (cid:48) of B uv . If B uv − { e, e (cid:48) } has a uv -path P , then D e restricts to aplanar embedding of C by using P to represent uv . But C is non-planar, so everyedge of B − uv is in an edge-cut of size at most 2 separating u and v . Combiningthis with the preceding paragraph shows that every edge of B uv is in an edge-cut ofsize exactly 2. It is an easy exercise to see that this implies B uv is a pair of digonalpaths.We conclude by showing that, for every edge e of (cid:101) C , cr( (cid:101) C − e ) ≤ 1. Supposefirst that e is not in a digon. Each B uv has a uv -path P uv that is clean in D e . Thus, D e [ G − e ] contains a subdivision of C − e in which no virtual edge (represented inthe subdivision by P uv ) is crossed. Therefore, the virtual edges may be replacedwith digons to give a 1-drawing of (cid:101) C − e , as claimed.Now suppose e is in the uv -digon. Let e (cid:48) be any edge of B uv . Then D e (cid:48) containsa 1-drawing of C , in which every other virtual edge wx is represented by a wx -path P wx in B wx that is clean in D e (cid:48) . All these other virtual edges may be replaced withdigons to give a 1-drawing of (cid:101) C − e , as required.HAPTER 15 On 3-connected graphs that are notperipherally-4-connected In this chapter, we reduce the problem of finding all 3-connected 2-crossing-critical graphs to the consideration of non-planar, peripherally-4-connected graphs.Our motivation for doing this is to use a known characterization of internally-4-connected graphs (a concept intimately related to peripherally-4-connected graphs)with no subdivision of V to find all the 3-connected, 2-crossing-critical graphs withno subdivision of V . Definition . A graph G is peripherally-4-connected if G is 3-connectedand, for any 3-cut S of G and any partition of the components of G − S into twonon-null subgraphs H and K , at least one of H and K has just one vertex.We begin this section by finding the four 3-connected, not peripherally-4-connected, 2-crossing-critical graphs that are not obtained from planar substitu-tions into a peripherally-4-connected graph. The bulk of the section is devotedto explaining in detail how to obtain the remaining 3-connected 2-crossing-criticalgraphs from peripherally-4-connected graphs. Finally, this theory is used to ex-plain how to find all the 3-connected 2-crossing-critical graphs that do not containa subdivision of V . In this section we find the four 3-connected, not peripherally-4-connected, 2-crossing-critical graphs that are not obtained by substituting planar pieces intodegree-3 vertices in a peripherally-4-connected graph (this substitution process be-ing the remainder of the section). We start by describing the four graphs andshowing that they are 2-crossing-critical. Definition . The graph K ∗ , is obtained from disjoint copies of K , byjoining the parts of the bipartition having three vertices in each of the copies by aperfect matching M .Observe that K , is obtained from K ∗ , by contracting all the edges of thematching M . The following generalizes the well-known fact that K , is 2-crossing-critical. Lemma . If H is obtained from K ∗ , by contracting some subset of M , then H is 2-crossing-critical. Proof. Suppose e is an edge of K ∗ , not in M . Then there is a 1-drawing of K ∗ , − e in which no edge of M is crossed. Thus, cr( H − e ) ≤ 1. If e ∈ M , then H − e is planar. It remains to show cr( H ) ≥ Suppose to the contrary that H has a 1-drawing D . Let H and H be the K , subgraphs of H contained in K ∗ , − M . For each vertex v of degree 3 in H ,there are three disjoint vH -paths in H ; adding v and these paths to H yields asubdivision H v of K , in H . Thus, D [ H v ] has a crossing, and, since there are twochoices for v , this crossing involves only edges of H and M .Interchanging the roles of H and H shows the crossing in D involves onlyedges of M . But then D [ H v ] has its only crossing on branches incident with v ,which is impossible.We remark that there are splits of K , that have crossing number 1 — splittwo of the degree 4 vertices so that the two partitions of the four neighbors aredifferent. Fortunately, they do not occur in our context.In order to show that these are the only four graphs with “non-planar 3-cuts”,we need to understand just what “non-planar 3-cuts” are. Definition . Let S be a 3-cut in a 2-connected graph, so there are sub-graphs H and K of G such that G = H ∪ K and H ∩ K = (cid:107) S (cid:107) . For L ∈ { H, K } , L + denotes the graph obtained from L by the addition of a new vertex adjacent toprecisely the vertices in S .We will see that, in the case G is 2-crossing-critical, with the exception of K , ,there are at most three non-trivial S -bridges, and so at least one of H and K isan S -bridge. Our next goal is to show that the four graphs in Lemma 15.3 are theonly four that have both H + and K + non-planar. We start with the following,which is likely well-known; however, we could not find a reference. It extends Hall’sTheorem [ ] that there is a subdivision of K , . Lemma . Let G be a 3-connected non-planar graph different from K andlet v be a vertex of G . Then G has a subdivision H of K , in which v is an H -node. Proof. Here is an outline of the easy, but tedious, proof. As a first step, weshow that there is a subdivision of K , containing v . By Hall’s Theorem [ ], G contains a subdivision L of K , . If v / ∈ L , then there are three disjoint vL -paths.There are three possibilities for the ends of these paths in L : two are in the sameclosed L -branch; two are in L -branches incident with a common L -node; and the L -branches containing the ends of the paths are pairwise disjoint. In the first case, v is incorporated into the interior of a branch of a new subdivision of K , , whilein the other cases, v is incorporated as a node of the new subidivision of K , .So now assume that v is in L , but not as a node. Then v is interior to some L -branch b with ends u and w . Let L (cid:48) = L − (cid:104) b (cid:105) — this is a subdivision of K , less an edge. Because there are, in G , disjoint L (cid:48) -avoiding v { u, w } -paths, standardproofs of Menger’s Theorem imply that there are three disjoint L (cid:48) -avoiding vL (cid:48) -paths, having u and w among their three L (cid:48) -ends. Therefore, we may assume notonly is v in the interior of b , but there is an L -avoiding vx -path from v to someother vertex x of L (cid:48) .Up to symmetry, there are three possibilities for x : it is a node of L other than u and w ; it is interior to an L -branch incident with u but not with w ; and it isinterior to an L -branch not incident with either u or w . Let y and z be nodes of L (cid:48) (note that u and w are not actually nodes of L (cid:48) ). We can assume x is either y ,or in the L -branch [ w, y ], or in the L -branch [ y, z ]. Let Y be a { u, w, x } -claw withcentre v , so that Y ∩ L (cid:48) is just u , w , and x . 28 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED If x is either z or in (cid:104) y, z (cid:105) , then ( L (cid:48) ∪ Y ) − (cid:104) w, x (cid:105) is a subdivision of K , with v as a node. If x is in (cid:104) y, z (cid:105) , then ( L (cid:48) ∪ Y ) − (cid:104) w, y (cid:105) is a subdivision of K , having v as a node.We are now ready for the classification of the 3-connected 2-crossing-criticalgraphs with two non-planar sides to a 3-cut. Theorem . Let G ∈ M have subgraphs H and K of G and a set S ofthree vertices of G such that:(1) G = H ∪ K ;(2) H ∩ K = (cid:107) S (cid:107) ;(3) H and K both have an (cid:107) S (cid:107) -bridge having all of S as attachments; and thetwo graphs H + and K + are both non-planar.Then G is one of the four graphs obtained from K ∗ , by contracting some subset of M . Proof. Let u , v , and w be the vertices in S . For L ∈ { H, K } , let v + L denotethe vertex in L + , but not in L . The graph L + is a subdivision of a 3-connectedgraph (the only possible vertices of degree 2 are u , v , and w ). Since L + is notplanar and has a vertex of degree 3, it is not a subdivision of K and, therefore,by Lemma 15.5 contains a subdivision L (cid:48) of K , in which v + L is a node. Now G (cid:48) = ( H (cid:48) − v + H ) ∪ ( K (cid:48) − v + K ) is a subdivision of K ∗ , , with some subset of M contracted. By Lemma 15.3, cr( G (cid:48) ) = 2, so G (cid:48) = G , as required. In this section, we discuss the general details of reducing a 3-connected graph toa peripherally-4-connected graph. These results apply in some generality and notjust in the context of 2-crossing-critical graphs. These are the first of several stepstoward finding all the 3-connected 2-crossing-critical graphs that do not contain asubdivision of V .These results are fairly technical but essential to this part of the theory. Definition . (1) A 3-cut S in a 3-connected graph is reducible if G − S has at most 3 components and they partition into two subgraphseach having at least two vertices.(2) The set K consists of those 3-connected graphs that do not contain asubdivision of K , .The following result is obvious from the definitions and begins to explain theappearance of K , in Definition 15.7 (2). Lemma . Let G be a 3-connected graph that is not peripherally-4-connected.Then either G has a reducible 3-cut or G has K , as a subgraph. The next result sets up the basic scenario that we will use throughout ourreduction to peripherally-4-connected graphs. Lemma . Let G ∈ K . Then there is a sequence G , G , . . . , G k of 3-connected graphs in K so that: G = G ; G k is peripherally-4-connected; and, foreach i = 1 , , . . . , k , there is a 3-cut S i in G i − and a non-trivial, planar S i -bridge B i so that Nuc( B i ) has at least two vertices and G i is obtained from G i − by con-tracting the nucleus of B i . Proof. Suppose G i − is 3-connected. Among all the choices of S i and S i -bridges B i so that Nuc( B i ) has at least two vertices, choose B i to be inclusion-wise maximal.We claim that the graph G i obtained from G i − by contracting Nuc( B i ) to a vertexis 3-connected.Otherwise, there is some pair { u, v } of vertices so that G i − { u, v } is not con-nected. If the vertex of contraction of Nuc( B i ) is neither u nor v , then { u, v } is a2-cut in G i − , a contradiction. Therefore, we can assume u is the contraction ofNuc( B i ).Let H and K be components of G i − { u, v } , with the labelling chosen so that | S i ∩ V ( H ) | ≥ | S i ∩ V ( K ) | ; in particular, | S i ∩ V ( K ) | ≤ 1. Let h ∈ V ( H ); if there isa vertex k ∈ V ( K ) \ S i , then { v } ∪ ( S i ∩ V ( K )) separates k from h in G i − , whichcontradicts the assumption that G i − is 3-connected.Therefore V ( K ) ⊆ S i , so there is a single vertex s in K , and s ∈ S i . It followsthat s is adjacent to only vertices in Nuc( B i ) and possibly to v . But this contradictsthe maximality of B i : let S (cid:48) = ( S \ { s } ) ∪ { v } . Observe that B i + s is a planar S (cid:48) -bridge, contradicting maximality of B i .Lastly, we show that if G i − does not have a subdivision of K , , then neitherdoes G i . Any subdivision of K , in G i must contain the vertex v i of contraction.Since v i has degree 3 in G i and B i − is an S -bridge, we can reroute the subdivisionof K , in G i into B i − to obtain a subdivision of K , in G i − . Definition . Let G ∈ K .(1) Then G reduces to G (cid:48) by 3-reductions if there is a sequence G , G , . . . , G k of 3-connected graphs so that G = G ; G k = G (cid:48) ; and, for each i =1 , , . . . , k , there is a 3-cut S i in G i − and an S i -bridge B i , whose nucleusat least two vertices, so that G i is obtained from G i − by contracting thenucleus of B i .(2) For each vertex v of G (cid:48) and each i = 0 , , . . . , k , K iv denotes the connectedsubgraph of G i that contracts to v . We also set K v = K v .(3) If v has just three neighbours x , y , and z in G (cid:48) , then G v is the graphobtained from K v by adding x , y , and z , and, for each t ∈ { x, y, z } andeach edge v (cid:48) t (cid:48) of G with v (cid:48) ∈ K v and t (cid:48) ∈ K t , adding the edge v (cid:48) t .We now commence a lengthy series of technical lemmas that all play vital rolesin usefully reducing the 3-connected graph 2-crossing-critical graph G to a smaller3-connected 2-crossing-critical graph G rep( v ) . The culmination of this part of thework is Theorem 15.25 in the next section, showing that G rep( v ) is 2-crossing-critical.This will lead to a program for determining all the 3-connected 2-crossing-criticalgraphs that reduce to a particular peripherally-4-connected graph. Lemma . Let G ∈ K and suppose G reduces by 3-reductions to the peri-pherally-4-connected graph G p4c . For any two vertices u, v of G p4c , there is a singlevertex in G incident with all edges having one end in K u and one end in K v . Proof. Let G = G , G , . . . , G k = G p4c be a sequence of 3-reductions. Choose i to be largest so that there are disjoint K i − u K i − v -edges ab and cd with a, c ∈ K i − u and b, d ∈ K i − v . In G i , either a and c have been identified or b and d have; bysymmetry, we may assume the former.The vertices b and d are obviously attachments of B i and so these are in S i .Let z i be the third vertex in S i . Since K i − u is connected and since, by Definition 30 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED u has three neighbours in G p4c , z i ∈ K u . Continue using the label a for thevertex obtained by contracting Nuc( B i ).At some point in the later 3-reductions, a and z i are identified and at anotherpoint b and d are identified. We show that neither can be done before the other,which is impossible.Suppose z i and a are identified first. When this identification occurs, a 3-cut S j and an S j -bridge B j so that z i and a are in Nuc( B j ). The vertices b and d areagain attachments of B j and so are in S j ; let z j be the third vertex in S j .Because i is largest so there are disjoint K i − u K i − v -edges, all edges between K ju and K jv at this moment are incident with a . It follows that { a, z j } is a 2-cut inthe current graph, separating z i from b . But this contradicts the fact that G j − is3-connected. Therefore, z i and a are not identified before b and d .On the other hand, suppose b and d are identified first, by the contraction ofNuc( B j ). When b and d are identified, the only neighbours of a are b , d , and z i .Following the identification of b and d , the only neighbours of a are z i and thevertex of identification, again contradicting 3-connection of G j .We need a slight variation on a standard definition. Definition . Let G be a connected graph.(1) An isthmus is a set I of parallel edges so that G − I is not connected.(2) A cut-edge is an edge e so that G − e is not connected.Obviously, e is a cut-edge of G if and only if { e } is an isthmus, but an isthmusmay have more than one edge. The distinction comes into play because at variouspoints we will consider edge-disjoint paths in certain subgraphs of our 2-crossing-critical graph; if there are not two edge-disjoint uv -paths, then there is a cut-edgeseparating u and v . On the other hand, the 3-connection of G does not precludethe possibility of parallel edges; at several points we will be able to identify thattwo vertices u and v have the property that they must be adjacent, but be unableto distinguish whether they are joined by 1 or 2 edges. A common scenario willhave the set of edges between them making an isthmus in some subgraph.In particular, the case that K v has an isthmus is a central one in reducing2-crossing-critical graphs. Lemma . Let G ∈ K reduce to the peripherally-4-connected graph G p4c bya sequence of 3-reductions. Suppose there is a vertex v of G p4c so that the graph K v has an isthmus I . Then, for each component K of K v − I , there are at leasttwo neighbours x and y of v in G p4c so that there are KK x - and KK y -edges in G . Proof. At some moment in the reduction of G , G i − has a 3-cut S i and B i is theplanar S i -bridge in G i − that contains I . Then B i − I is not connected; the ends u and w of the edge or edges in I are in different components K and L , respectively,of B i − I .Let x , y , and z be the neighbours of v in G p4c and let t be any vertex of G i − not in K iv ∪ K ix ∪ K iy ∪ K iz . (Since G p4c is not planar, it has at least five vertices.)In G i − there are three pairwise internally-disjoint ut -paths. These three pathsleave B i through distinct attachments of B i ; these are the vertices in S i . The sameargument applies for wt -paths.In particular, two of the ut -paths leave K on edges incident with vertices in S i . Likewise for L . Therefore, K and L are both joined by edges to the same attachment s ∈ S i . It follows that s is not in K iv , so s is in K ix , say. Moreover,since the K iv -ends of these two edges are not the same, Lemma 15.11 implies all theedges between K iv and K ix are incident with s .Since G i − is 3-connected, G i − − ( { s } ∪ I ) is connected. Therefore, there areedges of G i − leaving each of K and L ; each of these edges is also leaving K iv and,therefore, has its other end in one of K ix , K iy , and K iz . However, this other endcannot be s and, consequently, cannot be in K ix , as required.The connectivity of G has further implications about the structure of the K v . Lemma . Let G ∈ K reduce by 3-reductions to a peripherally-4-connectedgraph G p4c . Let v be a vertex of G p4c with just the three neighbours x , y , and z and suppose K v has at least two vertices. For each t ∈ { x, y, z } , let t (cid:48) be any vertexincident with all the K v K t -edges. Then x (cid:48) , y (cid:48) , and z (cid:48) are all distinct. Proof. Suppose x (cid:48) = y (cid:48) . Then x (cid:48) is in K v . Observe that no vertex of K v − { x (cid:48) , z (cid:48) } is adjacent to any vertex of of G − { x (cid:48) , z (cid:48) } not in K v . Since G is 3-connected, itfollows that K v consists of just x (cid:48) and z (cid:48) . In particular, z (cid:48) (cid:54) = x (cid:48) . Also, recall that K v contracts to a single vertex in the sequence of planar 3-reductions.At the moment of contraction of K v , G i − is 3-connected and x (cid:48) z (cid:48) is an isthmus.Therefore, Lemma 15.13 implies that z (cid:48) is joined to at least one of K ix and K iy ; thiscontradicts the fact that all edges from K iv to K ix ∪ K iy are incident with x (cid:48) .The vertices x (cid:48) , y (cid:48) , and z (cid:48) are not uniquely determined. It is possible that thereis only one vertex in each of K v and K x incident with all K v K x -edges; one obviousinstance is if there is only one K v K x -edge. We will follow up on this a little later.Here is a very simple and very useful observation. Lemma . Let H be a simple, non-planar, peripherally-4-connected graph.There is no 3-cycle of H having two vertices with just 3 neighbours. Proof. Suppose to the contrary there are three vertices x, y, z making a 3-cycle,with x and y having only three neighbours each. Let v and w be the other neigh-bours of x and y . Then x and y are the vertices of one component of H − { v, w, z } .Observe that H is non-planar, 3-connected, and has a vertex of degree 3. There-fore H is not K and so contains a subdivision of K , . It follows that H has atleast six vertices. Thus, there is another component of H − { v, w, z } .Since H is peripherally-4-connected, the only possibility is that there is exactlyone other component and it consists of a single vertex u , adjacent to all of v , w ,and z . The only other possible edges in H are between v , w , and z . However, theresulting graph is planar, a contradiction.The following result assures us that useful (and expected) paths exist in each K v . Lemma . Let:(1) G ∈ K reduce by 3-reductions to the peripherally-4-connected graph G p4c ;(2) G p4c have at least five vertices;(3) v be a vertex of G p4c so that K v has at least two vertices; and(4) x , y , and z be the neighbours of v in G p4c , with corresponding vertices x (cid:48) , y (cid:48) , and z (cid:48) in G as in Lemma 15.14. 32 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED Then: a) for any vertex w in K v − { x (cid:48) , y (cid:48) , z (cid:48) } , there are three w { x (cid:48) , y (cid:48) , z (cid:48) } -paths in G v that are pairwise disjoint except for w ; andb) if x (cid:48) ∈ K v , then there are x (cid:48) y (cid:48) - and x (cid:48) z (cid:48) -paths in G v − x that are disjointexcept for x (cid:48) . Proof. For a), let u be any vertex of G not in K v ∪ K x ∪ K y ∪ K t . Since G is 3-connected, there are three pairwise internally-disjoint wu -paths in G . Theresult follows from the observation that w and u are in different components of G − { x (cid:48) , y (cid:48) , z (cid:48) } .If b) fails, then there is a vertex w of G v − x that separates x (cid:48) from { y (cid:48) , z (cid:48) } .Since K v is an (cid:107){ x, y, z }(cid:107) -bridge in G v , w is in K v (possibly w = y (cid:48) or w = z (cid:48) ).Since { x (cid:48) , w } is not a 2-cut in G , x (cid:48) and w are adjacent in K v . But now they arejoined by an isthmus I in K v . Since x (cid:48) is a component of K v − I joined only to K x , we have a contradiction of Lemma 15.13. In this section we now turn our attention to the particular case G ∈ M .We want to show that the 3-reductions can be taken to be contractions of planarbridges. So suppose S is a non-peripheral 3-cut in G .If there are four or more non-trivial S -bridges (that is, having a nucleus), then G has a subdivision of K , and so is K , . In the remaining cases, there are atmost three non-trivial S -bridges. If there are three and B is one of them so that B + is not planar (as in Subsection 15.1), then the union K of the remaining S -bridgeshas K + not planar. Theorem 15.6 implies that G is one of four 2-crossing-criticalgraphs. Thus, if there are three non-trivial S -bridges, we may assume that, foreach one B , B + is planar. Finally, consider the case that there are precisely twonon-trivial S -bridges B and B . Since S is not peripheral, both B i have at leasttwo vertices. If both B + i are non-planar, then we are in the case dealt with inTheorem 15.6, so we may assume that one of them is planar. In summary, in everycase, we may assume that G p4c is obtained from 3-reductions in G in which thecontracting S i -bridge B i is always planar. Definition . Let G be a 3-connected graph and let G p4c be a peripherally-4-connected graph. Then G reduces to G p4c by planar 3-reductions if there is asequence G = G , G , G , . . . , G k = G p4c of 3-reductions so that, for each i =1 , , . . . , k , G i is obtained from G i − by contracting Nuc( B i − ) and B + i − is planar.We need two results about K v in the context of planar 3-reductions. Thisrequires further definitions. Definition . Let G be a 3-connected graph that reduces by 3-reductionsto the peripherally-4-connected graph G p4c . Suppose v is a vertex of G p4c havingonly the neighbours x , y , and z . For each t ∈ { x, y, z } , let m t denote the number ofvertices in K v adjacent to vertices in K t and let n t denote the number of verticesin K t adjacent to vertices in K v . (Lemma 15.11 implies that at least one of m t and n t is 1.)(1) The subgraph K max v induced by K v together with, for each t ∈ { x, y, z } with n t = 1, the vertex of K t adjacent to vertices in K v . (2) The subgraph K min v induced by K v together with, for each t ∈ { x, y, z } with m t > 1, the vertex of K t adjacent to vertices in K v .We remark that K v ⊆ K min v ⊆ K max v , and, for t ∈ { x, y, z } , K max v has a vertex t (cid:48) ∈ K t that is not in K min v precisely when n t = m t = 1. Lemma . Let G ∈ K reduce by 3-reductions to a peripherally-4-connectedgraph G p4c . Let v be a vertex of G p4c with just the three neighbours x , y , and z andsuppose K v has at least two vertices. Then there is a cycle C in K min v containingall of x (cid:48) , y (cid:48) and z (cid:48) . Proof. Suppose w is a cut-vertex of K min v , so there are subgraphs X and Y of K min v with X ∪ Y = K min v , X ∩ Y = (cid:107) w (cid:107) , and both X − w and Y − w are notempty. We may choose the labelling so that X has at least the two vertices x (cid:48) and z (cid:48) from { x (cid:48) , y (cid:48) , z (cid:48) } , while Y − w has at most one; we may further assume x (cid:48) (cid:54) = w . If y (cid:48) / ∈ Y − w , then w is a cut-vertex of G , contradicting the fact that G is 3-connected.Therefore, y (cid:48) ∈ Y − w .However, if y (cid:48) ∈ K v , then we have a contradiction to Lemma 15.16 (b). There-fore, y (cid:48) / ∈ K v . If there is a vertex in Y other than w and y (cid:48) , then we contradict3-connection of G , so y (cid:48) is adjacent only to w in G v . But then y (cid:48) / ∈ K min v .It follows that there is no cut-vertex in K min v . Thus, there is a cycle C in K min v containing x (cid:48) and y (cid:48) . Obviously, we are done if z (cid:48) ∈ C , so we assume z (cid:48) / ∈ C .Since there is no cut-vertex in K min v , there are two z (cid:48) C -paths P and P thatare disjoint except for z (cid:48) . If the C -ends of P and P are not both on the same x (cid:48) y (cid:48) -subpath of C , then G + v contains a subdivision of K , . This contradicts thefact that we are doing planar 3-reductions. Therefore, the C -ends of P and P areon the same x (cid:48) y (cid:48) -subpath of C and it is easy to find the desired cycle through allof x (cid:48) , y (cid:48) , and z (cid:48) .The following is the last lemma we need to get the main result of this section. Lemma . Let G ∈ M and suppose G reduces by planar 3-reductions tothe peripherally-4-connected graph G p4c . Let v and x be adjacent vertices in G p4c .Then there are at most two vertices in K v adjacent to vertices in K x . Proof. This is obvious if K v has at most one vertex. In the remaining case, v hasdegree 3 in G p4c ; let y and z be its other neighbours.Suppose by way of contradiction that s , t , and u are distinct vertices in K v alladjacent to vertices in K x . By Lemma 15.11, there is a vertex x (cid:48) incident will allthe K v K x -edges and, evidently, x (cid:48) ∈ K x .In the planar embedding D + v of G + v , letting w denote the new vertex adjacentto each of x , y , and z , we may choose the labelling so that the edges xw, xs, xt, xu occur in this cyclic order around x . Claim . There is an su -path in K v containing t . Proof. As K v is connected, there is an su -path P in K v . We are obviouslydone if t ∈ P , so we assume t / ∈ P . Let C be the cycle obtained by adding x (cid:48) to P and joining it to s and u .The rotation at x implies that t is on one side of D + v [ C ], while w , y , and,consequently, z , are on the other. Therefore, every t { y, z } -path in G + v goes througheither x or P . 34 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED If there is a cut-vertex r in K v separating t from P , then { r, x (cid:48) } is a 2-cut inthe 3-connected graph G , which is impossible. Therefore, there are tP -paths Q and R in K v that are disjoint except for t . We can now reroute P through t to obtainthe desired path. (cid:3) Since G is 2-crossing-critical, there is a 1-drawing D of G − x (cid:48) t . From Claim 1,there is an su -path P in K v containing t . Let C be the cycle obtained from P byadding x (cid:48) , x (cid:48) s and x (cid:48) u . Claim . All the vertices of G − ( K v ∪ K x ) are in the same face of D [ C ]. Proof. Suppose by way of contradiction that there are vertices in G − ( K v ∪ K x ) that are in different faces of D [ C ]. Case 1: there is a vertex p in G p4c so that K p contains vertices that are indifferent faces of D [ C ] . In this case there is an edge f of K p that crosses D [ C ]. As D has at most onecrossing, f is a cut-edge of K p . Lemma 15.13 implies each component of K p − f is adjacent to at least two different K n ’s. If one of them is adjacent to both K x and K v , then we have a 3-cycle pxv in G p4c in which both p and v have degree 3,contradicting Lemma 15.15.Therefore, we may assume each is adjacent to one, say K q and K r , that isneither K x nor K v . However, now { v, x, p } is a 3-cut in G p4c separating q and r in G p4c . Therefore one of them — say q — is adjacent to precisely these three verticesin G p4c , producing the 3-cycle { q, v, x } in G p4c that contradicts Lemma 15.15. Case 2: any two vertices of G − ( K v ∪ K x ) in different faces of D [ C ] are indifferent K p ’s. Since G − ( K v ∪ K x ) is connected, there is a path in G − ( K v ∪ K x ) joiningvertices in different faces of D [ C ]. Therefore, there is, for some vertices q and r of G p4c , a K q K r -edge f that crosses D [ C ]. It follows that D [ C ] has no self-crossings,so D [ C ] has only two faces.Clearly G p4c − { x, v, f } has K q and K r in different components. Since G p4c has at least six vertices, it has a vertex m different from all of v , x , q and r . Wemay choose the labelling so that D [ K q ] is in one face of D [ C ], while D [ K r ∪ K m ] iscontained in the other. It follows that { v, x, r } is a 3-cut in G p4c separating q from m . Since G p4c is peripherally-4-connected, one of q and m — say q — is adjacentprecisely to v , x , and r , yielding the 3-cycle { v, x, q } in G p4c that has two verticeswith only three neighbours, contradicting Lemma 15.15. (cid:3) We note that the crossing in D cannot involve two edges, each incident with avertex in K v , as otherwise G p4c is planar. In particular, D [ C ] is not self-crossing. Claim . OD G + v ( C ) is isomorphic to OD G ( C ). In particular, OD G ( C ) isbipartite. Proof. The main point is that there is a single C -bridge in G containing G − ( K v + x (cid:48) ). To prove this, we show that any two vertices in G − ( K v + x (cid:48) ) areconnected by a C -avoiding path. For vertices not in K v ∪ K x , this is easy: for anytwo vertices p and q in G p4c − { v, x } , there is a pq -path in G p4c − { v, x } , showingthat any two vertices in K p ∪ K q are joined by a path in G − ( K v ∪ K x ). If p ∈ K x − x (cid:48) , then Lemma 15.14 implies that the three vertices separating K x from its neighbours are distinct. For one of these vertices w (cid:48) that is not x (cid:48) , Lemma15.16 implies there is a pw (cid:48) -path in K x − x (cid:48) , completing the proof that there is asingle C -bridge B in G containing G − ( K v + x (cid:48) ).Every other C -bridge in G is contained in K v + x (cid:48) . These are all C -bridges in G + v ; the only other C -bridge in G + v is the one containing the vertex joined to x , y ,and z . This C -bridge has precisely the same attachments as B . This shows that OD G ( C ) and OD G + v ( C ) are isomorphic.Since G + v ( C ) is planar, OD G + v ( C ) is bipartite, yielding the fact that OD G ( C )is bipartite. (cid:3) Suppose first that C is clean in D . Since B is the unique non-planar C -bridgein G , D yields a 1-drawing of C ∪ B with C clean. Therefore, Corollary 4.7 impliescr( G ) ≤ 1, a contradiction.If, on the other hand, C is not clean in D , then C is crossed by an edge f . ByClaim 2, f is incident with a vertex in K v ∪ K x . If f is incident with a vertex in K v , then contract K v (with a vertex inserted at the crossing point, if necessary, toget a 1-drawing of G p4c so that both edges incident with the crossing are incidentwith v . This implies the contradiction that G p4c is planar.If f is not incident with x (cid:48) , then K x − x (cid:48) has vertices on both sides of D [ C ].One of these is in a component K x of K x − f that is on the side of D [ C ] that doesnot contain any vertex of G − ( K v ∪ K x ). Lemma 15.13 implies K x − x is joinedto a vertex in some other K w , w (cid:54) = v , which cannot happen without crossing D [ C ]a second time, a contradiction. It follows that f is incident with x (cid:48) . Furthermore,Lemmas 15.19 and 15.16 (a) imply that f is in a cycle C f in G − K v . The ends ofthe edge e v of K v crossed in D are separated by D [ C f ], so e v is a cut-edge of K v .Moreover, e v is in C .We now see that the C -bridges are B , those contained in one component of K v − e v , and those contained in the other component of K v − e v . Notice that B isa cut-vertex of OD G ( C ), and so it overlaps C -bridges of both the other types.Since OD G ( C ) is connected and bipartite, it follows that the C -bridges in eitherof the components of K v − e v occur on the same side of D [ C ] that they do in D + v .In particular, x (cid:48) t may be reintroduced to D to obtain a 1-drawing of G , which isimpossible. Strategy. The strategy now is to show that if we replace any K v with a smallestpossible representative subject to the preceding observations, then we produce a 2-crossing-critical graph. This is the last part of this section. This implies that G p4c turns into a 2-crossing-critical graph by choosing these smallest possible represen-tatives. From this, it is then possible to determine (although not in a theoreticalsense, but rather in a definite, finite — really manageable — way that we shall de-scribe) all the 3-connected 2-crossing-critical graphs that have these configurationsand reduce to G p4c by planar 3-reductions.There will remain the issue of determining all the possible G p4c . Of course, onecan list them all, but it is not clear at what point to stop. Fortunately, Theorem2.14 shows that we do not need to do this when G contains a subdivision of V ,as we already know what G looks like. When G does not contain a subdivisionof V , a theorem of Robertson plus some analysis implies that G p4c has at most9 vertices. We are left with the open question of finding the graphs in M that 36 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED contain a subdivision of V but do not contain a subdivision of V . In Section 16,we show that any such graph has at most about 4 million vertices. We next characterize certain properties of the graphs G v ; our goal is to showthat these (more or less) determine the crossing number of G . Definition . Let x , y , and z be vertices in a graph H so that H is an (cid:107){ x, y, z }(cid:107) -bridge. Then: • T is the set of vertices w ∈ { x, y, z } so that there are edge-disjoint w ( { x, y, z } \ { w } )-paths in H ; and • U is the set of vertices w ∈ { x, y, z } for which there are edge-disjoint pathsin H − w joining the two vertices in { x, y, z } \ { w } . • ( H, { x, y, z } ) is a ( T, U ) -configuration if the graph H + obtained from H by adding a new vertex adjacent just to x , y , and z is planar.Our entire argument depends on the fact, to be proved in the next section,that the pairs ( T, U ) effectively characterize 2-criticality. Theorem 15.24, the mainpoint of this section, shows that substituting one ( T, U )-configuration for anotherretains the fact that the crossing number is at least 2.For a ( T, U )-configuration, obviously there are only four possibilities for | T | . Itis a routine analysis of cut-edges to see that, if | T | ≤ 1, then U is empty, while if,for example, T = { x, y } , then U = { z } . Thus, for | T | ≤ U is determined by T .This is not the case for | T | = 3. In this instance, if z / ∈ U , then there is a cut-edgein G v − z separating x and y . From here and the fact that T = { x, y, z } , one easilysees that x, y ∈ U . Thus, if T = { x, y, z } , then | U | can be either 2 or 3. Therefore,there are in total five possibilities for the pair ( | T | , | U | ).We first show that replacing a ( T, U )-configuration with another ( T, U )-con-figuration does not lower the crossing number below 2. First the definition ofsubstitution. Definition . Let G reduce by planar 3-reductions to the peripherally-4-connected graph G p4c . Suppose v is a vertex of G p4c with neighbours x , y , and z sothat ( G v , { x, y, z } ) is, for some subsets T and U of { x, y, z } , a ( T, U )-configuration.Let N be the set of vertices t in { x, y, z } for which K max v ∩ K t is null. (See Definition15.18 for K max v .) Let ¯ N v denote the attachments of K max v : these are the verticesthat are of the form t (cid:48) , t ∈ { x, y, z } , chosen to be in K t whenever possible.(1) A ( T, U )-configuration ( H, { x, y, z } ) is ( G, K v ) -compatible if:(a) for each t ∈ N , then there is only one neighbour of t in H ;(b) the degrees of each t ∈ { x, y, z } are the same in both G v and H ; and(c) setting ¯ N H to consist of the union of the set of vertices of H in { x, y, z } \ N together with the neighbours in H of the vertices in N , H − N either has a single vertex or contains a cycle through all thevertices in ¯ N H .(2) The substitution of the K v -compatible ( T, U ) -configuration ( H, { x, y, z } ) for K v in G is the graph G Hv obtained from G by adding H − N byidentifying the vertices in ¯ N v with those in ¯ N H in the natural way, andthen deleting all vertices in K max v − ¯ N v .We are almost ready for a major plank in the theory. Our plan is to show that we can replace a “large” ( T, U )-configuration by a“small” ( T, U )-configuration and still be 3-connected and 2-crossing-critical. Thereis one special case that requires particular attention. Definition . A ( T, U )-configuration ( H, { x, y, z } ) is doglike with nose n if | T | = 3 and | U | = 2 and n is the vertex in T \ U . Theorem . Let G reduce by planar 3-reductions to the peripherally-4-connected graph G p4c . Suppose v is a vertex of G p4c with precisely the neighbours x , y , and z so that K v has at least two vertices so that ( G v , { x, y, z } ) is, for somesubsets T and U of { x, y, z } , a ( T, U ) -configuration. Let ( H, { x, y, z } ) be a ( G, K v ) -compatible ( T, U ) -configuration. If cr( G ) ≥ , then cr( G Hv ) ≥ . Proof. We remark that the non-planarity of G and the fact that we are doing planar G p4c is not planar. This fact will be used throughoutthe proof.Let H (cid:48) = H − { x, y, z } and let N be the set of vertices t in { x, y, z } so that K max v ∩ K t is null. By way of contradiction, we suppose G Hv has a 1-drawing D .We start with two simple observations. Claim . Some edge of H (cid:48) is crossed in D . Proof. If no edge of H (cid:48) is crossed in D , then Definition 15.22 (1b) implies wemay resubstitute K v for H (cid:48) to obtain a 1-drawing of G , a contradiction. (cid:3) Claim . There is no drawing D (cid:48) of G Hv in which each crossed edge is incidentwith a vertex in H (cid:48) . Proof. Otherwise, insert a vertex at each crossing point, and add this vertexto H (cid:48) . Then contract every edge in the new graph that has both ends in H (cid:48) , andalso contract all the K u to single vertices. The result is a planar embedding of G p4c , a contradiction. (cid:3) Therefore, we may assume the crossing edges are e v ∈ H (cid:48) with some other edge f not incident with any vertex in H (cid:48) . Observe that H (cid:48) cannot be a single vertex. Claim . f is not a cut-edge of G Hv − H (cid:48) . Proof. Suppose f is a cut-edge of G Hv − H (cid:48) . Since D [ G Hv − H (cid:48) ] has no crossing,it is planar. Therefore, the faces on each side of f in D [ G Hv − H (cid:48) ] are the same.Thus, the ends of e v are in the same face of D [ G Hv − H (cid:48) ].Consider now the planar embedding D [ G Hv − e v ]. The two ends of e v are inthe same face of the subembedding D [ G Hv − H (cid:48) ] and so may be joined by an arcthat is disjoint from D [ G Hv − H (cid:48) ]. This produces a drawing of G Hv in which allthe crossings involve e v and edges incident with at least one vertex in H (cid:48) . Thiscontradicts Claim 2. (cid:3) Since f is not a cut-edge of G Hv − H (cid:48) , there is a cycle C f of G Hv − H (cid:48) containing f . Moreover, D [ C f ] separates the two ends of e v , so e v is an cut-edge of H (cid:48) . Let H and H be the two components of H (cid:48) − e v .The next claim is central to the remainder of the argument. Claim . Let t ∈ { x, y, z } be a common neighbour of H and H . Then f isincident with t (cid:48) ∈ K t and one of the faces of H (cid:48) + t (cid:48) incident with both t (cid:48) and e v isempty except for the segment of f from t (cid:48) to the crossing with e v . 38 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED Proof. Let C be any cycle in H (cid:48) + t (cid:48) containing e v . Since e v is a cut-edge of H (cid:48) , t (cid:48) ∈ C . Since G p4c − { v, t } is connected, G − ( K v ∪ K t ) is connected.Suppose by way of contradiction that there are vertices u and w of G − ( K v ∪ K t )on both sides of D [ C ]. By the preceding paragraph, there is a uw -path P in G − ( K v ∪ K t ). Since P is graph-theoretically disjoint from C , but D [ u ] and D [ w ]are on different sides of D [ C ], D [ P ] crosses D [ C ]; this must be at the unique crossingof D , so f ∈ P and the crossing of D [ P ] with D [ C ] is the crossing of f with e v .Moreover, D [ C f ] crosses D [ C ] at the crossing of D and so they must crosssomewhere else. As C f and H (cid:48) are disjoint, the second crossing is at the vertex t (cid:48) .Since this is true of any cycle C f in G − K v , f is a cut-edge of ( G − K v ) − t (cid:48) .We now consider two cases. Case 1: there are distinct vertices t and t of G p4c − { t, v } so that D [ K t ] and D [ K t ] are on different sides of D [ C ] . In this case, either (i) for some vertex s of G p4c , f ∈ K s , in which case t and t are in different components of G p4c − { t, v, s } , or (ii) since G p4c is non-planarand so has at least five vertices, for some vertex s of G p4c that is an end of f , wemay choose t and t to again be in different components of G p4c − { t, v, s } .In either case, the internal 4-connection of G p4c implies that there is an i ∈{ , } so that t i is the only vertex in its component of G p4c − { t, v, s } . But then tvt i is a 3-cycle in G p4c having v and t i as degree 3 vertices, contradicting Lemma15.15. Case 2: there are not distinct vertices t and t of G p4c − { t, v } so that D [ K t ] and D [ K t ] are on different sides of D [ C ] . In this case, there is a vertex s of G p4c − { t, v } so that f ∈ K s and all thevertices of G − ( K v ∪ K x ) on one side of D [ C ] are in one component K s of K s − f ,while all the other vertices of G − ( K v ∪ K x ), including the other component K s of K s − f , are on the other side of D [ C ].Lemma 15.13 implies that K s has neighbours in two K r ’s. According to D ,these can only be K v and K t . But now the 3-cycle tvs has the two degree 3 vertices v and s , contradicting Lemma 15.15.Since f is on both sides of D [ C ], but one side has no vertex, it must be thatthe end of f on that side is in C . But f is disjoint from H (cid:48) , and so this end canonly be t (cid:48) . (cid:3) Our proof proceeds by considering how many common neighbours among K x , K y , and K z there are for H and H . We start by noting that there cannot bethree, since then the graph H + is not planar, contradicting Definition 15.21. Claim . H and H have exactly one common neighbour. Proof. We have already ruled out the possibility that H and H have threecommon neighbours.To rule out two common neighbours, suppose by way of contradiction that H and H have the two common neighbours K x and K y . By the preceding remark,at least one of H and H does not have a neighbour in K z . Since H (cid:48) does havea neighbour in K z , we may choose the labelling so that H has a neighbour in K z and H does not.Claim 4 implies f is incident with both x (cid:48) and y (cid:48) . But now D [ f ] can be reroutedalong the other side of the x (cid:48) H -edges, around H , and on to y (cid:48) so that G Hv has no crossings. This implies the contradiction that G p4c is planar. We conclude that H and H have at most one common neighbour.If they have no common neighbours, then H has neighbours just in K x , while H has neighbours in K y and K z , but not in K x . In this case, e v is a cut-edgein H separating x from { y, z } . It follows that x / ∈ T . Since G v is also a ( T, U )-configuration, there is an cut-edge e (cid:48) v of G v separating x from { y, z } . Now we canreplace H (cid:48) in T with K v in such a way that e (cid:48) v (in fact the only edge of G v incidentwith x ) is crossed by f to yield a 1-drawing of G . This contradiction completes theproof of the claim. (cid:3) We conclude from Claim 5 that H and H have precisely one common neigh-bour x (cid:48) . Claim 4 implies that f is incident with x (cid:48) .If, for some i ∈ { , } , H i has no other neighbour, then we may reroute f to goaround D [ H i ], yielding a planar embedding of G Hv and, therefore, of the non-planargraph G p4c , a contradiction.Thus, we may choose the labelling so that H has at least one neighbour in K y , while H has at least one neighbour in K z . If, say, H is joined to K y by onlyone edge, then y / ∈ T ; therefore, y is incident with a unique edge in G v and we canreplace D [ H ] with the planar embedding of K v so that it is the yK v -edge that iscrossed by f . This yields that contradiction that G has a 1-drawing.Thus, we may assume that T = { x, y, z } . However, there are not edge-disjoint yz -paths in H − x ( e v is a cut-edge separating y and z ). Therefore, U = { y, z } ,showing G v is doglike. It follows that G v − x has a cut-edge e (cid:48) v separating y and z . We may substitute the planar embedding of K v for D [ H ] so that e (cid:48) v crosses f ,yielding the final contradiction that G has a 1-drawing. In this section, we show that if G is a 3-connected 2-crossing-critical graph thatreduces by planar 3-reductions to a peripherally-4-connected graph, then there isa “basic” 3-connected 2-crossing-critical graph from which G is obtained by theregrowth mechanism of the preceding section. Theorem . Let G ∈ M reduce by planar 3-reductions to a peripherally-4-connected graph G p4c . Let v be a vertex of G p4c with just the three neighbours x , y , and z , so that ( G v , { x, y, z } ) is a ( T, U ) -configuration and K v has at least twovertices. Let G rep( v ) be the graph obtained from G by contracting as indicated inthe following cases.(1) If ( G v , { x, y, z } ) is doglike, then let e be the cut-edge of K v and contracteach component of K v − e to a vertex.(2) If ( G v , { x, y, z } ) is not doglike, then we have the following subcases.(a) If none of G x , G y , and G z is doglike, then contract K v to a vertex.(b) If ( | T | , | U | ) = (3 , , then contract K v to a vertex.(c) If G x is doglike and y / ∈ T , then let C be a cycle in G + v containing x (cid:48) , y (cid:48) , and z (cid:48) , delete everything in K v − E ( C ) and contract the edgesof C to the 3-cycle x (cid:48) y (cid:48) z (cid:48) .Then G rep( v ) ∈ M . There is one clarification that is required to understand one fine detail of G rep( v ) .If, for example, the vertex x (cid:48) is in K v , then we proceed precisely as described in the 40 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED statement. If, however, x (cid:48) is in K x and x ∈ T , then in G rep( v ) we retain only twoedges between x (cid:48) and the contracted vertex in K rep( v ) to which it is joined. Thisespecially applies in the case 2c: if z (cid:48) ∈ K z , then we keep only the two edges of C incident with z (cid:48) , while if z (cid:48) ∈ K v , then we keep all the z (cid:48) K z -edges.There is also an important remark to be made. We had long thought thatit was possible to reduce each K v to a single vertex and retain 2-criticality. Thismight be true in the particular cases of 3-connected 2-crossing-critical graphs withno subdivision of V , but it is certainly not true of all 3-connected 2-crossing-criticalgraphs.In Definition 2.10 we described the set S of all graphs that can be obtainedfrom the 13 tiles and the two frames. These graphs are all 3-connected and 2-cros-sing-critical. Consider any one of these that uses the right-hand frame in Figure2.1 and uses the second picture in the third row of Figure 2.2. With appropriatechoices of the neighbouring pictures, the 3-cycle in the upper half of the picture ispart of a doglike G v that contains the parallel edges in the picture and the paralleledges in the frame: the horizontal edge in the 3-cycle is K v . The vertical edge inthe other 3-cycle in the picture is a K x . When we do the planar 3-reductions inthis case, the contractions of K x and K v produce a pair of parallel edges not inthe rim. The conclusion is that the resulting peripherally-4-connected graph plusparallel edges is not 2-crossing-critical. Thus, the technicalities we must endure inthe statement of Theorem 15.25 seem to be unavoidable. Proof. We use the notation K rep( v ) for the contraction of K v in G rep( v ) . Phase 1: showing G rep( v ) is 3-connected. Let t and u be vertices of G rep( v ) . We show G rep( v ) − { t, u } is connected.Let w t and w u be the vertices of G p4c so that t ∈ K w t and u ∈ K w u (taking, forexample, K w t to be K rep( v ) if t ∈ K rep( v ) ). It follows from Lemma 15.16 that everyvertex of every K s has a path in G − { t, u } to at least one neighbour of K s that isnot one of K w t or K w u . This is also true of K rep( v ) , as may be seen by checkingthe analogues for K rep( v ) of Lemma 15.16 in the three cases for which K rep( v ) hasat least two vertices. (Note there are two possible outcomes for K rep( v ) in Case 2c,depending on whether z (cid:48) ∈ K v , in which case K rep( v ) is a 3-cycle, or z (cid:48) ∈ K z , inwhich case K v is an edge.)Since each K s is connected, G rep( v ) − { t, u } is connected. Phase 2: showing cr( G rep( v ) ) ≥ . The graph ¯ K rep( v ) obtained from K rep( v ) by adding x , y , and z is a ( G, K v )-compatible ( T, U )-configuration. Therefore, Phase 2 follows immediately from The-orem 15.24. Phase 3: showing that G rep( v ) is 2-crossing-critical. Let e be any edge of G rep( v ) . Then there is an edge e G in G naturally corre-sponding to e (in the sense that precisely the same contractions and deletions of G and G − e G can be used to obtain both G rep( v ) and G rep( v ) − e ). Special situation. There is one case where the choice of e G must be madewith special care. Suppose K v contracts down to the single vertex v and e is oneof two parallel edges vx . In the case K v has a cut-edge e (cid:48) , Lemma 15.13 implieseach component of K v − e (cid:48) is joined to two of the neighbours of v . Suppose that K x is the only common neighbour of these two components. Since G v is not doglike,some component L of K v − e (cid:48) is joined by exactly one edge to its other neighbour;choose e G to be an xL -edge. Definition . For each vertex w of K rep( v ) , L w denotes the subgraph of K v that contracts to w .Since G is 2-crossing-critical, there is a 1-drawing D of G − e G . If no edgeof any L w ⊆ K v is crossed in D , then these may each be contracted to obtain a1-drawing of G rep( v ) − e , and we are done. Claim . If there is a drawing of G − e G in which all the crossings are betweenedges incident with vertices in L w , then G rep( v ) − e is planar. Proof. Insert vertices at each crossing point and contract every edge in thenew graph that has both ends in some L u . The result is a planar embedding of G rep( v ) − e . (cid:3) Therefore, we may assume the crossing edges are e v ∈ L w ⊆ K v with someother edge f not incident with any vertex in L w . Case 1: f is a cut-edge of ( G − e G ) − L w . In this case, D [( G − e G ) − L w ] has no crossing, so it is planar. Therefore, thefaces on each side of f in D [( G − e G ) − L w ] are the same. Thus, the ends of e v arein the same face of D [( G − e G ) − L w ].Consider now the planar embedding D [( G − e G ) − e v ]. The two ends of e v arein the same face of the subembedding D [( G − e G ) − L w ] and so may be joined byan arc that is disjoint from D [( G − e G ) − L w ]. This produces a drawing of G − e G in which all the crossings involve e v and edges incident with at least one vertex in L w . Claim 1 implies G rep( v ) − e is planar, as required. Case 2: f is not a cut-edge of ( G − e G ) − L w . In this case, f is in a cycle C f of ( G − e G ) − L w . Moreover, D [ C f ] separatesthe two ends of e v , so e v is a cut-edge of L w . Let L w and L w be the componentsof L w − e v .We consider separately two cases for G v . Subcase 2.1: G v is doglike. In this subcase, K rep( v ) is two vertices w and ¯ w joined by a cut-edge e (cid:48) of G v − x , each joined by an edge to x (cid:48) , w is joined by at least two edges to K y and¯ w is joined by at least two edges to K z . Lemma 15.20 implies that K x has at mosttwo neighbours in K v . We already know there is one in each of L w and L ¯ w . Lemma15.11 now implies there is a vertex x (cid:48) ∈ K x incident with all the K v K x -edges in G .Thus, we may choose the labelling of L w and L w so that the neighbour of x (cid:48) in L w is in L w .We see that x (cid:48) and the end of e v in L w are neighbours of vertices in L w , andneither of these vertices is in L w . The only other possibilities for neighbours of L w outside of L w are in K y and L ¯ w , the latter being the end of e (cid:48) . A similar remarkholds for L w : it has the neighbour (via e v ) in L w , and can have at most neighboursin K y and L ¯ w (via e (cid:48) ).Since G is 3-connected, for each i = 1 , L iw has at least two neighbours outsideof L iw other than x (cid:48) . From the neighbour analysis of the preceding paragraph, there 42 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED are at most three in total: two to K y and one to L ¯ w . There are two ways this canhappen.In the first way, both edges from L w to K y have their ends in L w , while e (cid:48) hasan end in L w . But then e v is a cut-edge of K v that violates Lemma 15.13: the edge e G cannot connect L w to either x (cid:48) (Lemma 15.20 or K z (because e (cid:48) is a cut-edgeof G v − x ), so the component L w of K v − e v is joined only to K y .Therefore, e (cid:48) has one end in L w and the two K v K y edges have ends in differentones of L w and L w . It follows that y (cid:48) is incident with these edges, so Lemma 15.20implies y (cid:48) has precisely these neighbours in K v .Contract D [ e v ] so that L w is pulled across f and, if necessary, shrink D [ L w ] sothat we obtain a new drawing D of G − e G in which f crosses the edges from x (cid:48) and y (cid:48) to L w . Claim . f / ∈ L ¯ w . Proof. If f ∈ L ¯ w , then exactly the same analysis as for L w implies that L ¯ w − f has two components L w , from which there is an edge to x (cid:48) and an edge to z (cid:48) , and L w , from which there is an edge to z (cid:48) and L w . But now the graph-theoreticallydisjoint cycles in L w + y (cid:48) containing e v and L ¯ w + z (cid:48) containing f cross exactly oncein D , which is impossible. (cid:3) It follows from Claim 2 that f / ∈ L ¯ w . We contract the uncrossed D [ L w ] and D [ L ¯ w ] to obtain a drawing D of G rep( v ) − e , in which the only crossings are of f with the edges from x (cid:48) and y (cid:48) to L w . In D , there are parallel edges y (cid:48) w ; the onefrom y (cid:48) to L w is not crossed in D , so we may make all the others go alongside theuncrossed one. This yields a drawing D of G rep( v ) − e in which the only crossingis x (cid:48) w with f , so D is a 1-drawing of G rep( v ) − e , as required. Subcase 2: G v is not doglike. Subcubcase 2.1: there is a neighbour x of v in G p4c so that G x is doglike and x (cid:48) ∈ K v is the nose of G x . Let C be the cycle in G v that we contracted to the 3-cycle x (cid:48) y (cid:48) z (cid:48) . We let G C be the subgraph of G obtained by deleting all edges between the various L u exceptthe one or three edges in C . Choose the labelling so that y is a neighbour of v in G p4c so that there is exactly one K v K y -edge in G ; thus y (cid:48) ∈ K v .Let r be that element of { x, y, z } so that r (cid:48) ∈ L w . There are precisely twoedges e and e in G C coming out of L w in G v − r .Let L w be the component of L w − e v containing r (cid:48) and let L w be the other.Since C goes through r (cid:48) , at least one of e and e is incident with a vertex in L w .Therefore, at most one of e and e has an end in L w .We claim that L w is not joined to any other vertex in G C . The only possibilityis that there is an edge from L w to K x ∪ K y ∪ K z . Since all the K v K x - and K v K y -edges in G are incident with x (cid:48) and y (cid:48) , respectively and x (cid:48) and y (cid:48) are not in L w ,there are no edges in G from L w to K x ∪ K y .As for the possibility of an L w K z -edge, this can only exist if z (cid:48) ∈ K z . But z (cid:48) already has two known neighbours in K v , namely the K v -ends of the edges of C incident with z (cid:48) . Lemma 15.20 implies these are the only vertices of K v adjacent tovertices in K z . Therefore these known z (cid:48) -neighbours are the only ones; in particular, z (cid:48) has no neighbour in L w , as claimed. We obtain a 1-drawing of G rep( v ) − e by partially contracting D [ e v ] and, ifnecessary, scaling D [ L w ] down so that L w and L w are now drawn on the same sideof f . The only crossing in this new drawing is of the edge of D [ G C ], if it exists,that is not e v and joins L w to the rest of G C . Now we may contract all the L u tosingle vertices to obtain the required 1-drawing of G rep( v ) − e . Subsubcase 2: there is no neighbour x of v in G p4c so that G x is doglike and x (cid:48) ∈ K v is the nose of G x .. At this stage, K v contracts to a single vertex of G rep( v ) . In this case, K v − e v has two components K v and K v . Lemma 15.13 implies each of K v and K v areconnected in G to at least two of K x , K y and K z . Because G + v is planar, at mosttwo of K x , K y , and K z can be adjacent to both K v and K v .If both K x and K y have neighbours in both K v and K v , then there is an i ∈ { , } so that K iv has adjacencies only in those two. Now pull D [ K iv ] across f and, scaling D [ K iv ] if necessary, to obtain a planar embedding of G − e G . Thiscontracts to a planar embedding of G rep( v ) − e , as required.Thus, we may assume K v and K v have precisely one common neighbour in G .Each has its own neighbour. Since G v is not doglike, one of these, say K v , is joinedby a single edge to that unique neighbour and now we can drag K v across f . Thisworks unless e goes to K v and K v is joined to its unique neighbour by two edges.But this is the special situation, and e is joined to K v , not K v . The important corollary of Theorem 15.25 is that, if we replace each K v withits K rep( v ) , then we get a 2-crossing-critical model of G p4c with very simple re-placements for the vertices of G p4c . In this section, we explain how to obtain allthe 3-connected 2-crossing-critical graphs that reduce by planar 3-reductions to aparticular peripherally-4-connected graph.Let L be a non-planar peripherally-4-connected graph. For each vertex v of L having only three neighbours x , y , and z , we decide on the type of v ; thatis, we choose T v ⊆ { x, y, z } and, in the case | T v | = 3, we decide on U v : either U v = { x, y, z } , or U v consists of two of { x, y, z } . For each edge of L joining twovertices of degree at least 4, we decide whether the edge will be a single edge or aparallel pair.The choices must be made so that x ∈ T v if and only if v ∈ T x . If, for some v ,( | T v | , | U v | ) = (3 , 2) ( v is chosen to be doglike), then some other implications (as inTheorem 15.25) must be maintained. Choose the labelling so that x / ∈ U v . Then x is the nose of the dog, v is replaced with K v , so that K v is an edge y (cid:48) z (cid:48) , so that y (cid:48) incident with two edges going to K y , and likewise for z (cid:48) to K z . Each of y (cid:48) and z (cid:48) is also incident with an edge to x (cid:48) ∈ K x . Furthermore, K x can be either a vertex,or, if | T x | (cid:54) = 3, an edge, or a 3-cycle.Once all these choices have been made, the resulting graph is tested for 2-criticality. Thus, for a given peripherally-4-connected graph L , there will be manygraphs that require testing. If one of the resulting graphs L (cid:48) is found to be 2-cros-sing-critical, then there may be many other 3-connected 2-crossing-critical graphsthat arise from L (cid:48) . Recall that, for each vertex of L that has only three neighbours, 44 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED we have made a choice as to what type that vertex has. The following lemmaexplains what may replace the vertex of each type. Lemma . Suppose the peripherally-4-connected graph L has choices asexplained in the preceding paragraphs to produce a 3-connected 2-crossing-criticalgraph L (cid:48) . Suppose G is a 3-connected 2-crossing-critical graph that reduces by planar3-reductions to L so that L (cid:48) is the graph obtained from G by the replacementsdescribed in Theorem 15.25. Then, for each K v in L (cid:48) , K v is replaced by one of thepossibilities shown in Figures 15.1, depending on ( T v , U v ) . Proof. We only illustrate the tedious proof in a couple of cases. Case 1: ( T v , U v ) = ( { x, y, z } , { y, z } ).Let e be a cut-edge in G v − x separating y and z . Let K v − e have the two com-ponents K yv , containing the neighbour(s) of y , and K zv , containing the neighbour(s)of z . If K yv , for examples, is not just either a single vertex or an edge joining thetwo neighbours of y , then it contains a subdivision of one of these (either pick apath in K yv joining the neighbour of y to the K yv -end of e or pick a path joining thetwo neighbours of y ). It is easy to see that the subdivision (making a similar choiceon the z -side) is also a ( T v , U v )-configuration. By Theorem 15.24, the subgraph hascrossing number 2, and so is all of G . Thus, K v can be at most one of the threefigures in Figure 15.1 corresponding to ( | T | , | U | ) = (3 , Case 2: T v = { x, y, z } = U v .In this case, G v − x contains edge-disjoint yz -paths. Therefore, it contains twosuch paths P and Q that make a digonal pair. If P and Q are internally disjoint,then there is a ( P − { y, z } )( Q − { y, z } )-path R . If P and Q are not internallydisjoint, then set R = ∅ . In either case, set M = P ∪ Q ∪ R . There are two x ( M − { y, z } )-paths R and R in G v .If the ends of P and Q are in the same digon of P ∪ Q , then planarity of G + v implies R and R have their ends in the same one of P and Q . It follows that M ∪ R ∪ R is a ( T v , U v )-configuration, and so is G v by 2-criticality and Theorem15.24.The fact that G is 3-connected implies that there cannot be more than fourcommon internal vertices to P and Q , as if there were six digons, then some twoconsecutive ones would not contain an end of either R or R . This would readilyyield a 2-cut in G , which is impossible. This is why the number of possibilities for G v in this case is finite.In some of the larger ( T, U )-configurations, there are edges that are not requiredto produce the relevant paths between s , t , and u , but, rather, are there to maintainthe connectedness of the configuration. These edges might be deletable withoutreducing the crossing number below 2. Thus, each candidate 3-connected graphproduced by the method described needs to have its criticality checked. In order to find the 2-crossing-critical graphs that do not contain V , we wishto use the characterization by Robertson of V -free graphs. This characterization,described in the next section, is in terms of internally-4-connected graphs . These Figure 15.1. The possible ( T, U )-configurations.graphs are very closely related to peripherally-4-connected graphs and it is thepurpose of this section to describe the reduction of a peripherally-4-connected graphto an internally-4-connected graph, and back again. Definition . A hug in a graph G is an edge e in a triangle T whosevertex v not incident with e has degree 3. The triangle T is the e -triangle , v is the head of the hug and the two edges of T other than e are the arms of the hug. Definition . A G is internally-4-connected if it is peripherally-4-connec-ted and has no hugs. 46 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED utv zyxw Figure 15.2. The thick edge is a bear hug. The dotted edges tw and vz might be subdivided, and the dashed edge uw need not bepresent. If uw is not present, then { ux, uy } is a simultaneouslydeletable pair of bear hugs.It is not correct that simply deleting (successively) the hugs from a peripherally-4-connected graph produces an internally-4-connected graph. There is a particularsituation that arises that needs special care. Definition . (1) A hug e with head v is a bear hug if there is anend u of e , incident with a second hug uy whose head t is different from v ,and so that, with w the other end of e , the neighbours of u are containedin the union of { t, v, w } and the set of neighbours of t . (See Figure 15.2.)(2) A hug is deletable if it is not a bear hug.(3) A pair of bear hugs having a common end is simultaneously deletable .We are now in a position to reduce a peripherally-4-connected graph to aninternally-4-connected graph. Theorem . Let G be a non-planar peripherally-4-connected graph and let G = G , G , . . . , G k be a sequence of graphs so that, for each i = 1 , , . . . , k , thereis either a hug h i or a simultaneously deletable pair h i of bear hugs in G i − so that G i = G i − − h i . Then, for i = 0 , , , . . . , k :(1) G i is a subdivision of a non-planar peripherally-4-connected graph;(2) if v has degree 2 in G i but not in G i − , then h i is a simultaneously deletablepair of bear hugs in G i − , both incident with v ; and(3) every degree 2 vertex in G i has two degree 3 neighbours in G i .Furthermore, if the sequence G , G , . . . , G k is maximal, then G k is a subdivi-sion of an internally-4-connected graph. We emphasize that, in the reduction process described in the statement, G i isobtained from G i − by the deletion of either one or two edges. Proof. Suppose by way of contradiction that i is least so that G i is planar. Since G is not planar, i > 0, so G i = G i − − h i . Each edge in h i joins two neighboursof a degree 3 vertex in G i and so may be added to the planar embedding of G i to produce a planar embedding of G i together with that edge of h i . In the case | h i | = 2, the heads of the hugs are not adjacent. Thus, both hugs may be addedsimultaneously, while preserving planarity. Thus, G i − is planar, contradicting thechoice of i .By way of contradiction, we may let i be least so that G i is not a subdivisionof a peripherally-4-connected graph. Thus, i ≥ 1. Throughout the proof, whenwe refer to the vertices t, u, v, w, x, y, z , we are always referring to the labelling inFigure 15.2. In each of the three cases, there are two possibilities for h i to beconsidered.It will be helpful to notice that, in the case h i consists of a simultaneouslydeletable pair of bear hugs, the vertex u is not a node of G i and is incident withboth deleted edges. Claim . G i is a subdivision of a 3-connected graph. Proof. Let a and b be distinct nodes of G i . Then a and b are distinct nodesof G i − , so there are three internally disjoint ab -paths P , P , P in G i − .If e ∈ h i , then the head c of the e -triangle has degree 3. If e is in some P i and T is the triangle containing e and its head, then we may replace P i ∩ T with thepath in T complementary to P i ∩ T . The at most two modifications result in threeinternally disjoint paths that are also paths in G i . (cid:3) Claim . If a has degree at least three in G i − and degree 2 in G i , then:(1) | h i | = 2;(2) a is incident with both edges in h i ; and(3) both neighbours of a have degree 3 in G i . Proof. Let e ∈ h i . The head b of the e -triangle has degree 3 in G i − and,since G i − is a subdivision of a peripherally-4-connected graph, no other vertex ofthe e -triangle has degree 3, so Lemma 15.15 shows they both have degree at least4. It follows that if e is the only edge in h i , then the ends of e have degree at least3 in G i and no new vertex of degree 2 is introduced in G i .Therefore h i is a deletable pair. The only new vertex of degree 2 in G i is u , so a = u . Also, the only neighbours of u in G i have degree 3 in G i . (cid:3) The remaining possibility is that there is a set { a, b, c } of nodes of G i and a3-separation ( H, J ) of G i so that H ∩ J = (cid:107) a, b, c (cid:107) and both H − { a, b, c } and J − { a, b, c } have at least two nodes of G i .Because G i − is a subdivision of a peripherally-4-connected graph, there is anedge e ∈ h i having one end r H in H − { a, b, c } and one end r J in J − { a, b, c } .Suppose for the moment that h i has a second edge. Since G i − is a subdivisionof a peripherally-4-connected graph, not all the neighbours of u in G i − can be inthe same one of H and J . We may choose the labelling so that x = r J . As t is acommon neighbour of u = r H and x = r J , we conclude that t ∈ { a, b, c } , say t = a .It follows that at least one of v and y (the other two neighbours of u ) is in H −{ t, b, c } . Since v and y are adjacent, it follows that both are in H and, furthermore, uy is also in H . In particular, there is a unique edge in h i that has one end in H − { a, b, c } and one end in J − { a, b, c } .Now the two possibilities for h i are merged: e is the unique edge in h i havingone end r H in H − { a, b, c } and one end r J in J − { a, b, c } . The head q of the e -triangle must be in { a, b, c } , say q = a . 48 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED b cr J r H q Figure 15.3. When s = b , G i − is a subgraph of the illustratedplanar graph.Since q has degree 3, we may choose the labelling so that r H is the only neigh-bour of q in H − { q, b, c } . The neighbour r J of q is in J − { q, b, c } . Note that r H and r J are both nodes of G i − .The third neighbour s of q is in J , so { r H , b, c } is a 3-cut in G i − . Since G i − is peripherally-4-connected, there is a unique node p in H − r H , which is joined bybranches in G i − to all of r H , b , and c .If s ∈ { b, c } , then the discussion in the preceding paragraph applies with r J and J in place of r H and H , respectively. The nodes of G i − are now all known(there are only 7), and the edges are almost completely determined. In particular, G i − is a subgraph of the planar graph shown in Figure 15.3, contradicting the factthat G i − is non-planar. Therefore, s is in J − { q, b, c } .The vertex r H is the only candidate for the second branch vertex (after p ) of G i in H − { q, b, c } , so it must be joined by a G i -branch to at least one of b and c ;choose the labelling so that b is an end of such a G i -branch.If b has only one neighbour in J − { q, b, c } , then p and b are both degree 3vertices in a triangle in G i − ; since G i − is a subdivision of a peripherally-4-con-nected graph, this contradicts Lemma 15.15. The same reasoning implies that both r H and b have degree at least 4 in G i − . These imply that r H p , r H b , and pb are alledges of G i − .Because r H r J is in h i and q is the head of the r H r J -triangle, we know that r H r J , qr H , and qr J are all edges of G i − . Furthermore, r H s is not a G i − -branch(it would yield a second edge with one end in each of H −{ a, b, c } and J −{ a, b, c } ).The triangles pr H b and qr H r J show that r H r J is a bear hug. Since it wasdeleted, it must be in a simultaneously deletable pair of bear hugs. This impliesthat r H b is the other edge in that pair. Thus, H − { a, b, c } has only one nodein G i , a contradiction that completes the proof that each G i is a subdivision of aperipherally-4-connected graph.We move on to showing that a maximal sequence ends in a subdivision of aninternally-4-connected graph. So suppose G i is not a subdivision of an internally-4-connected graph. Since it is a subdivision of a peripherally-4-connected graph V -FREE 2-CROSSING-CRITICAL GRAPHS 149 H , there is a 3-cut { a, b, c } in H so that ab is an edge of G i . Since H is periphe-rally-4-connected, there is a vertex p adjacent in H to all of a, b, c and with noother neighbours in H . Lemma 15.15 shows that the triangle p, a, b has at mostone vertex of degree 3; since p is such a vertex, a and b have degree at least 4 in H . It follows that pa and pb are edges of G i and, therefore, ab is a hug in G i .It is evident from the definitions that, as soon as G i has a hug, then either G i has a hug that is not a bear hug or G i has a pair of simultaneously deletable bearhugs. In either case, G i is not the last in a maximal sequence.We conclude this section with a brief discussion of the reverse process: howto generate all the peripherally-4-connected graphs that reduce to a given non-planar internally-4-connected graph G . Every graph created through iterating thefollowing procedure is peripherally-4-connected and non-planar. We choose eithertwo non-adjacent neighbours of a degree 3 vertex and add the edge between them,or we choose an edge e joining degree 3 vertices and a neighbour of each vertexincident with e , subdivide e once, and join both the chosen neighbours to thevertex of subdivision.Every internally-4-connected graph produces only finitely many peripherally-4-connected graphs through this process, as the number of possible additions isinitially finite and strictly decreasing. V -free 2-crossing-critical graphs In this section, we complete our analysis of peripherally-4-connected 2-crossing-critical graphs by considering the case of 3-connected 2-crossing-critical graphs thatdo not contain a subdivision of V . This is the whole reason for studying periphe-rally-4-connected graphs, since there is a characterization of the closely relatedinternally-4-connected graphs that do not contain a subdivision of V .Two important classes of graphs in this context are the following. Definition . (1) A bicycle wheel is a graph consisting of a rim ,which is a cycle C , and an axle , which is consists of two adjacent vertices x and y not in the rim, together with spokes , which are edges from { x, y } to C .(2) A is a graph G containing a set W of four vertices so that G − W has no edges.Maharry and Robertson [ ] prove Robertson’s Theorem that an internally-4-connected graph with no subdivision of V is one of the following:(1) a planar graph;(2) a non-planar graph with at most seven vertices;(3) C (cid:50) C ;(4) a bicycle wheel; and(5) a 4-covered graph.Suppose G is a 3-connected graph that does not contain a subdivision of V and G reduces by planar 3-reductions to the peripherally-4-connected graph G p4c .It follows that G p4c has no V . Eliminating hugs as described in Theorem 15.31produces an internally-4-connected graph G i4c . Deleting hugs does not affect theplanarity of the graph; since G p4c is not planar, so is G i4c . By Robertson’s Theorem,one of the following happens: 50 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED (1) G i4c is not planar and has at most seven vertices;(2) G i4c is C (cid:50) C ;(3) G i4c is a bicycle wheel; and(4) G i4c is a 4-covered graph.Our ambition in the remainder of this section is to explain how to determineall the peripherally-4-connected graphs G p4c that can be the outcome of a sequenceof planar 3-reductions starting from a 3-connected, 2-crossing-critical graph G thathas no subdivision of V . Any peripherally-4-connected graph with no subdivisionof V that either has crossing number exactly 1 or is itself 2-crossing-critical needsto be tested. Those with crossing number 1 might extend to a 2-crossing-criticalexample by duplication of edges and/or replacing vertices of degree 3 by one of thebasic ( T, U )-configurations, as explained in the preceding section.The first two items arising from Robertson’s Theorem are easily dealt with. Acomputer program can easily find all internally-4-connected graphs with at most 7vertices and determine which ones either have crossing number 1 or are 2-crossing-critical. The graph C (cid:50) C is itself 2-crossing-critical, so this is one of the 3-connected, 2-crossing-critical graphs that do not contain a subdivision of V . Definition . Let G p4c be a peripherally-4-connected graph and let G i4c be the internally 4-connected graph obtained from G p4c by simplifying (that is,leaving only one edge in each parallel class) and eliminating hugs. Then G p4c is a peripherally-4-connected extension of G i4c .We conclude this section by showing how to which bicycle wheels and 4-coveredgraphs G i4c can have such a 2-crossing-critical G p4c as an extension. In particular, G i4c must either have crossing number 1 or itself be 2-crossing-critical; in the lattercase G p4c = G i4c . CASE 1: the bicycle wheels. Let x and y be the adjacent vertices making the axle of the bicycle wheel G i4c ,and let C be the cycle that is the rim. Our goal is to provide sufficient limitationson C to show that the computation is feasible. Here is our first limitation, whichcan very likely be improved. Lemma . Suppose G ∈ M reduces by planar 3-reductions to the graph G p4c that is a peripherally-4-connected extension of G i4c . If G i4c is a bicycle wheelwith axle xy and rim C , then x is not adjacent in G i4c to six consecutive verticeson C , none of which is adjacent to y . Proof. Suppose by way of contradiction that x , x , x , x , x , x are six consecutive(in this order) vertices of C adjacent to x but not y . Lemma 15.15 implies no twoconsecutive ones of these vertices have only three neighbours in G p4c . By symmetry,we may assume x has a neighbour u that is not adjacent to x in G i4c .Because G p4c is a peripherally-4-connected extension of G i4c , there are vertices w and z so that x , u , and w are the neighbours (in both graphs) of z and no othervertex has just these three neighbours. Since y is not adjacent to x and x has morethan 3 neighbours, z ∈ C . If follows that x and u are the C -neighbours of z and w isthe neighbour of z that is in { x, y } . In particular, z , being a neighbour of x is either x or x , so w = x . In either case, three consecutive vertices from x , x , . . . , x are V -FREE 2-CROSSING-CRITICAL GRAPHS 151 such that the outer two are adjacent by a chord in G p4c ; if necessary, we relabel sothese are x , x , x . In particular, x has just three neighbours in G p4c .Let D be a 1-drawing of G p4c − xx and let K be the subgraph of G p4c − xx induced by x , x , x , and x . Claim . K is clean in D . Proof. In G p4c − xx , x has only two neighbours, so the edge x x and thepath ( x , x , x ) make a pair of parallel edges. Therefore, we may assume neitherof these is crossed in D .It suffices to prove that xx is not crossed in D , as the proof for xx is sym-metric. Suppose by way of contradiction that xx is crossed in D and consider theplanar embedding of G p4c − { xx , xx } induced by D . Since G i4c − { xx , xx } is asubgraph, it is also planar, embedded in the plane by D .Since x has only three neighbours in G i4c − { xx , xx } , we can add the edge xx alongside the path ( x, x , x ) to obtain a planar embedding of G i4c − xx . Thenwe may add the edge xx alongside the path ( x, x , x ) to get a planar embeddingof G i4c . However, this contradicts the fact that G i4c is not planar. (cid:3) Now let K be the subgraph of G p4c − xx induced by x , x , x , and x . Because x , x , and x are consecutive along C , there is a unique K -bridge B in G p4c − xx .The claim shows K is clean in D , so D [ B ] is contained in one face F of D [ K ].Adjusting which of D [ x x ] and D [( x , x , x )] is which, if necessary, we mayarrange D so that both x and x are incident with a face of D [ K ] that is not F .This permits us to add xx to D without additional crossings, to obtain a 1-drawingof G . This final contradiction yields the result.Along the same lines, we have the following limitation. Lemma . Suppose G ∈ M reduces by planar 3-reductions to the graph G p4c that is a peripherally-4-connected extension of G i4c . If G i4c is a bicycle wheelwith axle xy and rim C , and there are four distinct vertices of C adjacent to both x and y , then these are the only six vertices of G i4c . Proof. Suppose to the contrary that u , u , u , and u are distinct vertices of C adjacent to both x and y in G p4c and there is another vertex u . We may choosethe labelling of x and y so that xu ∈ G i4c . Let D be a 1-drawing of G p4c − xu .Let K be the subgraph of G p4c − xu consisting of C and all edges between x and vertices of C . (We do not include any chords of C that might exist in G p4c .)If x and y are both in the same face of D [ C ], then y is in some face F of D [ C ]and at least two of u , u , u , and u are not incident with F . This implies thecontradiction that D has at least two crossings.We conclude that y is not in the same face of D [ C ] with x . It follows that xy crosses C in D and this is the only crossing. We claim we can add the edge xu to D to obtain a 1-drawing of G p4c .Let F be the unique face of D [ K ] incident with both u and u and let C (cid:48) bethe cycle bounding F . If we cannot add xu in F , then there is an edge e of G p4c that has an end in each of the two components of C (cid:48) − { x, u } . Since C (cid:48) − x ⊆ C ,it follows that both ends w and w of e are in C .Since e is not an edge of G i4c , there are vertices w and z of G p4c so that z has just the neighbours w , w , and w . Since both x and y have at least four 52 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED neighbours, z / ∈ { x, y } . Since one of x and y is a neighbour of z , w ∈ { x, y } .Finally, z has at least two neighbours in C , so these are w and w . We concludethat z = u .We note that xy cannot cross the 3-cycle u w w in D . Therefore, we can move w w to the face of D [ C ] that contains y ; in this new 1-drawing of G p4c − xu , x and u are incident with the same face, giving the contradiction that G p4c has a1-drawing.The final limitation is the following. Lemma . Suppose G ∈ M reduces by planar 3-reductions to the graph G p4c that is a peripherally-4-connected extension of G i4c . Suppose G i4c is a bicyclewheel with axle xy and rim C , and there are six distinct vertices x , y , x , y , x , y in this cyclic order on C , so that, for i = 1 , , , , x i is adjacent to x and y i isadjacent to y . Then these are the only six vertices of C . We remark that we allow for the possibility that some (or all) of the x i are alsoadjacent to y and, likewise, some of the y i can be adjacent to x . Proof. Suppose to the contrary that there is another vertex u in C . If possible,choose the x i , y i and u so that u is adjacent to only one of x and y . We may assumethat u occurs between x and y in the cyclic order on C . By the choice of the x i , y i , and u , if u is adjacent to both x and y , then so are x and y and all verticesbetween them on C .Let D be a 1-drawing of G p4c − xu . Let K be the subgraph of G p4c − xu consisting of C and all edges between x and vertices of C . (We do not include anychords of C that might exist in G p4c .) If x and y are on the same side of D [ C ],then at most one of the y i is incident with the face of D [ K ] containing y , showing D has at least two crossings, a contradiction. Therefore, the crossing of D is of xy with an edge of C .There is a face of D [ K ] incident with both x and u ; let C (cid:48) be its boundingcycle. If we cannot add xu to D , it is because there is an edge e of G p4c − xu withan end in each of the components of C (cid:48) − { x, u } . Since C (cid:48) − x ⊆ C , it follows thatthe ends w and w of e are both in C . Because G p4c is a peripherally-4-connectedextension of a bicycle wheel, there are vertices z and w so that z has only theneighbours w , w , and w .Both x and y have at least four neighbours in G i4c , so z / ∈ { x, y } ; thus, z ∈ C .Since z has two neighbours in C and at least one in { x, y } , it follows that w ∈{ x, y } , while w and w are the two C -neighbours of z . Therefore, z = u . As u isadjacent to x , we conclude that u is not also adjacent to y . But now we can movethe edge w w to the other side of C so that the resulting 1-drawing of G p4c − xu extends to a 1-drawing of G , a contradiction.Lemmas 15.34, 15.35, and 15.36 effectively limit the possibilities for G i4c . Eachof these must be checked for either having crossing number 1 or being 2-crossing-critical. Those with crossing number 1 must have their peripherally-4-connectedextensions tested for 2-criticality. No matter what improvement is made to Lemma15.34, this will require computer work to complete. CASE 2: the 4-covered graphs. V -FREE 2-CROSSING-CRITICAL GRAPHS 153 We begin our analysis by describing three particular internally-4-connected 2-crossing-critical graphs that are 4-covered. Definition . (1) The 3-cube Q is the 3-regular, 3-connected, pla-nar, bipartite graph with 8 vertices.(2) The graph Q v is the bipartite graph obtained from Q by adding one newvertex joined to all four vertices on one side of the bipartition of Q .(3) The graph Q e is the bipartite graph obtained from Q by adding two ofthe four missing (bipartite-preserving) edges.(4) The graph Q t is the graph obtained from Q by adding a 3-cycle abc onone side of the bipartition of Q together with one edge joining the fourthvertex d of the same part to the non-adjacent vertex in the other part ofthe bipartition. Lemma . The graphs Q v , Q e , and Q t are all 2-crossing-critical. Proof. We start with the following observation. Claim . If D is a 1-drawing of Q , then D is the unique planar embeddingof Q . Proof. If e and f are two non-adjacent edges of Q , then it is easy to see thatthey are in disjoint cycles. Therefore, no two edges of Q cross in D . (cid:3) We use Claim 1 to show that cr( Q v ) ≥ 2, cr( Q ee ) ≥ 2, and cr( Q t ) ≥ Q v ) ≥ Q e , each of the two new edges joins vertices not on the same face of Q and so each has a crossing with Q . Thus, cr( Q e ) ≥ Q t , the new edge e incident with d must cross Q in any drawing D of Q t for which D [ Q t ] has no crossings. If the 3-cycle D [ abc ] also has a crossing with Q , then D has two crossings. Otherwise, D [ abc ] separates the two ends of D [ e ],so D [ e ] crosses D [ abc ]. Thus, cr( Q t ) ≥ Q v , deleting any edge of the 3-cube makes a face incident with 3 of thefour vertices and so yields a 1-drawing. Likewise deleting one of the edges incidentwith the new vertex yields a 1-drawing.For Q e , obviously deleting either of the edges not in Q yields a 1-drawing.On the other hand, if e is an edge of Q incident with at most one of the vertices of Q e of degree 4, then deleting e makes one of the newly adjacent pairs now lie onthe same face, yielding the required 1-drawing. If e is one the remaining two edgesof Q , there is a 1-drawing of Q − e with one crossing that extends to a 1-drawingof Q e − e .For Q t , criticality of all the edges not incident with d is obvious, as it is thenew edge e incident with d . The remaining three edges are symmetric. Deletingany one of these results in a subgraph that has crossing number 1 (we may movethe other end of e to the other side of abc to get a 1-drawing). 54 15. ON 3-CONNECTED GRAPHS THAT ARE NOT PERIPHERALLY-4-CONNECTED Lemma . Suppose G ∈ M reduces by planar 3-reductions to a peripherally-4-connected G p4c with at least 8 vertices that is an extension of the internally-4-connected 4-covered graph G i4c . Then either G is one of the graphs Q v , Q e , or G p4c has exactly 8 vertices. Proof. Let a, b, c, d be the four vertices so that G i4c − { a, b, c, d } is an independentset I . For each x ∈ { a, b, c, d } , let X be the set of vertices in I adjacent to everythingin { a, b, c, d } \ { x } , and let R be the remaining vertices in I ; a vertex in R is joinedto all of { a, b, c, d } .Note that a vertex in R has degree 4 in G i4c , so it is also a vertex of G ; itcannot be the outcome of any 3-reductions. If | R | ≥ 3, then G contains K , andso G = K , , a contradiction. Thus, | R | ≤ x ∈ { a, b, c, d } , | X | ≥ 2, then { a, b, c, d } \ { x } is a 3-cut in G i4c that separates any two vertices v, w in X from all the other vertices in I \ { v, w } ,of which there are at least two. This contradicts the fact that G i4c is internally4-connected. Thus, | X | ≤ G p4c has at most 10 vertices, but we can proceed a littlefurther.If R = ∅ , then G i4c is planar (adding the K on { a, b, c, d } does not affectplanarity), which is a contradiction. Thus, | R | > x ∈ { a, b, c, d } , | X | = 1, then the bipartite subgraph of G i4c consist-ing of { a, b, c, d } and the four vertices in A ∪ B ∪ C ∪ D is the 3-dimensional cube Q . Adding one of the vertices in R to Q produces Q v . That is, if all of A , B , C ,and D are not empty, | R | = 1 and G = Q v .Thus, we may assume R (cid:54) = ∅ and D = ∅ .If | R | = 2, then for G i4c to have at least 8 vertices, at least two of A , B , and C are not empty. Thus, Q e ⊆ G p4c , so G p4c = Q e .In the final situation, we have | R | = 1 and, because G p4c has at least 8 vertices,all of A , B , and C are not empty. In particular, G p4c has exactly 8 vertices, asrequired.A computer search can find all the peripherally-4-connected graphs having 8vertices. These will include all the examples that are peripherally-4-connectedextensions of internally-4-connected, 4-covered graphs having 8 vertices. This com-pletes our analysis of 3-connected, 2-crossing-critical graphs with no subdivision of V .HAPTER 16 Finiteness of 3-connected 2-crossing-critical graphswith no V n This section is devoted to showing that, for each n ≥ 3, there are only finitelymany 3-connected 2-crossing-critical graphs that do not contain a subdivision of V n . In particular, Theorem 16.14 asserts that if G has a subdivision of V n but nosubdivision of V n +2 , then | V ( G ) | = O ( n ).The finiteness has been proved previously by completely different methods in[ ]. In our particular context, this shows that there are only finitely many 3-connected 2-crossing-critical graphs that have a subdivision of V but do not have asubdivision of V ; these are the only ones missing from a complete determinationof the 2-crossing-critical graphs.The first subsection shows that, if G is a 3-connected 2-crossing-critical graphthat does not contain a subdivision of V n +2 , then, for any V n ∼ = H ⊆ G , each H -bridge in G has at most 88 vertices. The second subsection shows that, for aparticular subdivision H of V n , there are only O ( n ) H -bridges having a vertexthat is not an H -node. These easily combine to give the O ( n ) bound of Theorem16.14. V n -bridges are small The main result of this subsection is to show that if G ∈ M and V n ∼ = H ⊆ G ,then any H -bridge B is a tree with a bounded number of leaves, so that | V ( B ) | ≤ O ( n ) non-trivial H -bridges.The next lemma will have as a corollary the first main result of this subsection. Lemma . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ , and B an H -bridge. Then | att( B ) | ≤ n + 12 . Proof. Let e be an edge of B incident with x ∈ att( B ) and y ∈ Nuc( B ). Then D e [ B − e ] is contained in a face F of D e [ H ]. Because we know the 1-drawings of V n , we know that each face of D e [ H ] is incident with at most n + 1 H -branches.Moreover, B − e is an H -bridge in G − e and att G − e ( B − e ) is either att G ( B ) oratt G ( B ) \ { x } .If B has at least 11( n + 1) + 2 attachments, then some H -branch b containsat least 12 attachments of B − e . Let a . . . a be any 12 distinct attachments of B − e occurring in this order in b . Let T ⊆ B be a minimal tree that meets att( B )at a , a , a , a , a , a , a , and a , so that these a i are the leaves of T , and let Q = [ a , b, a ]. Set Y = T ∪ Q .For i = 1 , , , 10, there is a unique cycle C i ⊆ Y that meets b precisely in a i Qa i +2 . Let I ⊆ { , , , } be the subset such that, for i ∈ I , x / ∈ C i ; clearly | I | ≥ V n For each i ∈ I , let M i be the C i -bridge in G − e with H ⊆ M i ∪ C i . As x / ∈ C i , x ∈ Nuc( M i ). Let B i be the C i -bridge in G − e containing y or B i = y if y ∈ C i .Let P i be a minimal subpath of C i containing B i ∩ C i , so that a i Qa i +2 (cid:54)⊆ P i . Claim . Let i, j, k ∈ I be distinct. If y / ∈ M i ∪ M j , then: • B i = B j ; • P i = P j ⊆ C i ∩ C j ; and • y ∈ M k . Proof. If u and v are vertices in C i ∩ C j , then u and v are not in b and thereis a unique uv -path P in T . We note that P ⊆ C i ∩ C j . Thus, C i ∩ C j is a path.If there were a yC i -path disjoint from C j , then y ∈ M i , a contradiction. There-fore, every yC i -path meets C j and, symmetrically, every yC j -path meets C i . Thus,every y ( C i ∪ C j )-path has one end in C i ∩ C j . It follows that if y ∈ C i ∪ C j , then y ∈ C i ∩ C j , so in this case B i = B j = (cid:107) y (cid:107) .In the case y / ∈ C i ∪ C j , let B be the ( C i ∪ C j )-bridge containing y . Thepreceding paragraphs show that att( B ) ⊆ C i ∩ C j , so that in fact B is also both a C i - and a C j -bridge. In particular, B i = B j = B .For the last part, we assume y / ∈ M k and note that B = B i = B j = B k and C i ∩ C j ∩ C k is a non-null path P (cid:48) . If P (cid:48) has length at least one, then P (cid:48) ∪ C i ∪ C j ∪ C k contains a subdivision of K , and yet has all three of the vertices on one sideincident with a common face, which is impossible. Therefore, P (cid:48) consists of a singlevertex z .If z is not y , B has only z as an attachment in G − e . It follows that either z or { z, x } is a cut-set of G , contradicting the fact that G is 3-connected. Thus, z = y ,and so, for some t ∈ { i, j, k } , y is an attachment of M t ; in particular, y ∈ M t , acontradiction. (cid:3) By Claim 1, there is an i ∈ I such that y ∈ M i . For such an i , set C = C i andnote that x ∈ M i − att( M i ), so that M = M i + e is a C -bridge in G . Furthermore,att G ( M ) = att G − e ( M − e ).Notice that D e [ C ] is clean, since the crossing of D e is between disjoint H -branches. Thus, C has BOD in G − e . Also, any C -bridge B (cid:48) (cid:54) = M has C ∪ B (cid:48) planar. As att G ( M ) = att G − e ( M − e ), C has BOD in G .Recall that the H -bridge B has a i , a i +1 , and a i +2 as attachments. For anyvertex u of B not in b , there is an H -avoiding ua i +2 -path, whose edge e (cid:48) incidentwith u is in some C -bridge B (cid:48) . Since x and y are on the same side of D e [ C ], M iscontained on that side of D e [ C ] and e (cid:48) is on the other side. Therefore, B (cid:48) (cid:54) = M .In D e (cid:48) , the crossing is in H and D e (cid:48) [ C ] is clean. That is, D e (cid:48) [ C ∪ M ] is a1-drawing with C clean. Corollary 4.7 shows cr( G ) ≤ 1, the final contradiction.The following corollary is the first main result of this section. Corollary . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ , B an H -bridge. Then | att( B ) | ≤ . Proof. If n = 3, then the result is an immediate consequence of Lemma 16.1. Thus,we may assume n ≥ 4. If B has attachments in the interiors of non-consecutivespokes, then G is the Petersen graph and the result clearly holds.Otherwise, B has attachments in at most two consecutive spokes. Thus, thereis a subdivision H (cid:48) of V contained in H that contains all the attachments of B .Applying Lemma 16.1 to H (cid:48) , we again see that | att( B ) | ≤ V n -BRIDGES ARE SMALL 157 We now turn to the other half of the argument that bounds the number ofvertices in an H -bridge, namely, that the bridge is a tree. We need a new notion. Definition . Let T ∗ be a graph consisting of subdivision of a K , togetherwith three pendant edges, one incident with each of the three degree 2 vertices inthe K , . A tripod is any graph T obtained from T ∗ by contracting any subset ofthe pendant edges; if all three pendant edges are contracted, then an edge is addedbetween the two copies of K , , but not having a vertex of contraction as an end— this may be done in any of three essentially different ways. The attachments ofthe tripod are the degree 1 and 2 vertices in T .We are now ready for the second half of the main result of this section. Lemma . Suppose G ∈ M , V n ∼ = H ⊆ G , n ≥ , G has no subdivision of V n +1) , and B is an H -bridge. Then either B is a tree or B has a tripod, n = 3 and | V ( G ) | ≤ . Proof. By way of contradiction, suppose B has a cycle C . If att( B ) ∩ C (cid:54) = ∅ ,let e be an edge of C incident with u ∈ att( B ). If C ∩ att( B ) = ∅ , then let e beany edge of C . The choice of e shows that B − e is an H -bridge in G − e and thatatt G − e ( B − e ) = att G ( B ). Since D e [ H ] contains the crossing in D e [ G − e ], D e [ B − e ]is contained in a face F of D e [ H ].Let C (cid:48) = ∂F × , so C (cid:48) is a cycle in G (cid:48) = ( G − e ) × . Since G (cid:48) is planar, C (cid:48) hasBOD in G (cid:48) and C (cid:48) ∪ B (cid:48) is planar for each C (cid:48) -bridge B (cid:48) in G (cid:48) . If C (cid:48) ∪ B were planar,then G (cid:48) + e would be planar, in which case cr( G ) ≤ 1, a contradiction. Therefore, C (cid:48) ∪ B is not planar.We now introduce a convenient notion. Definition . Let G be a graph. The graph G t is the graph whose verticesare the G -nodes and whose edges are the G -branches. Claim . ( C (cid:48) ∪ B ) t is 3-connected. Proof. Let L = ( C (cid:48) ∪ B ) t . If | V (Nuc( B )) | = 1, then L is a wheel and theclaim follows. So assume | V (Nuc( B )) | ≥ 2. We show that any two vertices of L are joined by three internally disjoint paths. For u, w ∈ Nuc( B ), this is true in G ,so let P , P , P be such paths in G . If at least one P i is contained in B − C (cid:48) , thenwe can easily modify the others to use C (cid:48) rather than G − B to get three paths in L . If all three intersect C e , then B ∩ ( P ∪ P ∪ P ) is two claws Y u and Y w . Thereis a Y u Y w -path in Nuc( B ), which returns us to the previous case.If u ∈ Nuc( B ) and w ∈ C (cid:48) , then w is an attachment of B . Let Y be a claw in B with centre u and talons on C (cid:48) . Using a C (cid:48) -avoiding wY -path in B , if necessary,we can assume w is a talon of Y . It is then easy to use C (cid:48) to extend the other twopaths in Y to w .Finally, if u, w ∈ C (cid:48) , then both u and w are attachments of B , so there is a C (cid:48) -avoiding path joining them. This path and the two uw -paths in C (cid:48) yield therequired three paths. (cid:3) Definition . Let C be a cycle in a graph G and let P and P be disjoint C -avoiding paths in G . Then P and P are C -skew paths if the two C -bridges in C ∪ P ∪ P overlap.As C (cid:48) ∪ B has no planar embedding, [ ] implies B has either a tripod whoseattachments are in C (cid:48) or two C (cid:48) -skew paths. 58 16. FINITENESS OF 3-CONNECTED 2-CROSSING-CRITICAL GRAPHS WITH NO V n Claim . If B has a tripod T , then n = 3, G = H ∪ T and | V ( G ) | ≤ Proof. Let S be the attachments of T . As H ∪ T is 2-connected and, relativeto the cut S , both H (cid:48) + (taking H (cid:48) to be any V containing S ) and T + are non-planar. By Theorem 15.6, cr( H (cid:48) ∪ T ) ≥ 2. Thus, G = H (cid:48) ∪ T , so n = 3 and, againby Theorem 15.6, | V ( G ) | ≤ (cid:3) Thus, we can assume B has no tripod. Then B has C (cid:48) -skew paths, say P and P . Since these do not exist in B − e , e is in one of them. If C ∩ att( B ) = ∅ , choose e (cid:48) any edge of C not in P ∪ P . If C ∩ att( B ) (cid:54) = ∅ , choose e (cid:48) to be the other edgeof C incident with the same attachment as e .Repeat with G − e (cid:48) . This yields C (cid:48)(cid:48) so that B has C (cid:48)(cid:48) -skew paths u (cid:48) u (cid:48) and w (cid:48) w (cid:48) ( e (cid:48) incident with u (cid:48) ). Since u u ∪ w w ⊆ B − e (cid:48) , they are not C (cid:48)(cid:48) -skew. In C (cid:48) , we have the cyclic order u , w , u , w , say. In C (cid:48)(cid:48) we have u u w w . Likewisein C (cid:48) we have u (cid:48) u (cid:48) w (cid:48) w (cid:48) , while in C (cid:48)(cid:48) we have u (cid:48) w (cid:48) u (cid:48) w (cid:48) .Let D and D (cid:48) be 1-drawings of H having all attachments of B on faces F, F (cid:48) ,respectively, so that the cyclic orders of att( B ) are different in ∂F and ∂F (cid:48) . Claim . n ≥ Proof. Let H be a subdivision of V in G . We remark that if f and f (cid:48) are anydisjoint H -branches having internal vertices that are ends of an H -avoiding path P in G , then H ∪ P is a subdivision of V in G .We consider first the case that att( B ) is not contained in any 4-cycle of H .Because we know the 1-drawings of H and att( B ) is contained in the boundary ∂F of a face F of such a 1-drawing, ∂F is × v v v × . If B has attachments in both (cid:104)× v (cid:105) and (cid:104) v ×(cid:105) , then G has a subdivision of V , as required. Thus, we may assumethat att( B ) is contained in a 4-cycle Q of H , which we may take to be [ v v v v v ].In at least one of D and D (cid:48) , Q is self-crossed (otherwise the cyclic orders ofatt( B ) are the same) and B is drawn in the face × v v v × . However, in this caseatt( B ) ⊆ (cid:104)× , v ] ∪ [ v , ×(cid:105) and at least two attachments of B are in each. In thiscase, we again have a subdivision of V in G , as required. (cid:3) Claim . B has no (interior) spoke attachment. Proof. From Claim 3, we know that n ≥ 4. By way of contradiction, weassume B has an attachment in (cid:104) s (cid:105) . From the listing of the faces of 1-drawings of V n , the only possibilities for each of ∂F and ∂F (cid:48) are: : (1) [ v , r , v , s , v n +1 , r n , v n , s , v ]; : (1’) [ v , r − , v − , s − , v n − , r n − , v n , s , v ]; : (2) (cid:104) v , r , v , s , v n , r n , v n +1 , r n +1 , v n +2 (cid:105) ; : (2’) (cid:104) v − , r − , v , s , v n , r n − , v n − , r n − , v n − (cid:105) ; : (3) (cid:104) v n − , r n − , v n , s , v , r − , v − , r − , v − (cid:105) ; : (3’) (cid:104) v n +1 , r n , v n , s , v , r , v , r , v ]; : (4) (cid:104) v − , r − , v , s , v n , r n , v n +1 (cid:105) ; : (4’) (cid:104) v n − , r n − , v n , s , v , r , v (cid:105) ; : (5) [ v , v , v , . . . , v n , s , v ]; : (5’) [ v , s , v n , v n +1 , v n +2 , . . . , v − , v ].We now consider these possibilities in pairs. In every case, the ends of the skewpaths will occur in the same cyclic order on the boundaries of the two faces, whichis impossible. V n -BRIDGES ARE SMALL 159 : (1,1’) att( B ) ⊆ s ; same cyclic order, a contradiction. : (2,2’) att( B ) ⊆ s ; same cyclic order, a contradiction. : (3,3’) att( B ) ⊆ s ; same cyclic order, a contradiction. : (4,4’) att( B ) ⊆ s ; same cyclic order, a contradiction. : (5,5’) att( B ) ⊆ s ; same cyclic order, a contradiction. : (1,2) att( B ) ⊆ (cid:104) v , r , v , s , v n , r n , v n + 1]; same cyclic order, a contradiction. : (1,2’) att( B ) ⊆ [ v , s , v n ]; same cyclic order, a contradiction. : (1,3) att( B ) ⊆ s ; same cyclic order, a contradiction. : (1,3’) att( B ) ⊆ (cid:104) v n +1 , r n , v n , s , v , r , v ]; same cyclic order, a contradiction. : (1,4) att( B ) ⊆ (cid:104) v , r , v , s , v n ]; same cyclic order, a contradiction. : (1,4’) att( B ) ⊆ [ v , r , v , s , v n ]; same cyclic order, a contradiction. : (1,5) att( B ) ⊆ [ v , r , v , s , v n ]; same cyclic order, a contradiction. : (1,5’) att( B ) ⊆ [ v n +1 , r n , v n , s , v ]; same cyclic order, a contradiction. : (2,3) att( B ) ⊆ s ; same cyclic order, a contradiction. : (2,3’) att( B ) ⊆ (cid:104) v n +1 , r n , v n , s , v , r , v ]; same cyclic order, a contradiction. : (2,4) att( B ) ⊆ (cid:104) v n +1 , r n , v n , s , v ]; same cyclic order, a contradiction. : (2,4’) att( B ) ⊆ (cid:104) v , r , v , s , r n ]; same cyclic order, a contradiction. : (2,5) att( B ) ⊆ (cid:104) v , r , v , s , v n ]; same cyclic order, a contradiction. : (2,5’) att( B ) ⊆ [ v , s , v n , r n , v n +1 , r n +1 , v n +2 (cid:105) ; same cyclic order, a contradic-tion. : (3,4) att( B ) ⊆ (cid:104) v − , r − , v , s , v n ]; same cyclic order, a contradiction. : (3,4’) att( B ) ⊆ (cid:104) v n − , r n − , v n , s , v ]; same cyclic order, a contradiction. : (3,5) att( B ) ⊆ (cid:104) v n − , r n − , v n , s , v ]; same cyclic order, a contradiction. : (3,5’) att( B ) ⊆ (cid:104) v − , r − , v − , r − , v , s , v n ]; same cyclic order, a contradiction. : (4,5) att( B ) ⊆ [ v , s , v n ]; same cyclic order, a contradiction. : (4,5’) att( B ) ⊆ (cid:104) v − , r − , v , s , v n , r n , v n +1 (cid:105) ; same cyclic order, a contradiction.As any pair gives the same cyclic order, we always get a contradiction. (cid:3) Claim . B is not a local H -bridge. Proof. Suppose B is local, with att( B ) ⊆ Q . From Claims 3 and 4, we mayassume n ≥ B has no spoke attachment. Thus, att( B ) ⊆ r ∪ r n . Moreover, B cannot have attachments in both (cid:104) r (cid:105) and (cid:104) r n (cid:105) because G has no subdivision of V n +1) . On the other hand, B has at least two attachments in both r and r n orelse the cyclic order of the ends of the skew paths is always the same. So we mayassume att( B ) ∩ r = { v , v } . We need two attachments in r n . From the listingof faces in 1-drawings of V n , the only possibilities for ∂F and ∂F (cid:48) occur when Q is not self-crossed and so the cyclic orders of the attachments of B are the same inboth cases, a contradiction. (cid:3) Claim . For some i , att( B ) ⊆ r i ∪ r i + n +1 . Proof. By Claims 3, 4, and 5, n ≥ B has no spoke attachments, and B isnot local.We consider in turn the possibilities for the face of D e [ H ] that contains B − e .We know B is not local, so it can only be contained in a face whose boundary hasone of the following forms:(1) [ × , r i , v i , s i , v i + n , r i + n − , × ];(2) [ × , r i , v i , r i − , v i − , s i − , v n + i − , r n + i − , × ];(3) [ × , r i , v i +1 , r i +2 , . . . , v i + n − , r i + n − , × ]; 60 16. FINITENESS OF 3-CONNECTED 2-CROSSING-CRITICAL GRAPHS WITH NO V n (4) [ v i , s i , v n + i , r n + i , v n + i +1 , . . . , r i − , v i ]; or(5) [ × , r i , v i +1 , r i +1 , . . . , r n + i − , v n + i , r n + i , × ].As in the proof of Claim 4, the faces of D e [ H ] and D e (cid:48) [ H ] containing B − e and B − e (cid:48) , respectively, cannot both be of one of the types (3, 4, 5): the verticesof att( B ) will occur in the same order in both cases.If one of the drawings has B − e or B − e (cid:48) in a face of type (1), then we aredone: att( B ) ⊆ r i ∪ r i + n − . The remaining case is that one of the drawings has B − e or B − e (cid:48) drawn in a face of type (2).All other possibilities having been eliminated, we may assume (taking i = n +1)att( B ) ⊆ [ × , r , v , r , v , s , v n , r n , × ] . Because B is not local, att( B ) ∩ (cid:104) r (cid:105) (cid:54) = ∅ . Because att( B ) occurs in different ordersin ∂F and ∂F (cid:48) , att( B ) ∩ r n (cid:54) = ∅ . By way of contradiction, we suppose B also hasan attachment in [ v , r , v (cid:105) . The only other face which could allow these threeattachments is [ × , r , v , r , . . . , v i − , r i − , v n , r n , × ]. Notice v is not in this secondboundary, so one attachment is in (cid:104) r (cid:105) . Because V n +1) (cid:54)⊆ G , no attachment is in (cid:104) r n (cid:105) . Thus att( B ) ∩ r n = { v n } . But then, once again, the attachments of B occurin the same cyclic orders in ∂F and ∂F (cid:48) , a contradiction. (cid:3) As we have seen above, the alternative to “ B is neither a tree nor contains atripod” is that B has the C (cid:48) -skew paths P and P , as well as the C (cid:48)(cid:48) -skew paths P (cid:48) and P (cid:48) . Claim 6 shows the four ends of P and P are in r ∪ r n +1 . If threeof them are in r , say, then they occur in the same cyclic order in ∂F and ∂F (cid:48) , acontradiction. So two are in r and two in r n +1 . If P has both ends in r , say,then the ends of P and P can never interlace, a contradiction as they interlace in ∂F . So each has one end in each of r and r n +1 . Likewise for P (cid:48) , P (cid:48) .Adding at most 3 paths in B − att( B ) to P ∪ P ∪ P (cid:48) ∪ P (cid:48) , we obtain B (cid:48) ⊆ B containing P ∪ P ∪ P (cid:48) ∪ P (cid:48) so that B (cid:48) is an H -bridge in H ∪ B (cid:48) .Recall that n ≥ B (cid:48) are in H − (cid:104) s (cid:105) .Suppose D (cid:48)(cid:48) is a 1-drawing of ( H ∪ B (cid:48) ) − (cid:104) s (cid:105) . Then D (cid:48)(cid:48) [ B (cid:48) ] is in a face F (cid:48)(cid:48) of D (cid:48)(cid:48) [ H − (cid:104) s (cid:105) ]. Since r and r n +1 both have at least two attachments of B (cid:48) , theyare both incident with F (cid:48)(cid:48) . Thus one of the pairs P , P and P (cid:48) , P (cid:48) is a ∂F (cid:48)(cid:48) -skew pair. Therefore, cr(( H ∪ B (cid:48) ) − (cid:104) s (cid:105) ) ≥ 2, contradicting the fact that G is2-crossing-critical.Combining Corollary 16.2 and Lemma 16.4, we immediately have the mainresult of this section. Theorem . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ , and suppose G has nosubdivision of V n +1) . If B is an H -bridge, then | V ( B ) | ≤ . This completes the first main step of our effort to show that 3-connected, 2-crossing-critical graphs with no subdivision of V n have bounded size. This subsection, the final leg of this work, is devoted to showing that there isa particular subdivision H of V n in G so that there are at most O ( n ) H -bridgesin G that have a vertex that is not an H -node. Theorem 16.7 shows that, for any V n ∼ = H ⊆ G , all H -bridges have at most 88 vertices (when there is no subdivisionof V n +1) ). The combination easily implies G has at most O ( n ) vertices. Definition . Let G be a graph and let n be an integer, n ≥ 3. A subdivi-sion H of V n in G is smooth if, whenever B is an H -bridge with all its attachmentsin the same H -branch, B is just an edge that is in a digon with an edge of H .We begin by showing that every G ∈ M with a subdivision V n has a smoothsubdivision H of V n . For such an H , every vertex of G either is an H -node or isin an H -bridge that does not have all its attachments in the same H -branch. So itwill be enough to show that the number of these H -bridges is O ( n ).This analysis is completed in three parts. We start with the result that thereare not many H -bridges having an attachment in a particular vertex of H and anattachment in the interior of some H -branch. This is useful for H -bridges havingboth node and branch attachments, but is also used in the second part, which isto bound the number of H -bridges having attachments in the interiors of the sametwo H -branches. The final part puts these together with those H -bridges havingattachments in three or more H -nodes.We start by showing that every G ∈ M with a subdivision of V n has a smoothsubdivision of V n . Lemma . Let G ∈ M and suppose G contains a subdivision of V n , with n ≥ . Then G has a smooth subdivision of V n . Proof. Choose H to be a subdivision of V n in G that minimizes the number ofedges of G that are in H . We claim H is smooth.To this end, let B be an H -bridge with all attachments in the same H -branch b and let P be a minimal subpath of b containing att( B ). Set K = B ∪ P and noticethat K is both H -close and 2-connected. By Lemma 5.13, K is a cycle, so B is justa path and, since G is 3-connected, just an edge. It remains to prove that P is justan edge as well.Let H (cid:48) = ( H ∪ B ) − (cid:104) P (cid:105) . Evidently H (cid:48) is a subdivision of V n in G and | E ( H (cid:48) ) | = | E ( H ) | − | E ( P ) | + 1. Since | E ( H ) | ≤ | E ( H (cid:48) ) | by the choice of H , we seethat | E ( P ) | ≤ 1, and, therefore, P is just an edge, as required.We now turn our attention to the H -bridges of a smooth subdivision H of V n .There are three main steps. Step 1: Bridges attaching to a particular vertex and branch. The first step in bounding the number of H -bridges is to bound the number ofthem that can have an attachment at a particular vertex of H and in the interiorof a particular H -branch. This is the content of this step. Lemma . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ and suppose H is smooth.For a vertex (not necesssarily a node) u of H and an H -branch b , there are at most H -bridges with an attachment at u and an attachment in (cid:104) b (cid:105) − u . Proof. Suppose there are 42 such H -bridges. Let B be one of them, let e ∈ E ( B )and let D be a 1-drawing of G − e . If u / ∈ (cid:104) b (cid:105) , then at most 4 faces of D [ H ] areincident with (cid:104) b (cid:105) , and therefore at least 11 of these H -bridges (other than B )are in the same face F of D [ H ]. If u ∈ (cid:104) b (cid:105) , then precisely two faces of D [ H ] areincident with u , so at least 21 of these bridges are in the same face F of D [ H ] andof these at least 11 have an attachment in the same component of D [ b − u ] ∩ ( ∂F ) × .In both cases, let B be the set of 11 bridges, contained in F , having u as an 62 16. FINITENESS OF 3-CONNECTED 2-CROSSING-CRITICAL GRAPHS WITH NO V n attachment and an attachment in the same component b (cid:48) of D [ b − u ] ∩ ( ∂F ) × . As D [( ∂F ) × ∪ ( ∪ B ∈B B )] is planar with ( ∂F ) × bounding a face, no two ( ∂F ) × -bridgesin B overlap.Let P = b (cid:48) and Q = ( ∂F ) × − (cid:104) P (cid:105) . Lemma 4.8 applies to ( ∂F ) × , P , Q , B . Asthere are no digons disjoint from H , there is a unique (up to inversion) ordering B , . . . , B of B so that P = P B . . . P B and Q = Q B . . . Q B .Because u ∈ Q B ∩ Q B ∩· · ·∩ Q B and the Q B i are internally disjoint subpathsof Q , all of Q B , . . . , Q B are just u . For i = 1 , . . . , 11, let a i and a (cid:48) i be the ends of P i , so that P = ( . . . , a , . . . , a (cid:48) , . . . , a , . . . , a (cid:48) , . . . , a , . . . , a (cid:48) , . . . ). Claim . For i ∈ { , . . . , } , a i (cid:54) = a (cid:48) i +1 . Proof. Otherwise, a i = a (cid:48) i = a i +1 = a (cid:48) i +1 , implying that B i and B i +1 consti-tute a digon disjoint from H , which is impossible. (cid:3) For i, j ∈ { , , . . . , } with i < j , set K ij = ( ∪ jk = i B k ) ∪ a i P a (cid:48) j . Claim . For i, j ∈ { , . . . , } with i < j , K ij is 2-connected. Proof. Let R i be an H -avoiding ua i -path in B i , and R j an H -avoiding ua (cid:48) j -path in B j . Then C ij := R i ∪ R j ∪ a i P a (cid:48) j ⊆ K ij is a cycle containing u and a i P a (cid:48) j .For x ∈ B k , i ≤ k ≤ j , x / ∈ H , for any H -node w (cid:54) = u , G has 3 internallydisjoint xw -paths; at least two of these leave B k in a k P a (cid:48) k , and so no cut vertex of K ij separates x from C ij . (cid:3) Since b (cid:48) is not crossed in D , D [ K i,i +2 ] is clean and is contained in F ∪ ∂F .There is a unique face F i of D [ K i,i +2 ] so that F i (cid:54)⊆ F ; since K i,i +2 is 2-connected, F i is bounded by a cycle C i . As D [ K i,i +2 ] ⊆ F ∪ ∂F , ∂F ⊆ F i ∪ ∂F i . As D [ u ] ∈ ∂F ∩ D [ K i,i +2 ], D [ u ] ∈ ∂F i . Likewise D [ a i P a (cid:48) i +2 ] ⊆ ∂F i .Thus, u ∈ C i and a i P a (cid:48) i +2 ⊆ C i . Therefore, C i ∩ H is u and a i P a (cid:48) i +2 , fromwhich we deduce that there is a C i -bridge M i so that H ⊆ C i ∪ M i . Observe that B i +1 is a C i -bridge different from M i .For i = 2 , , 8, let e i be an edge of B i +1 incident with u , and let D i be a1-drawing of G − e i . Claim . For i ∈ { , , } , C i has BOD in G and D i [ C i ] is not clean. Proof. At most one of D [ C i ], i ∈ { , , } is crossed, so for at least one i ∈ { , } , D e [ C i ] is clean. It follows that C i has BOD in G − e .By Claim 1, a (cid:54) = a i , whence B ⊆ M i , and B − e ⊆ M i − e . Furthermore, u ∈ H , so u ∈ att( M i − e ). Thus att G − e ( M i − e ) = att G ( M i ) and M i − e is a C i -bridge in G − e . We conclude that the overlap diagrams for C i in G − e and G are isomorphic and, therefore, C i has BOD in G .We now show that all three C j , j ∈ { , , } , have BOD in G . If D i [ C i ] is clean,then D i [ C i ∪ M i ] is a 1-drawing of C i ∪ M i , implying via Corollary 4.7 that cr( G ) ≤ D i [ C i ] is not clean, and, therefore, for j ∈ { , , } \ { i } , D i [ C j ]is clean. Thus, C j has BOD in G − e i , and, following the argument above for C i ,we deduce that C j has BOD in G . (cid:3) Claim . For i ∈ { , , } , one face of D i [ C i ] contains all H -nodes, other than(possibly) u . Proof. Let e (cid:48) i be the edge of H so that D i [ e (cid:48) i ] crosses D i [ a i ba (cid:48) i +2 ] and let b (cid:48) i be the H -branch containing e (cid:48) i . If n = 3, let R be a hexagon in H containing b and b (cid:48) i . For n ≥ 4, both b and b (cid:48) i are in the rim R of H .Since b and b (cid:48) i are disjoint, for n ≥ R − ( (cid:104) b (cid:105) ∪ (cid:104) b (cid:48) (cid:105) ) has two components, eachwith at least two nodes of H . Either of these with ≤ n nodes has all its nodesadjacent by spokes to the other component. Obviously, there is at least one such.Observe that if A is any path in R − ( (cid:104) b (cid:105) ∪ (cid:104) b (cid:48) i (cid:105) ) such that D i [ A ] has a vertexin each face of D i [ C i ], then u ∈ V ( A ) and the two paths P, P (cid:48) in A having u as anend are such that D i [ P ] and D i [ P (cid:48) ] are in different faces of D i [ C i ].Let K be a component of R − ( (cid:104) b (cid:105)∪(cid:104) b (cid:48) i (cid:105) ) not containing u and let L be the other.Then D i [ K ] is in the closure of a face F i of D i [ C i ]. We claim that D i [ L ] ⊆ F i ∪ { u } .Any H -node w in L that is joined by a spoke to an H -node w (cid:48) in K has D i [ w ] ⊆ F i ∪ D i [ u ], since otherwise D i [ ww (cid:48) ] crosses D i [ C i ].If there is an H -node w in L that is not adjacent by a spoke to any vertex in K , then w is adjacent by a spoke to another H -node w (cid:48) in L and, moreover, w and w (cid:48) are the first and last nodes of L . As D i [ ww (cid:48) ] is disjoint from D i [ C i ], we deducethat there is a face F of D i [ C i ] so that D i [ w ] and D i [ w (cid:48) ] are both in F ∪ D i [ u ].Therefore, D i [ L ] is contained in that face. As at least one H -node in L is adjacentby a spoke to an H -node in K , we conclude that D i [ L ] ⊆ F i ∪ D i [ u ]. (cid:3) Let F i be the face of D i [ C i ] containing all the H -nodes and let F (cid:48) i be the otherface of D i [ C i ]. Claim . For i ∈ { , , } , the crossing in D i is not in (cid:10) a i +1 , b, a (cid:48) i +1 (cid:11) . Proof. Suppose by way of contradiction that e (cid:48) i is an edge of G − e i so that D i [ e (cid:48) i ] crosses (cid:10) a i +1 , b, a (cid:48) i +1 (cid:11) . Clearly, a i +1 (cid:54) = a (cid:48) i +1 . Since H − (cid:104) b (cid:105) is 2-connected,there is a cycle C (cid:48) ⊆ H containing e (cid:48) i . Let P be an H -avoiding a i +1 a (cid:48) i +1 -path in B i +1 and let C be the cycle P ∪ [ a i +1 , b, a (cid:48) i +1 ]. Then C and C (cid:48) are graph-theoreticallydisjoint and D i [ C ] ∩ D i [ C (cid:48) ] contains the crossing of D i . But then D i [ C ] and D i [ C (cid:48) ]must cross a second time, a contradiction. (cid:3) Claim . The only C i -bridge that overlaps B i +1 is M i . Proof. Let B be a C i -bridge different from M i overlapping B i +1 . Thenatt( B ) ⊆ [ a i ba (cid:48) i +2 ] ∪ { u } . As H is smooth, u ∈ att( B ). We claim both B i +1 and B overlap M i .By Claim 1, a i (cid:54) = a (cid:48) i +1 , so B i +1 either has an attachment in (cid:10) a i , a (cid:48) i +2 (cid:11) or ithas both a i and a (cid:48) i +2 as attachments. In either case, B i +1 overlaps M i (which hasattachments at u, a i , a (cid:48) i +2 ).Likewise B either has two attachments in (cid:2) a i , a (cid:48) i +2 (cid:3) or at least one attachmentin (cid:10) a i +1 , a (cid:48) i +1 (cid:11) ⊆ (cid:10) a i , a (cid:48) i +2 (cid:11) , so B overlaps M i . But now B i +1 , B i , and M i make atriangle in OD ( C i ), contradicting Claim 3. (cid:3) Let b (cid:48) be the H -branch that crosses C i in D i and let x be the H -node so thatthe crossing is in [ x, b (cid:48) , u ]. Claim . Let L be the graph [ D i [ G − e i ] ∩ (cl( F (cid:48) i ))] × ∪ B i +1 . Then the C i -bridgecontaining [ × , b (cid:48) , u ] overlaps B i +1 in L . Proof. If L embeds in the plane with C i bounding a face, then this embeddingcombines with D i restricted to the closure of F to yield a 1-drawing of G , which 64 16. FINITENESS OF 3-CONNECTED 2-CROSSING-CRITICAL GRAPHS WITH NO V n is impossible. As each individual C i -bridge B in L has C i ∪ B planar, there areoverlapping C i -bridges in L .By definition, L is planar with all C i -bridges other than B i +1 on the same sideof C i . Therefore B i +1 overlaps some other C i -bridge in L . By Claim 6, this is notany C i -bridge other than D i [ M i ] × ∩ D i [ L ], that is, the one containing [ × , b (cid:48) , u ]. (cid:3) By Claim 4, [ a i , b, a (cid:48) i +2 ] − × has a component A containing att( B i +1 ) − u . Let z be the one of a i and a i +2 that is an end of A and let Q be the minimal subpathof A containing all of z, a i +1 , a (cid:48) i +1 . By Claim 7, M i has an attachment w i ∈ [ zQ (cid:105) and an H -avoiding path Q i from w i to a vertex x i ∈ (cid:104)× , b (cid:48) , u (cid:105) . Notice that, if j ∈ { , , } \ { i } , then Q i ∩ C j = ∅ .There are at most two H -branches (or subpaths thereof) incident with u thatcan cross b . Thus for some i, j ∈ { , , } , b (cid:48) i = b (cid:48) j . Choose the labelling so that x i is no further in b (cid:48) i from u than x j is. Since xb (cid:48) j u contains x i , D j [ x i ] ⊆ F (cid:48) j but D j [ w i ] ⊆ F j . Since Q i ∩ C j = ∅ , D j [ Q i ] crosses C j , the final contradiction.The other steps in the argument are to show that a smooth subdivision H of V n in G has few bridges with attachments in the interiors of distinct H -branches.There are two parts to this: either the branches do or do not have a node incommon. We first deal with the latter case. Step 2: H -bridges joining interiors of disjoint H -branches. Lemma . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ , H smooth and suppose G has no subdivision of V n +1) . If b , b are disjoint H -branches, then there are atmost n + 9 H -bridges having attachments in both (cid:104) b (cid:105) and (cid:104) b (cid:105) . Proof. Suppose there is a set B of 164 n + 10 H -bridges having attachments inboth (cid:104) b (cid:105) and (cid:104) b (cid:105) . Let B ∈ B and let e ∈ B . In D e , at most 4 faces are incidentwith (cid:104) b (cid:105) , so there is a set B (cid:48) consisting of 41 n + 3 elements of B \ { B } in the sameface of D e [ H ]. By Lemma 4.8, there is a unique ordering ( B , . . . , B n +3 ) of theelements of B (cid:48) so they appear in this order in both (cid:104) b (cid:105) and (cid:104) b (cid:105) . It follows that B , . . . , B n +2 have all attachments in (cid:104) b (cid:105) ∪ (cid:104) b (cid:105) . By Lemmas 4.8 and 16.10, B i and B i +41 are totally disjoint. So there are n + 1 totally disjoint (cid:104) b (cid:105) (cid:104) b (cid:105) -pathswith their ends having the same relative orders on both.We aim to use these disjoint paths to find a subdivision of V n +1) in G . Weneed the following new notion. Definition . Let e = uw and f = xy be edges in a graph G . Twocycles C and C (cid:48) in G are ef -twisting if C = ( u, e, w, . . . , x, f, y, . . . ) and C (cid:48) =( u, e, w, . . . , y, f, x, . . . ), i.e., C and C (cid:48) traverse the edges e and f in opposite ways.We note that V has edge-twisting cycles: if e = uw and f = xy are disjointedges in V , with u, x not adjacent, then the 4-cycle ( u, w, x, y, u ) and the 6-cycle( u, w, z, y, x, z (cid:48) , u ) are ef -twisting.Next suppose n ≥ 4. There are three possibilities for b and b . : Case 1 : Both b and b are in R . We may assume without loss of generality(recall that b and b are not adjacent) that b = r , b = r i , 2 ≤ i ≤ n .Set H (cid:48) = R ∪ s ∪ s ∪ s , so H (cid:48) ∼ = V . Then b and b are in disjoint H (cid:48) -branches and so H (cid:48) , and therefore H , contains b b -twisting cycles. : Case 2 : One is in R , the other is a spoke. We may assume without loss ofgenerality that b = r , b = s i , i / ∈ { , } . Set H (cid:48) = R ∪ s ∪ s ∪ s i . Then b and b are in disjoint H (cid:48) -branches, so H (cid:48) , and therefore H , contains b b -twisting cycles. : Case 3 : Both b and b are spokes. We may assume without any loss of generalitythat b = s , b = s i . Then there exists j ∈ { , . . . , n − } \ { , i } . Set H (cid:48) = R ∪ s ∪ s i ∪ s j . Then b and b are in disjoint H (cid:48) -branches and so H (cid:48) , and therefore H , contains b b -twisting cycles.Choose the cycle C in the twisting pair in H for b and b so that C traverses b and b in order so that the ends u i , w i of the n + 1 disjoint paths occur in C as u , u , . . . , u n +1 , . . . , w , . . . , w n +1 . Then C and these paths are a subdivision of V n +1) in G , contradicting the assumption that G has no subdivision of V n +1) .Next is the third and final consideration. Step 3: H -bridges joining interiors of H -branches having a common node. Lemma . Let G ∈ M , V n ∼ = H ⊆ G , n ≥ , and let b , b be adjacent H -branches. Then at most H -bridges have attachments in both (cid:104) b (cid:105) and (cid:104) b (cid:105) . Proof. By way of contradiction, suppose there is a set { B , B , B } of 3 such H -bridges. For each i ∈ { , , } , let e i ∈ B i . There is precisely one face F i , of a1-drawing D i of G − e i , that is incident with both (cid:104) b (cid:105) and (cid:104) b (cid:105) . Thus, for each B j , j (cid:54) = i , D i [ B j ] ⊆ F i . Clearly for { j, k } = { , , } \ { i } , B j and B k do not overlapon F i . In particular, their attachments in b and b are in the same order as wetraverse them from their common end u . Thus we may assume B , B , B appearin this order from u on both b and b .Notice that att( B ) (cid:54) = att( B ). Therefore, there is a cycle C ⊆ B ∪ b ∪ b consisting of a (cid:104) b (cid:105) (cid:104) b (cid:105) -path in B and a subpath of b ∪ b containing u , such that C does not contain some attachment w of B . Reselect e ∈ B to be incident with w . Let M C be the C -bridge so that H ⊆ C ∪ M C .Then w ∈ Nuc( M C ), so B ⊆ M C . Furthermore, if e is incident with anattachment x of M C , then x is contained in R . In particular, it is incident withanother edge of M C . Thus, M C − e is a C -bridge in G − e having the sameattachments as M C has in G . Because C is H -close, D [ C ] is clean; furthermore, D [ C ∪ M C ] is a 1-drawing of C ∪ M C . Since D [ C ] is also clean, C has BOD in G − e and hence in G . Corollary 4.7 implies the contradiction that cr( G ) ≤ V n +2 . Theorem . Suppose G ∈ M and there is an n ≥ so that G has asubdivision of V n , but no subdivision of V n +1) . Then | V ( G ) | = O ( n ) . Proof. By Lemma 16.9, G has a smooth subdivision H of V n . We may assumeno H -bridge contains a tripod, as otherwise | V ( G ) | ≤ 14 by Lemma 16.4.We first claim that a vertex u of H that is not an H -node is an attachmentof some H -bridge B not having all its attachments in the same H -branch. Since u has degree 2 in H and degree greater than 2 in G , u is an attachment of some H -bridge. Because H is smooth, an H -bridge that has all its attachments in the 66 16. FINITENESS OF 3-CONNECTED 2-CROSSING-CRITICAL GRAPHS WITH NO V n same H -branch is an edge in a digon. If all the H -bridges attaching at u are suchedges, then u has only two neighbours and G is not 3-connected, a contradiction.Thus, every vertex of G is either an H -node or is in some H -bridge that doesnot have all its attachments in the same H -branch. We bound the number of these H -bridges as follows.We claim that, for any three H -nodes u, v, w , at most two H -bridges have allthree of u, v, w as attachments. To see this, suppose three nontrivial H -bridges B i , i = 1 , , 3, all have all of u, v, w as attachments. Each B i contains a claw Y i having u, v, w as talons. Then Y ∪ Y ∪ Y ∪ H contains a subdivision of K , ,in which case 2-criticality implies G is K , . Thus, at most two H -bridges haveattachments in any three nodes. So there are at most 2 (cid:0) n (cid:1) nontrivial H -bridgeswith only node attachments.Every other H -bridge of concern has an attachment in the interior of some H -branch and at some vertex of H not in that H -branch. Lemma 16.10 impliesthat there are at most (2 n )(3 n )41 H -bridges with an attachment in an H -node andin an open H -branch.Lemma 16.11 implies there are at most ( (cid:0) n (cid:1) − n )(164 n + 9) H -bridges havingattachments in the interiors of disjoint H -branches.Lemma 16.13 implies there are at most 2 H -bridges with attachments on twogiven adjacent H -branches and so there are at most 6 n (2) H -bridges with attach-ments on two adjacent H -branches.Every H -bridge has at most 88 vertices, and every vertex of G is either an H -node or in one of these enumerated H -bridges. Therefore, | V ( G ) | ≤ (cid:26) (cid:18) n (cid:19) + 2 n · n · 41 + 6 n (2) + (cid:20)(cid:18) n (cid:19) − n (cid:21) (cid:20) n + 9 (cid:21)(cid:27) . HAPTER 17 Summary This short section provides a single theorem and some remarks summarizingthe current state of knowledge about 2-crossing-critical graphs. Theorem . Let G be a 2-crossing-critical graph.(1) Then G has minimum degree at least two and is a subdivision of a 2-cros-sing-critical graph with minimum degree at least three.Thus, we henceforth assume G has minimum degree at least three.(2) If G is 3-connected and contains a subdivision of V , then G ∈ T ( S ) (Definition 2.12). That is, G is a twisted circular sequence of tiles, eachtile being one of the 42 elements of S (Definition 2.10).(3) If G is 3-connected and does not have a subdivision of V , then G has atmost three million vertices (so there are only finitely many such examples).Each of these examples either • has a subdivision of V or • is either one of the four graphs described in Theorem 15.6 or obtainedfrom a 2-crossing-critical peripherally-4-connected graph with at mostten vertices by replacing each vertex v having precisely three neighborswith one of at most twenty patches, each patch having at most sixvertices (so G has at most sixty vertices).(4) If G is not 3-connected, then either • G is one of 13 examples that are not 2-connected, or • G is 2-connected, has two nonplanar cleavage units, and is one of 36graphs, or • G is 2-connected, has one nonplanar cleavage unit, and is obtainedfrom a 3-connected 2-crossing-critical graph by replacing digons withdigonal paths. We conclude with some remarks on what remains to be done to find all 2-crossing-critical graphs. Remark . In Section 15.7, we provided a method for finding all 3-connected,2-crossing-critical graphs not containing a subdivision of V . It would be desirablefor this program to be completed. Remark . The remaining unclassified 3-connected, 2-crossing-critical graphshave a subdivision of V but not of V . The works of Urrutia [ ] and Austin [ ]have found many of these, but more work is needed to find a complete set. Itmay be helpful to note that we have found all such examples that do not have arepresentativity 2 embedding in the projective plane. The known instances are all quite small, so it is reasonable to expect that each of these has at most 60 verticesor so. ACKNOWLEDGEMENTS Initial impetus to this project came through Shengjun Pan, who describedmechanisms for proving a version of Theorem 2.14 (for G containing a subdivisionof V n , with 2 n likely somewhat larger than 10).We are grateful to CIMAT for hosting us on multiple occasions for work on thisproject. In particular, we appreciate the support of Jose Carlos G´omez Larra˜naga,then director of CIMAT. ibliography 1. D. Archdeacon, A Kuratowski theorem for the projective plane, Ph.D. thesis, The Ohio StateUniversity, 1980.2. D. Archdeacon, A Kuratowski theorem for the projective plane, J. Graph Theory 5 (1981),no. 3, 243–246.3. E. Austin, 2-crossing critical graphs with a V -minor, MMath thesis, U. Waterloo, 2011,http://uwspace.uwaterloo.ca/handle/10012/6464.4. D. W. Barnette, Generating projective plane polyhedral maps, J. Combin. Theory Ser. B 51(1991), no. 2, 277–291.5. L. Beaudou and D. Bokal, On the sharpness of some results relating cuts and crossing numbers,Electron. J. Combin. 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