Big mapping class groups and the co-Hopfian property
BBIG MAPPING CLASS GROUPS AND THE CO-HOPFIANPROPERTY
JAVIER ARAMAYONA, CHRISTOPHER J. LEININGER, AND ALAN MCLEAY
Abstract.
We study injective homomorphisms between big mapping class groups ofinfinite-type surfaces. First, we construct (uncountably many) examples of surfaceswithout boundary whose (pure) mapping class groups are not co-Hopfian; these are thefirst such examples of injective endomorphisms of mapping class groups that fail to besurjective.We then prove that, subject to some topological conditions on the domain surface,any continuous injective homomorphism between (arbitrary) big mapping class groupsthat sends Dehn twists to Dehn twists is induced by a subsurface embedding.Finally, we explore the extent to which, in stark contrast to the finite-type case,superinjective maps between curve graphs impose no topological restrictions on theunderlying surfaces. Introduction and main results
Throughout this article, all surfaces will be assumed to be connected, orientable andsecond-countable, unless otherwise specified. A surface S has finite type if its fundamentalgroup is finitely generated, and has infinite type otherwise.The mapping class group Map( S ) is the group of orientation-preserving homeomor-phisms of S , up to homotopy; if ∂S (cid:54) = ∅ , then we require that all homeomorphisms andhomotopies fix ∂S pointwise. In the case when S has finite type, Map( S ) is well-knownto be finitely presented. If, on the contrary, S has infinite type, then Map( S ) becomes anuncountable, totally disconnected, non-locally compact topological group with respect tothe quotient topology stemming from the compact-open topology on the homeomorphismgroup of S . We refer the reader to Section 2 for expanded definitions, and to the recentsurvey [5] for a detailed treatment of mapping class groups of infinite-type surfaces, nowcommonly known as big mapping class groups .1.1. Algebraic rigidity for mapping class groups.
A seminal result of Ivanov [23]states that if S is a (sufficiently complicated) finite type surface S , then every automor-phism of Map( S ) is a conjugation by an element of the extended mapping class group Map ± ( S ), namely the group of all homeomorphisms of S up to isotopy. In other words,every isomorphism Map( S ) → Map( S ) is induced by a homeomorphism S → S . Theanalog for infinite-type surfaces was recently obtained by Bavard-Dowdall-Rafi [8]; seeTheorem 2.3 below.The proofs in the finite and infinite type settings proceed along similar lines, based onthe work of Ivanov [23]. First, one proves an algebraic characterization of Dehn twistsand uses it to deduce that any isomorphism φ : Map( S ) → Map( S ) must send (pow-ers of) Dehn twists to (powers of) Dehn twists. In particular, φ induces a simplicialautomorphism φ ∗ : C ( S ) → C ( S ) of the curve complex . At this point, the argument J.A. was supported by grant PGC2018-101179-B-I00. He acknowledges financial support from the Span-ish Ministry of Science and Innovation, through the “Severo Ochoa Programme for Centres of Excellencein R&D” (CEX2019-000904-S). C.L. was partially supported by NSF grants DMS-1811518 and DMS-2106419. A.M was supported by the Luxembourg National Research Fund OPEN grant O19/13865598. a r X i v : . [ m a t h . G T ] M a r JAVIER ARAMAYONA, CHRISTOPHER J. LEININGER, AND ALAN MCLEAY boils down to showing that every simplicial automorphism of C ( S ) is induced by a home-omorphism of S [23]. In fact, this approach applies, with a finite-list of exceptionalsurfaces, to any isomorphism between finite-index subgroups of Map( S ) or the pure map-ping class group PMap( S ), the subgroup of Map( S ) whose elements fix every end of S [23, 9, 10, 21, 22, 20, 31]1.2. The co-Hopfian property.
The combination of results of Ivanov-McCarthy [24]and Bell-Margalit [10] imply that, with the possible exception of the twice-puncturedtorus, if S has finite type then every injective homomorphism Map( S ) → Map( S ) isinduced by a homeomorphism of S . In particular, the mapping class group of a (sufficientlycomplicated) finite-type surface is co-Hopfian: every injective endomorphism is surjective.For infinite-type surfaces, Question 4.5 of the AIM Problem List on Surfaces of InfiniteType [1] (see also [5, Question 5.2]) asks: Question 1.1.
Are big mapping class groups co-Hopfian?
Remark.
The answer to the above question is immediately “no” for surfaces with non-empty boundary. For instance, if a surface S has non-empty boundary and its space ofends is homeomorphic to a Cantor set, then there exists a proper π -injective continuousmap S → S that is not surjective. In turn, this map induces an injective homomorphismMap( S ) → Map( S ) that is not surjective; compare with Theorem 2 below.Our first result states that the answer to Question 1.1 is also negative for surfaces with-out boundary. Recall that the Loch Ness Monster is the unique (up to homeomorphism)connected orientable surface of infinite genus and exactly one end. We will prove:
Theorem 1.
Let S be either the once-punctured Loch Ness Monster or a torus minusthe union of a Cantor set and an isolated point. Then there exists a homomorphism φ : Map( S ) → Map( S ) such that:(1) φ is continuous and injective, but not surjective,(2) There is a Dehn twist t such that φ ( t ) is not supported on any finite-type subsurfaceof S . In the particular case when S is the Loch Ness Monster, no power of φ ( t ) is supported on a finite-type subsurface of S .(3) There exists a partial pseudo-Anosov f ∈ Map( S ) such that φ ( f ) is a multitwist. Remark.
Part (2) of the above theorem yields a negative answer to Problem 4.75 of theAIM Problem List on Surfaces of Infinite Type [1].The construction behind Theorem 1 is inspired by the construction of the non-geometric injective homomorphism of the first two authors with Souto [2, Theorem 2], building on awell-known construction of homomorphisms from covers (see e.g. [12, 24, 34]). Namely, weconstruct a covering map S (cid:48) → S which induces an injective homomorphism Map( S ) → Map( S (cid:48) ), in such a way that the surface S (cid:48)(cid:48) obtained by filling in all but one of thepunctures of S (cid:48) is homeomorphic to S , and yet the homomorphism Map( S ) → Map( S (cid:48)(cid:48) )remains injective. Once this has been done, the resulting homomorphism is easily seento not be surjective in light of Bavard-Dowdall-Rafi’s result [8] mentioned above, since φ sends some finitely supported elements to elements which are not finitely supported.Motivated by this construction, we also observe the following. Theorem 1.2.
Let S denote the closed surface of genus g ≥ minus the union of aCantor set and an isolated point. Then there exists a continuous injective homomorphism Map( S ) → Map( R (cid:114) C ) , where C denotes a Cantor set. As far as we know, these are the first examples of injective homomorphisms betweenmapping class groups where the genus decreases from domain to codomain.
IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 3
Not co-Hopfian pure mapping class groups.
If we restrict Question 1.1 to puremapping class groups, we will see that there is a much richer palette of examples ofinjective, but not surjective, homomorphisms. We say that a surface S is self-doubling ifthere exists a multicurve B such that;(1) S \ B has two connected components L and R ,(2) L and R are both homeomorphic to S , and(3) there exists orientation-reversing homeomorphism ι : S → S of order two suchthat ι ( L ) = R and ι ( b ) = b for each b ∈ B . Theorem 2.
For every self-doubling surface S there exists a continuous injective homo-morphism PMap( S ) → PMap( S ) that is not surjective. Examples.
We now list some concrete examples of surfaces to which Theorem 2 applies.(1) A first example of a self-doubling surface S is the sphere minus the union of aCantor set and the north and south pole: any essential curve b that separates thepoles splits S into two subsurfaces L and R with interiors homeomorphic to S .(2) Alternatively, suppose S has infinite genus such that every planar end is isolated,and for every non-planar end there is a sequence of planar ends converging to it.(We remark that there are uncountably many surfaces with this property.) Then S is also self-doubling; see Theorem 4.2.We give an equivalent definition of self-doubling surfaces, along with more examples,in Section 4.1.4. Twist-preserving homomorphisms.
In light of the discussion above, we nowfocus on homomorphisms that send Dehn twists to Dehn twists; we call these twist-preserving homomorphisms . In this section we will allow surfaces to have a non-emptyboundary, though we assume it is compact (and is thus a finite union of circles). We willprove the following result; recall that a map between topological spaces is proper if thepreimage of every compact set is compact:
Theorem 3.
Let S and S (cid:48) be surfaces of infinite type, where S has positive genus. Assumefurther that either the boundary of S is empty, or else S has at most one end accummu-lated by genus. If φ : PMap( S ) → PMap( S (cid:48) ) is a continuous injective twist-preservinghomomorphism, then there is a proper π -injective embedding S → S (cid:48) that induces φ . Remark.
Theorem 3 no longer holds if ∂S (cid:54) = ∅ and S has more than one end accumulatedby genus; see the final remark of Section 5. Also, in the same remark we will see thatthe result we will prove in fact applies to a larger class of homomorphisms than injectiveones.Continuing with our discussion, observe that if ∂S = ∅ , any proper π -injective embed-ding S → S (cid:48) is homotopic to a homeomorphism. Hence we obtain: Corollary 1.3.
Let S and S (cid:48) be surfaces of infinite type, where S has positive genusand no boundary. If φ : PMap( S ) → PMap( S (cid:48) ) is a continuous injective twist-preservinghomomorphism, then it is induced by a homeomorphism S → S (cid:48) ; in particular, it issurjective. To prove Theorem 3, one first observes that φ induces a simplicial map C ( S ) → C ( S (cid:48) )that preserves intersection number one. Once this has been shown, the result followsconsidering exhaustions and a result of Souto and the first named author [4], plus acontinuity argument. We conjecture that, although continuity is needed in our proof,it is in fact not necessary for Theorem 3 to hold. In this direction, a result of Mann[26] states that the mapping class group of certain surfaces have the automatic continuity property. Specifically, this automatic continuity holds when the surface is a closed surfacepunctured along the union of a Cantor set and a finite set; for these surfaces we obtain: JAVIER ARAMAYONA, CHRISTOPHER J. LEININGER, AND ALAN MCLEAY
Corollary 1.4.
Let S and S (cid:48) be surfaces of infinite type, where S is obtained from aclosed surface of positive genus by removing the union of a Cantor set and a finite set. If φ : PMap( S ) → PMap( S (cid:48) ) is an injective twist-preserving homomorphism, then there isa homeomorphism S → S (cid:48) that induces φ . Injections between curve graphs.
In the finite-type setting, an important stepfor establishing the co-Hopfian property for finite-index subgroups of mapping class groupsis that superinjective self-maps of the curve graph are induced by homeomorphisms; see[21], for instance. We recall that a simplicial map between curve graphs is superinjective if it maps pairs of curves with non-zero intersection number to pairs of curves with thesame property.While it is known that every automorphism of the curve graph of an infinite-type surfaceis induced by a homeomorphism, [17, 8], it is easy to see that this is no longer true if onedrops the requirement that the map be surjective; see [19, 5] for concrete examples. Ourfinal result highlights the extent of this failure:
Theorem 4.
Let S and S (cid:48) be infinite-type surfaces, where S (cid:48) has infinite genus. Then:(1) There exists a superinjective simplicial map C ( S ) → C ( S ) that is not surjective.(2) There exists a superinjective simplicial map C ( S ) → C ( S (cid:48) ) . In the case of infinite-genus surfaces, a proof of part (1) Theorem 4 was previouslyobtained by Hern´andez-Valdez [19] using different methods.
Remark. If S and S (cid:48) are arbitrary infinite-type surfaces, then there is always an injective simplicial map C ( S ) → C ( S (cid:48) ). To see this, simply observe that the curve graph of anyinfinite-type surface contains the complete graph on countably many vertices, and thatthe curve graph of any surface has a countable set of vertices.Theorem 4 should be compared with Theorem 3, which implies that every superin-jective map between curve graphs that comes from an injective homomorphism betweenthe corresponding mapping class groups is induced by an embedding of the underlyingsurfaces. Plan of the paper.
In Section 2 we will give all the necessary background needed inthe article. Sections 3.2 and 3.3 are devoted to the proof of Theorem 1 in the cases,respectively, of the one-holed torus minus a Cantor set and the once-punctured Loch NessMonster. In Section 4 we deal with Theorem 2. Section 5 is devoted to the proof Theorem3. Finally, in Section 6 we establish Theorem 4.
Acknowledgements.
We are grateful to the organizers of the AIM workshop “Surfacesof Infinite Type” for the discussions that are the origins of this paper. Thanks also to AIMfor its hospitality and financial support. We would like to thank Nick Vlamis and HenryWilton for enlightening conversations about Lemma 6.2, and Vlamis for the reference[14]. We also thank Israel Morales for bringing to our attention the refererence [6], andTy Ghaswala for comments on an earlier version.2.
Preliminaries
In this section we will briefly introduce all the background material needed for ourresults. We refer the reader to the survey paper [5] for a detailed account on these topics.2.1.
Surfaces.
In what follows, by a surface we will mean a orientable second-countabletopological surface, with a finite number (possibly zero) of boundary components, all ofthem assumed to be compact. If the fundamental group of a surface S is finitely generated,we will say that S has finite type ; otherwise it has infinite type . The space of ends of S isEnds( S ) := lim ← π ( S (cid:114) K ) , IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 5 where K denotes a compact subset of S . The space of ends becomes a topological spacewith respect to the final topology obtained from endowing each set defining the inversesystem with the discrete topology. It is well-known that Ends( S ) is a closed subset of aCantor set. We say that an end is planar if it has a planar neighborhood; otherwise wewill say that the end is non-planar or that it is accumulated by genus . An isolated planarend is usually called a puncture ; as is customary, we will treat punctures both as ends andas marked points on the surface, switching between the two viewpoints without furthermention. Finally, we will denote by Ends ∞ ( S ) the subspace of non-planar ends, which isclosed in Ends( S ).A classical result [29] asserts that any surface S is uniquely determined by the quadru-ple ( g, b, Ends( S ) , Ends ∞ ( S )), where g ∈ N ∪ {∞} is the genus, b ∈ N is the number ofboundary components. More concretely, two surfaces are homeomorphic if and only iftheir genera and number of boundary components are equal, and there is a homeomor-phism between the spaces of ends that restricts to a homeomorphism between the spacesof non-planar ends.2.2. Curves and multicurves.
A simple closed curve on S is said to be inessential if itbounds a disk, a once-punctured disk, or an annulus whose other boundary is a boundarycomponent of S ; otherwise we say the curve is essential . By a curve we mean the isotopyclass of an essential simple closed curve on S . A multicurve is a set of curves on the surfacethat have pairwise disjoint representatives. Given a multicurve Q , we will write S (cid:114) Q to mean the (possibly disconnected) surface obtained from S by cutting along (pairwisedisjoint representatives of) each element of Q .2.3. Mapping class group.
Let Homeo + ( S ) be the group of all orientation-preservingself-homeomorphisms of S , equipped with the compact-open topology. We will denote byHomeo ( S ) the connected component of the identity in Homeo + ( S ), noting it is a normalsubgroup. The mapping class group of S isMap( S ) := Homeo + ( S ) / Homeo ( S ) . As is customary, if ∂S (cid:54) = ∅ , in this definition we implicitly require that all homeomor-phisms fix ∂S pointwise.The mapping class group is naturally a topological group with respect to the quotienttopology coming from the compact-open topology on Homeo + ( S ). It is a classical factthat Map( S ) is a finitely presented group if S has finite type, while it is easy to see thatit is uncountable whenever S has infinite type. Moreover, in this latter case, Map( S ) istotally disconnected and not locally compact; see [5] for more details.One of the motivating results for us is the following theorem of Bavard-Dowdall-Rafi[8] mentioned in the introduction. Theorem ([8]) . For any infinite-type surface S , any isomorphism Map( S ) → Map( S ) isinduced by a homeomorphism S → S . In particular, any such isomorphism is continuous.The pure mapping class group PMap( S ) is the subgroup of Map( S ) whose elements fixevery end of S . If S has finite type, it is well-known that PMap( S ) is generated by Dehntwists along a finite set of curves (see [16, Section 4.4]). When S has infinite type, this isno longer true in general; instead, we have the following result; see [28] for the definitionof a handle shift : Theorem 2.1 ([28]) . PMap( S ) is topologically generated by Dehn twists if and only if S has at most one end accumulated by genus. Otherwise, it is topologically generated byDehn twists and handle shifts. JAVIER ARAMAYONA, CHRISTOPHER J. LEININGER, AND ALAN MCLEAY
Here, being topologically generated by Dehn twists means that the subgroup generatedby Dehn twists (i.e. the compactly-supported mapping class group, see below) is dense inPMap( S ).2.3.1. Some important subgroups.
Following [8], we will say that an element of Map( S )has finite support if it represented by a homeomorphism which is the identity outsidesome finite-type subsurface of S . We will write Map f ( S ) for the subgroup consistingof elements with finite support. Although not directly relevant to our arguments, werecord the following beautiful result of Bavard-Dowdall-Rafi [8], which gives an algebraiccharacterization of elements with finite support. In particular, it serves to shed light onthe potential differences between isomorphisms and injective homomorphisms betweenbig mapping class groups. Proposition 2.2 ([8]) . An element of
Map( S ) has finite support if and only if its conju-gacy class in Map( S ) is countable. The compactly-supported mapping class group Map c ( S ) is the subgroup of Map( S )whose elements are represented by homeomorphisms which are the identity outside somecompact subsurface of S . Observe that Map c ( S ) is a subgroup of Map f ( S ), proper when S has more than one puncture, and of PMap( S ). On the other hand, Map f ( S ) is a subgroupof PMap( S ) if and only if S has at most one puncture. Before ending this section, werecord the following immediate observation for future use: Lemma 2.3.
For every infinite-type surface S , we have Map c ( S ) = lim −→ Map( X ) , where the direct limit is taken over all compact subsurfaces X ⊂ S , directed with respectto inclusion. Curve graph.
The curve graph C ( S ) is the simplicial graph whose vertex set is theset of curves on S , and where two vertices are deemed to be adjacent if the correspondingcurves may be realized disjointly on S . In what follows we will not distinguish betweenvertices of the curve graph and the curves representing them.Observe that Map( S ) acts on C ( S ) by simplicial automorphisms. A classical fact,discovered initially by Ivanov [23] is that every simplicial automorphism of C ( S ) is inducedby a homeomorphism of S , provided S is not the twice-punctured torus. The analog forinfinite-type surfaces has been obtained independently by Hern´andez-Morales-Valdez [18]and Bavard-Dowdall-Rafi [8]. A crucial step in their proofs is the following so-called Alexander method for infinite-type surfaces [18], which we record for future use:
Theorem 2.4.
Let S be a connected orientable infinite-type surface with empty boundary.The natural action of Map( S ) on C ( S ) has trivial kernel; in other words, if f ∈ Map( S ) induces the identity transformation on C ( S ) , then it is the identity in Map( S ) . Covers and forgetting
In this section we prove Theorem 1 and Theorem 1.2. These theorems are abouthomomorphisms between mapping class groups of surfaces with a single puncture, andthe proofs exploit their relationship with automorphism groups and the Birman exactsequence. We start with some generalities, then focus on the specific cases of the theorems.3.1.
Covers and automorphism groups.
In what follows, suppose S is an orientable,connected surface without punctures such that π ( S ) is non-abelian. Given z ∈ S anypoint, let S z = S (cid:114) { z } denote the surface obtained from S by removing z (thus producing apuncture we call the z –puncture ). Any homeomorphism f of S z induces a homeomorphism IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 7 of S that fixes the point z , which by an abuse of notation we also denote by f . This definesa homomorphism Homeo + ( S z ) → Homeo + ( S ) . This homomorphism sends Homeo ( S z ) into Homeo ( S ) and so induces a homomorphismMap( S z ) → Map( S ) . Given a loop γ representing an element of π ( S, z ), one associates a homeomorphism f γ : S z → S z by “point pushing along γ ”. More precisely, f γ : S → S is a homeomorphismisotopic to the identity by an isotopy f t with f = f γ , f = id, and f t ( z ) = γ ( t ), for all t ∈ [0 , S has finite type, Birman proved that [ γ ] (cid:55)→ [ f γ ] defines an injectivehomomorphism π ( S, z ) → Map( S z ) in such a way that the sequence(1) 1 (cid:47) (cid:47) π ( S, z ) (cid:47) (cid:47) Map( S z ) (cid:47) (cid:47) Map( S ) (cid:47) (cid:47) Proposition 3.1.
For any connected, orientable surface S without punctures and non-abelian fundamental group, there is an injection of π ( S, z ) → Map( S z ) given by [ γ ] (cid:55)→ [ f γ ] making (1) exact. (cid:4) For the remainder of this section, we use this theorem to identify π ( S, z ) with its imagein Map( S z ). Remark.
The theorem also holds when S has punctures, replacing Map( S z ) with thesubgroup preserving the z –puncture.Given f ∈ Homeo + ( S z ), after extending over z we have an induced automorphism f ∗ ∈ Aut( π ( S, z )), which descends to a homomorphism ι : Map( S z ) → Aut( π ( S, z ))given by ι ([ f ]) = f ∗ . This further descends to a homomorphism Map( S ) → Out( π ( S ))which is injective by Theorem 2.4. The inclusion of π ( S, z ) into Map( S z ), composed with ι is precisely the isomorphism onto the group of inner automorphisms, and thus we get ahomomorphism of short exact sequences:1 (cid:47) (cid:47) π ( S, z ) (cid:47) (cid:47) (cid:15) (cid:15) Map( S z ) (cid:47) (cid:47) ι (cid:15) (cid:15) Map( S ) (cid:47) (cid:47) (cid:15) (cid:15) (cid:47) (cid:47) π ( S, z ) (cid:47) (cid:47) Aut( π ( S, z )) (cid:47) (cid:47) Out( π ( S, z )) (cid:47) (cid:47) . The first vertical map is the identity and the last is injective, from which it follows thatthe middle is also injective.A regular covering space p : (cid:101) S → S with the property that every homeomorphism of S lifts to a homeomorphism of (cid:101) S will be called a geometrically characteristic cover. If p ∗ ( π ( (cid:101) S, (cid:101) z )) is characteristic, general map lifting implies that the cover is geometricallycharacteristic, but this is not a necessary condition in general; see the examples below.Now suppose p : (cid:101) S → S is geometrically characteristic, let z ∈ S be a point, andfix any (cid:101) z ∈ p − ( z ). Being regular and geometrically characteristic implies that for any f ∈ Homeo + ( S z ), after extending over z , there is a unique lift (cid:101) f : (cid:101) S → (cid:101) S that fixes (cid:101) z , andthus determines a homeomorphism of the same name (cid:101) f ∈ Homeo + ( (cid:101) S (cid:101) z ). The general factwe need is the following (compare with [2, Theorem 2]). Proposition 3.2. If p : (cid:101) S → S is a geometrically characteristic covering space, π ( S ) isnon-abelian, and π ( (cid:101) S ) is non-trivial, then the assignment f (cid:55)→ (cid:101) f described above descendsto a continuous, injective homomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ) . Moreover, via the inclusions from the Birman exact sequence we have φ ◦ p ∗ = id | π ( (cid:101) S, (cid:101) z ) . JAVIER ARAMAYONA, CHRISTOPHER J. LEININGER, AND ALAN MCLEAY
Proof.
Continuity of Homeo + ( S z ) → Homeo + ( (cid:101) S (cid:101) z ) is clear. Since f (cid:55)→ (cid:101) f maps Homeo ( S z )to Homeo ( (cid:101) S, (cid:101) z ) (by lifting isotopies), we get a well-defined, continuous homomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ). Since the homomorphisms ι : Map( S z ) → Aut( π ( S, z )) and˜ ι : Map( (cid:101) S (cid:101) z ) → Aut( π ( (cid:101) S, (cid:101) z )) are injective, to prove the first part of the proposition itsuffices to prove injectivity of the homomorphism φ ∗ : ι (Map( S z )) → (cid:101) ι (Map( (cid:101) S (cid:101) z )) definedby φ ∗ ◦ ι = (cid:101) ι ◦ φ , or more explicitly ι ([ f ]) = f ∗ φ ∗ (cid:55)−→ ˜ f ∗ = ˜ ι ([ ˜ f ]) = ˜ ι ( φ ([ f ])) . Consequently, we need only show that the kernel of this homomorphism is trivial.To this end, suppose that f ∈ Homeo + ( S z ) is any element such that ˜ f ∗ is the identity.Let G = p ∗ ( π ( (cid:101) S, (cid:101) z )) (cid:47) π ( S, z ) , so that p ∗ is an isomorphism from π ( (cid:101) S, (cid:101) z ) onto the normal subgroup G (normality followsfrom regularity of p ). Since (cid:101) f is a lift of f that preserves (cid:101) z , we have that p ∗ ◦ ˜ f ∗ = f ∗ ◦ p ∗ . In particular, this means that f ∗ preserves G and p ∗ conjugates (cid:101) f ∗ to f ∗ | G . Since ˜ f ∗ is theidentity, this implies that f ∗ | G is the identity. We claim that f ∗ is the identity. To provethis, first observe that for all x ∈ G and a ∈ π ( S, z ), we have axa − ∈ G , and thus axa − = f ∗ ( axa − ) = f ∗ ( a ) f ∗ ( x ) f ∗ ( a ) − = f ∗ ( a ) xf ∗ ( a ) − . This implies that for all a ∈ π ( S, z ) and all x ∈ G , a − f ∗ ( a ) commutes with x .Since π ( S, z ) is a non-abelian surface group (either free or a closed surface group), weknow that centralizers of elements are cyclic, and since G is a non-trivial normal subgroupof π ( S, z ), we can find two elements x, y ∈ G who centralizers intersect trivially. Nowfor any a ∈ π ( S, z ), a − f ∗ ( a ) is in the centralizer of x and y , and thus a − f ∗ ( a ) = 1.Therefore, f ∗ ( a ) = a , and hence f ∗ is the identity proving that φ ∗ , and hence φ , isinjective.To prove the last statement, given any element g of a group, let c g denote the innerautomorphism determined by conjugating by g . From the inclusions in the Birman exactsequence as noted above, given any [ γ ] ∈ π ( (cid:101) S, (cid:101) z ) we have (cid:101) ι ([ γ ]) = c [ γ ] and ι ( p ∗ ([ γ ])) = c p ∗ ([ γ ]) . Since φ ∗ ◦ ι = (cid:101) ι ◦ φ , we have (cid:101) ι ( φ ( p ∗ ([ γ ]))) = φ ∗ ( ι ( p ∗ ([ γ ]))) = φ ∗ ( c p ∗ ([ γ ]) ) = c [ γ ] = (cid:101) ι ([ γ ]) , where the second-to-last equality comes from the fact that the isomorphism describedabove, p ∗ : π ( (cid:101) S, (cid:101) z ) → G < π ( S, z ), conjugates c [ γ ] to c p ∗ ([ γ ]) , while the argument aboveshows that it conjugates φ ∗ ( c p ∗ ([ γ ]) ) to this as well, hence c [ γ ] = φ ∗ ( c p ∗ ([ γ ]) ). Therefore, (cid:101) ι ◦ φ ◦ p ∗ = (cid:101) ι and since (cid:101) ι is injective, it follows that φ ◦ p ∗ is the identity on π ( (cid:101) S, (cid:101) z ), asrequired. (cid:4) The once-punctured torus minus a Cantor set.
Here we prove Theorem 1 inone of the two cases.
Proof of Theorem 1 for the once-punctured torus minus a Cantor set.
Let T be a torusand choose, once and for all, a Cantor set C ⊂ T , set S = T (cid:114) C , fix z ∈ S , and write S z = S (cid:114) z as above. Then C ∪ { z } is canonically homeomorphic to Ends( S z ) and T isthe end-compactification of S z . Consider the surjective homomorphism π ( S, z ) → π ( T, z ) ∼ = Z × Z IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 9 given by “filling in” every element of C . Fixing any integer m ≥ m , we get a surjective homomorphism ρ : π ( T, z ) → Z m × Z m . Let p : (cid:101) T → T be the regular cover corresponding to ker( ρ ), and fix some ˜ z ∈ p − ( z ).Write (cid:101) C = p − ( C ), and observe that the surface (cid:101) S = (cid:101) T (cid:114) (cid:101) C is homeomorphic to S , and (cid:101) S (cid:101) z = (cid:101) S (cid:114) { (cid:101) z } is homeomorphic to S z .Every homeomorphism of T lifts to (cid:101) T since the kernel of π ( T, z ) → Z m × Z m is characteristic. Moreover, every homeomorphism of S extends uniquely to a homeo-morphism of T and hence the restricted covering (with the same name) p : (cid:101) S → S isgeometrically characteristic. Proposition 3.2 therefore provides an injective, continuoushomomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ). Since (cid:101) S (cid:101) z and S z are homeomorphic surfaces,by choosing a homeomorphism between them we can view φ as a continuous homomor-phism φ : Map( S z ) → Map( S z ). We have already shown that φ is injective. We nowverify the remaining properties claimed in the statement: Compact to non-finite support.
To prove part (2) of Theorem 1 we will show that the φ –image of the Dehn twist t a ∈ Map c ( S z ) about a non-separating curve a is not supportedon any finite-type subsurface of S z .In order to achieve this it suffices to find a point of ˜ C that is not fixed by our chosenlift (cid:101) t a = φ ( t a ). Let β ⊂ S z be an arc which begins at z and ends at a point e ∈ C so that β essentially intersects a exactly once. If ˜ β is the lift of β starting at (cid:101) z then (cid:101) t a ( ˜ β ) alsostarts at (cid:101) z , but these two arcs terminate at distinct points in the preimage of e . Therefore φ ( t a ) does not have finite support. (Observe, however, that there is a nontrivial powerof φ ( t a ) that has finite support; this will not happen in the case of the Loch Ness Monster.) Not co-Hopfian.
Suppose φ were surjective, and thus an automorphism. Theorem 2.3implies that φ is induced by a homeomorphism of S z , and in particular preserves theproperty of having finite support, which is a contradiction to the previous paragraph.This proves part (1) Non-geometric.
To prove part (3), and thus complete the proof of Theoreom 1, wemust show that there are partial pseudo-Anosovs in Map( S z ) that map to multitwists inMap( (cid:101) S (cid:101) z ). The proof is essentially the same as that of [2, Theorem 2(1)]; we sketch it forcompleteness.As in [2], one can find a loop γ representing an element in π ( (cid:101) S, (cid:101) z ) which is simple,but for which p ∗ ([ γ ]) is not represented by any simple closed curve. The mapping classassociated to [ γ ] via the Birman exact sequence is a multi-twist about the boundary ofa regular neighborhood of γ , while by a result of Kra [25], p ∗ ([ γ ]) is a pseudo-Anosovon X (cid:114) { z } , where X is the subsurface filled by a loop representing p ∗ ([ γ ]) with minimalself-intersection. Since φ ◦ p ∗ ([ γ ]) = [ γ ] by Proposition 3.2, it follows that p ∗ ([ γ ]) is pseudo-Anosov on a proper subsurface while φ ( p ∗ ([ γ ])) is a multi-twist. This completes the proofof the theorem in this case. (cid:4) The Loch Ness Monster.
In this section we prove Theorem 1 for the once-punctured Loch Ness Monster. In what follows, S will denote the Loch Ness Monster ,that is, the connected orientable surface of infinite genus and exactly one end. As inthe case of a torus minus a Cantor set, we fix once and for all, a point z ∈ S . We willagain apply Proposition 3.2 for an appropriate cover p : (cid:101) S → S to induces an injectivehomomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ). This time the cover will be of infinite degree which, in turn, will be the key to the existence of compactly-supported elements withimage for which no non-trivial power has compact support. Consider the homomorphism ρ : π ( S, z ) → H ( S, Z )obtained by first abelianizing and then reducing modulo 2. Observe that H := ker( m ) isa characteristic subgroup of π ( S, z ). Let p : ˜ S → S be the cover associated to H , whichis usually called the mod-2 homology cover of S . We claim: Proposition 3.3. (cid:101) S is homeomorphic to S .Proof. First, observe that the Loch Ness Monster S is a characteristic cover of the closedsurface Σ of genus 2; more precisely, it is the covering surface associated to the commutatorsubgroup of π (Σ). Since (cid:101) S is a characteristic cover of S , it follows that (cid:101) S is also acharacteristic cover of Σ.Let H be the characteristic subgroup of π (Σ) corresponding to π ( (cid:101) S ) and D = π (Σ) /H the group of deck transformations of ˜ S → Σ. Since this action is properlydiscontinuous and cocompact, the ˘Svarc-Milnor Lemma (see e.g. [13]) implies that D is quasi-isometric to (cid:101) S . By Stallings’ Theorem [33] D , or equivalently ˜ S , has one end,two ends, or infinitely many ends. By the classification of infinite-type surfaces [29], itfollows that (cid:101) S is homeomorphism to either the Loch-Ness Monster S ; Jacob’s Ladder L ;the Cantor tree T ; or the Blooming Cantor Tree B ; see [7] for details. Since S is theonly one of these that has one end, we suppose (cid:101) S has more than one end and derive acontradiction.To this end, we appeal to Stallings Theorem again and note that D admits a non-trivial action on a tree A with finite edge stabilizers and without edge inversions. Fromthis action construct an equivariant map f : (cid:101) S → A , which we can assume is transverseto the union of midpoints X ⊂ T of every edge, so that f − ( X ) is a properly embedded1–submanifold which is D –invariant (c.f. [32]). This descends to a closed 1–submanifoldin Σ, and via the action of π (Σ) → D on A , any loop in the complement of the 1–submanifold fixes a vertex. Observe that there is a non-separating simple closed curvein a component of this complement (this is true for any 1–submanifold in the genus 2surface Σ) which, when viewed as an element of π (Σ), fixes a vertex. Since the cover ischaracteristic, every non-separating simple closed curve fixes a vertex of A .Now choose a generating set a , . . . , a so that all a i are simple as are a i a j , for all i (cid:54) = j .Then all a i and a i a j act elliptically, in which case Serre’s criterion [30] implies π (Σ) fixesa point of A , which is a contradiction. (cid:4) Proof of Theorem 1 for the once-punctured Loch Ness Monster.
Fixing (cid:101) z ∈ p − ( z ), Propo-sition 3.2 implies that p induces an injective homomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ).Since p has infinite degree, there are Dehn twists whose image does not have compactsupport, even up to taking powers. For the same reason, φ is not surjective. The factthat there exist partial pseudo-Anosovs whose image is a Dehn twist follows along theexact same lines as in the torus case. Finally, since (cid:101) S and S are homeomorphic, so are (cid:101) S (cid:101) z and S z , and thus we may view the homomorphism φ as an injective, but not surjective,endomorphism of Map( S z ). This finishes the proof of the Theorem in the case of theLoch Ness Monster. (cid:4) Remark.
After this article was finished, we learned that Proposition 3 had already beenestablished in [6, Proposition 4.1].3.4.
Decreasing genus.
Our final application of Proposition 3.2 is the following.
Proof of Theorem 1.2.
For g ≥
1, let Σ g denote the surface of genus g ≥
1, let C g ⊂ Σ g be a Cantor set, and S = Σ g (cid:114) C g . Let p : (cid:101) Σ → Σ g be the universal cover. Choosing a IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 11 disk D ⊂ Σ that contains C g , we note that p − ( C g ) is a disjoint union of Cantor sets thatonly accumulate at infinity of (cid:101) Σ. Consequently, since the one-point compactification of (cid:101)
Σis the sphere, it follows that (cid:101) S = (cid:101) Σ (cid:114) p − ( C g )is homeomorphic to the 2–sphere minus a Cantor set.Now observe that p : (cid:101) S → S is a geometrically characteristic cover, and so for anybasepoint z ∈ S and choice of (cid:101) z ∈ p − ( z ), Proposition 3.2 implies that p induces aninjective homomorphism φ : Map( S z ) → Map( (cid:101) S (cid:101) z ). Observe that S z is a surface of genus g minus a Cantor set and an isolated point, while (cid:101) S (cid:101) z is a 2–sphere minus a Cantor setand an isolated point, which is also homeomorphic to the plane R minus a Cantor set.This completes the proof. (cid:4) Self-Doubling
In this section we will prove Theorem 2 and provide a large class of self-doublingsurfaces. We will use an argument similar to the example in [24, Section 2], althoughadapted to the case of surfaces without boundary.4.1.
The construction.
Let ¯ S be an orientable connected surface with non-empty bound-ary. In this section, we still require each boundary component of ¯ S be compact, althoughthis time we allow the set of boundary components to be countable. Let B = { b , b , . . . } be a (possibly finite) subset of the set of boundary components, and S be the surface thatresults from ¯ S by gluing disks with marked points z i onto b i . This operation gives riseto a boundary deleting homomorphism PMap( ¯ S ) → PMap( S ), which fits in a short exactsequence(2) 1 → Π i (cid:104) T b i (cid:105) → PMap( S ) → PMap( S ) → , see [16] for details. Now suppose d B S is the surface obtained by gluing two disjoint copiesof ¯ S along the boundary components in B . This operation induces two natural inclusionmaps ψ : ¯ S (cid:44) → d B S and ψ : ¯ S (cid:44) → d B S such that int( ψ ( ¯ S )) ∩ int( ψ ( ¯ S )) = ∅ and ψ ( b i ) = ψ ( b i ) is an essential curve. We abusenotation by writing b i and B for the images of b i and B , respectively. Furthermore, thereis an induced orientation-reversing homeomorphism ι : d B S → d B S that swaps the images of ψ and ψ and, in particular, fixes the set B ⊂ d B S . On thelevel of mapping class groups, we have two injective homomorphismsΨ : PMap( ¯ S ) (cid:44) → PMap( d B S ) and Ψ : PMap( ¯ S ) (cid:44) → PMap( d B S ) , such that for all f ∈ PMap( S ), we have Ψ ( f ) = f , Ψ ( f ) = f , where f = ιf ι − ; herewe consider f , f , and ι as elements of Map ± ( d B S ). In particular, a left twist about b i is sent to a left twist by Ψ , and to a right twist by Ψ . Consider now the map d : PMap( ¯ S ) → PMap( d B S )such that d ( f ) = f f . Note that d is indeed a homomorphism as the images of Ψ andΨ commute. Furthermore, the kernel of d is equal to Π i (cid:104) T b i (cid:105) , so from the short exactsequence (2) we have an induced injective homomorphismPMap( S ) (cid:44) → PMap( d B S ) . Self-doubling surfaces.
With the doubling construction to hand, we now prove Theo-rem 2. To this end, we are tasked with finding a surface ¯ S with boundary components B such that the surfaces S and d B S constructed above are homeomorphic. Note that thisis equivalent to the definition of self-doubling from the introduction. Proof of Theorem 2.
Suppose the surface S is self-doubling, and let L and R be the sub-surfaces whose interiors are homeomorphic to S . We then define B = ∂L . Following thedoubling construction above, we have the injective homomorphismPMap( S ) ∼ = PMap( L )Π i (cid:104) T b i (cid:105) (cid:44) → PMap( S ) , where b i ∈ B . (cid:4) In the remainder of this section, we give conditions on the space of ends giving rise toself-doubling surfaces.4.2.
Ends and orbits.
A surface S is stable if for any end e ∈ Ends( S ) there exists asequence of nested subsurfaces U ⊃ U ⊃ . . . defining e such that U i ∼ = U i +1 [15]. Here,we call each U i a stable neighborhood of e .We say that an end e ∈ Ends( S ) is of higher rank than e (cid:48) ∈ Ends( S ) if each stableneighborhood of e contains an element of the Map( S )-orbit of e (cid:48) . Finally, we define F ⊂
Ends( S ) to be the set of ends whose Map( S )-orbit is finite [15].We can now prove the first case of Theorem 2. Theorem 4.1.
Let S be a stable surface with infinitely many punctures, and suppose thegenus of S is either zero or infinite. If F = ∅ then S is self-doubling.Proof. Let S be the surface obtained by removing an open disk surrounding one puncture z . Let d b S be the doubled surface of S along the sole boundary component b . The inducedmap on the space of ends gives us the homeomorphism(Ends( d b S ) , Ends ∞ ( d b S )) = (Ends( S ) , Ends ∞ ( S )) ∪ (Ends( S ) , Ends ∞ ( S )) ∼ = (Ends( S ) , Ends ∞ ( S )) ∪ (Ends( S ) , Ends ∞ ( S )) . Now, suppose e ∈ Ends( S ) has highest rank and denote the Map( S )-orbit of e by O ( e ).Since O ( e ) is infinite, there must be an accumulation point of its elements in Ends( S ).However, as each element of O ( e ) is of highest rank this accumulation point must alsobelong to O ( e ); it follows that O ( e ) is a Cantor set, see [27, Proposition 4.7] for moredetails.Since S is stable, there exists a maximal finite set of highest rank ends e , e , . . . , e n suchthat O ( e i ) (cid:54) = O ( e j ). Indeed, if there exist infinitely many highest rank ends with differentorbits then any accumulation point of such ends would not have a stable neighborhood.Write E ( e i ) for the end space of a stable neighborhood of e i and E ∞ ( e i ) = E ( e i ) ∩ Ends ∞ ( S ). Since each E ( e i ) is a Cantor set we have that(Ends( S ) , Ends ∞ ( S )) ∼ = (cid:91) i ( E ( e i ) , E ∞ ( e i )) ∼ = (cid:91) i ( E ( e i ) , E ∞ ( e i )) ∪ ( E ( e i ) , E ∞ ( e i )) . We can now conclude that(Ends( d b S ) , Ends ∞ ( d b S )) ∼ = (Ends( S ) , Ends ∞ ( S )) ∪ (Ends( S ) , Ends ∞ ( S )) ∼ = (cid:91) i ( E ( e i ) , E ∞ ( e i )) ∪ ( E ( e i ) , E ∞ ( e i )) ∼ = (Ends( S ) , Ends ∞ ( S )) , so S and d B S are homeomorphic. (cid:4) IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 13
Figure 1.
The figure shows two copies of the surface S . The doubleobtained using half the punctures is also homeomorphic to S . There is anatural orientation reversing homeomorphism ι which fixes the boundarycomponents of S . Doubling along infinitely many boundary components.
We define an end to be truly non-planar if it is non-planar and has a neighborhood that does not contain apuncture.
Theorem 4.2.
Suppose S is a stable surface of infinite genus with infinitely many punc-tures. If F contains no planar nor truly non-planar ends then S is self-doubling.Proof. If F is empty then we the result follows from Theorem 4.1. As in the previoussubsection, since S is stable we have that |F | = n is finite as otherwise there would be anaccumulation point without a stable neighborhood [15]. Therefore we have a partition ofEnds( S ) Ends( S ) = P ∪ P ∪ · · · ∪ P n such that P contains no elements of F , and all other P k contain a single element of F .Moreover, we may choose each P k to be the end space of a stable neighborhood of thecorresponding element of F .Our strategy is to choose an infinite collection of punctures in each set P k , where k ≥ S with boundary B ,then prove that the resulting surface d B S is homeomorphic to S . Note that since noelement of F is truly non-planar, each P k does indeed contain infinitely many punctures,when k ≥ k ≥ e the unique end in P k ∩ F . Let U ⊃ U ⊃ . . . be a nested sequence of homeomorphic subsurfaces with connected boundary such thatEnds( U ) = P k . Define X i = U i \ U i +1 . Without loss of generality we may assume that X i ∼ = X i +1 and that X i contains at least two punctures (it may contain infinitely manypunctures). Now, remove an open neighborhood of one puncture in each X i . Repeatingthis process for each P k (where k ≥
1) results in a surface S and a set of boundarycomponents B . We define d B S as usual, and write d B ( U i ) ⊂ d B S for the subsurfaceobtained by doubling U i .Note that d B ( U ) ⊃ d B ( U ) ⊃ . . . is a nested sequence of homeomorphic subsurfaces,in particular, it defines a unique end of d B S . We will now show that the end spaces of d B ( U ) and U are homeomorphic. Indeed, if each X i has infinitely many punctures, thenby construction we have that (cid:0) Ends( d B ( X i )) , Ends ∞ ( d B ( X i )) (cid:1) ∼ = (cid:0) Ends( X i ∪ X i +1 ) , Ends ∞ ( X i ∪ X i +1 ) (cid:1) . However, since U ⊃ U ⊃ U ⊃ . . . also defines the end e , it follows that the end spacesof d B ( U ) and U are homeomorphic. This shows that d B ( U ) is homeomorpic to U withan open disk removed, that is, d B ( U ) has two boundary components. α α α α α α γ γ γ γ γ γ β Figure 2.
A spanning chain for the surface S , .If each X i contains finitely many punctures then the argument above follows similarlyas both d B ( U ) and U contain infinitely many punctures with a single accumulation point.Finally, since P contains no elements of F , the argument from Theorem 4.1 implies that P ∼ = P ∪ P . It follows that(Ends( d B S ) , Ends ∞ ( d B S )) ∼ = ( P ∪ P ) ∪ P ∪ . . . P n ∼ = P ∪ P ∪ . . . P n ∼ = (Ends( S ) , Ends ∞ ( S ))Since d B S has infinite genus and no boundary components, it is homeomorphic to S . (cid:4) Twist-pair preserving homomorphisms
The purpose of this section is to prove Theorem 3. The key ingredient of our argumentsis that a twist-pair preserving homomorphism preserves certain combinatorial configura-tions of curves that we term spanning chains , and which we now define; the reader shouldkeep Figure 2 in mind:
Definition.
Let S g,p be the connected orientable surface of genus g and for which thenumber of punctures plus boundary components is p . A spanning chain is a set { β, α , α , . . . , α g − , α g , γ , . . . , γ p +1 } of non-separating curves on S g,p such that: • i ( β, α ) = 1 and i ( β, α j ) = i ( β, γ i ) = 0 for all i and all j (cid:54) = 3. • i ( α i , α i +1 ) = 1, and i ( α i , α j ) = 0 otherwise. • i ( γ i , α g ) = 1, for all i , and i ( γ i , α j ) = 0 otherwise. • i ( γ i , γ j ) = 0 for all i, j .Observe that any surface filled by a collection of distinct curves satisfying the conditionsin the definition of a spanning chain above must have genus g since a regular neighborhoodof the union of the curves can be reconstructed from the data, up to homeomorphism(not necessarily preserving the names of the curves); see again Figure 2. We will need thefollowing well-known fact; see [16, Section 4.4.4.] for instance: Proposition 5.1.
Suppose g ≥ . If C is a spanning chain on S g,p , then the Dehn twistsalong the elements of C generate PMap( S g,p ) . We will also make use of the braid relation between Dehn twists; see e. g. [16, Section3.5.1]:
IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 15
Lemma 5.2.
Let α, β be curves on a surface. Then t α t β t α = t β t α t β if and only if i ( a, b ) = 1 . We are now ready to prove Theorem 3:
Proof of Theorem 3.
Let φ : Map( S ) → Map( S (cid:48) ) be a continuous injective twist-preservinghomomorphism. In particular, φ induces a simplicial map φ ∗ : C ( S ) → C ( S (cid:48) )by the rule φ ∗ ( α ) = β ⇐⇒ φ ( t α ) = t β . Moreover, φ ∗ is superinjective , meaning that i ( α, β ) = 0 if and only if i ( φ ∗ ( α ) , φ ∗ ( β )) (cid:54) = 0.Finally, Lemma 5.2 implies that i ( φ ∗ ( α ) , φ ∗ ( β )) = 1 if and only if i ( α, β ) = 1.Now, let Z ⊂ Z ⊂ . . . be an exhaustion of S by connected, properly embedded, π –injective, finite type subsurfaces for which the inclusion of Z i into S induces an injectionPMap( Z i ) → PMap( S ). Without loss of generality, we may assume each component of ∂Z i is contained in the interior of Z i +1 or is a component of ∂S . Note that since each Z i isproperly embedded and PMap( Z i ) injects into PMap( S ), any puncture of S is a punctureof Z i for some i . Since the genus of S is positive and S has infinite type, without loss ofgenerality we may assume that the same is true for the genus of Z i for all i , and that Z i is not a torus with one puncture/boundary component for any i .For each i , choose a spanning chain C i of Z i . By the discussion above, φ ∗ ( C i ) is a set ofcurves on S (cid:48) of the same cardinality as C i , and whose elements have the same intersectionpattern as the curves in C i . Let Z (cid:48) i be the surface obtained from the regular neighborhoodof φ ∗ ( C i ) by adding any complementary once-punctured disks or peripheral annuli whicha boundary component of the neighborhood may bound. By construction Z (cid:48) i is filled by φ ∗ ( C i ), and thus Z i and Z (cid:48) i have the same genus. Moreover, PMap( Z (cid:48) i ) < PMap( S (cid:48) ), andProposition 5.1 implies that the homomorphism φ restricts to an injective homomorphism φ i : PMap( Z i ) → PMap( Z (cid:48) i ) . Since the genera of Z i and Z (cid:48) i are equal and the homomorphism φ i is injective, [4, Theorem1.1] implies that φ i is induced by a (unique) proper π -injective embedding h i : Z i → Z (cid:48) i , which is in fact a homeomorphism since Z i and Z (cid:48) i are each filled by a spanning chain ofthe same cardinality and combinatorial type. Remark.
Although the main result of [4] is stated for surfaces of genus at least four(see the remark below the statement of Theorem 1.1 in [4]), it is also true for twist-preserving injective homomorphisms between pure mapping class groups of surfaces ofthe same positive genus. The reader can check that (after a suitable reduction of thetarget surface), the standing assumption of [4, Section 10] holds, and that every argumentgoes through without modification from then on.We also note that the main result of [4] deals with arbitrary non-trivial homomorphismsbetween pure mapping class groups, and under suitable genus bounds, any such homo-morphism is induced by an embedding between the underlying surfaces. In the case of aninjective homomorphism, the reader will quickly verify that the definition of embedding of [4] simply means “proper topological embedding”.Fix a complete hyperbolic metric with geodesic boundary on S (cid:48) . We view h i as a properembedding from Z i to S (cid:48) , and note that h i and φ ∗ agree on every curve in Z i . Since Z i ismore complicated than a torus with one boundary/puncture, h i is uniquely determined,up to isotopy, by this agreement with φ ∗ on curves. As the punctures of Z (cid:48) i were punctures of S (cid:48) (by construction), it follows that h i maps punctures of Z i to punctures of S (cid:48) . Now Z i ⊂ Z i +1 , and it follows from the uniqueness statement that h i and h i +1 agree on Z i ,up to isotopy. Starting with h , and adjusting h by isotopy if necessary, we may assumethat h and h agree on Z . Continuing inductively and adjusting h i +1 to agree with h i on Z i , we get a well-defined injective continuous function h : S → S (cid:48) , that agrees with h i on Z i for all i . Without loss of generality, by arranging it to be thecase at every step, we may assume that h ( ∂Z i ) = h i ( ∂Z i ) is a union of closed geodesics inthe fixed hyperbolic metric. Since any curve in S is contained in Z i for some i , it followsthat h ( δ ) = φ ∗ ( δ ) for every curve δ on S .We now claim that h is proper. To see this, suppose for contradiction that this were notthe case. Then there exists a compact set K of S (cid:48) such that h − ( K ) is not compact. Wemay enlarge K if necessary to a finite type, connected, properly embedded subsurface of W ⊂ S (cid:48) with geodesic boundary (in our fixed hyperbolic metric). Then h − ( W ) is a non-empty, non-compact, closed subset of S . Now we observe that h − ( W ) ∩ ∂Z i (cid:54) = ∅ for all i sufficiently large: otherwise for some i , W would be entirely contained in h ( Z i ) = h i ( Z i ),and hence so would K , implying that h − ( K ) = h − i ( K ) is compact (by properness of h i ),a contradiction. Therefore, we can find components α i ⊂ ∂Z i so that for all i sufficientlylarge, h ( α i ) transversely and essentially intersects W . In particular, the sequence { t α i } of mapping classes converges to the identity in Map( S ). In contrast, the image sequence h ( α i ) is a sequence of pairwise distinct curves, all of which intersect W . As a consequence,the sequence { t h ( α i ) } cannot converge to the identity in Map( S (cid:48) ). This contradicts thefact that φ is continuous, as φ ( t α i ) = t φ ∗ ( α i ) = t h ( α i ) . Since h is a proper embedding it induces h (cid:93) : PMap( S ) → PMap( S (cid:48) ), an injectivehomomorphism.By hypothesis, S either has empty boundary or else has at most one end accumulatedby genus. In the former case, it follows that h is in fact a homeomorphism, and thus h (cid:93) and φ are equal by Theorem 2.4. In the latter case, we know that h and φ ∗ agree on everyisotopy class of simple closed curve, and hence h (cid:93) and φ agree on every Dehn twist. SinceDehn twists topologically generate PMap( S ) [28] and φ is continuous, it follows that φ and h (cid:93) are equal, as required. (cid:4) Remark. (1) As mentioned in the introduction, Theorem 3 no longer holds if ∂S (cid:54) = ∅ and S has more than one end accumulated by genus. Indeed, suppose that S has at leasttwo ends accumulated by genus, in which case there exists a surjective homomorphism ρ : PMap( S ) → Z by [3]. Since ∂S (cid:54) = ∅ , we may find a surface S (cid:48) for which there existsa π -injective embedding ι : S → S (cid:48) and such that the mapping class group of S (cid:48) \ S isinfinite.Consider the injective homomorphism φ : PMap( S ) → PMap( S (cid:48) ) induced by thisembedding, noting that its image is supported on PMap( ι ( S )). Now choose an infinite-order element g ∈ PMap( S (cid:48) ) supported on S (cid:48) \ ι ( S ). Using the homomorphism ρ weconstruct a homomorphism ρ g : PMap( S ) → PMap( S (cid:48) ) whose image is the cyclic groupgenerated by g ; in particular, the image of ρ g is supported on PMap( S (cid:48) \ ι ( S ))As the images of φ and ρ g commute, we may consider the homomorphism ( φ, ρ g ) :PMap( S ) → PMap( S (cid:48) ), whose image is contained inPMap( ι ( S )) × PMap( S (cid:48) \ ι ( S )) < PMap( S (cid:48) ) . Observe that this homomorphism is continuous, injective (since φ is) and twist-preserving(since the ρ -image of every Dehn twist is trivial, by [3]). However, it is not induced byany subsurface embedding S → S (cid:48) , as required. IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 17 (2) As may have become apparent in the proof of Theorem 3, the requirement that thehomomorphism φ be injective may be relaxed in various ways. For instance, say that atwist-preserving homomorphism preserves twist pairs if two Dehn twists commute if andonly if their images commute. With this definition, a minor modification of the proofabove yields: Let S and S (cid:48) be surfaces of infinite type, and assume S has positive genus. Assume furtherthat either the boundary of S is empty, or else S has at most one end accummulated bygenus. If φ : PMap( S ) → PMap( S (cid:48) ) a continuous twist-pair preserving homomorphism,then there is a proper π -injective embedding h : S → S (cid:48) that induces φ . Indeed, the only subtlety when mimicking the proof above occurs in justifying whythe restriction homomorphism φ i : PMap( Z i ) → PMap( Z (cid:48) i ) is injective. To this end, if anon-central element f ∈ PMap( Z i ) is in the kernel of φ i , we may use its Nielsen-Thurstonnormal form to find a curve α ⊂ Z i such that i ( f ( α ) , α ) >
0. However, since f ∈ ker( φ i )we have that i ( φ ∗ ( f ( α )) , φ ∗ ( α )) = 0, which contradicts that φ ∗ is superinjective.If on the other hand, there is f ∈ ker( φ i ) which is central, then we have an inducedinjective homomorphism between the pure mapping class groups modulo (the relevantpart of) their centers, at which point the main result of [4] applies.6. Superinjective maps of curve graphs
Finally, in this section we give a proof of Theorem 4. We separate the proof into twoparts.
Proof of part (1) of Theorem 4.
Equip S with a fixed hyperbolic metric, and realize everyvertex of C ( S ) by its unique geodesic representative in its homotopy class.Suppose first that S has infinite genus. By choosing two points p, q in the complementof the union of all simple closed geodesics on S , we obtain a (non-surjective) superinjectivemap C ( S ) → C ( S (cid:114) { p, q } ). Now, observe that C ( S (cid:114) { p, q } ) ∼ = C ( S (cid:48) ), where S (cid:48) is obtainedby removing two open discs (with disjoint closures) from S . Finally, we may glue a cylinderto S (cid:48) along its boundary components, obtaining a surface S (cid:48)(cid:48) ∼ = S and the inclusion S (cid:48) → S (cid:48)(cid:48) induces a superinjective map C ( S (cid:48) ) → C ( S (cid:48)(cid:48) ) ∼ = C ( S ). The composition of thesesuperinjective maps/isomorphisms C ( S ) → C ( S (cid:114) { p, q } ) ∼ = C ( S (cid:48) ) → C ( S (cid:48)(cid:48) ) ∼ = C ( S )is a superinjective map which is not surjective (since the first map is non-surjective).Suppose now that S has finite genus. Then either S has infinitely many punctures orits space of ends is a Cantor set union a finite set. In the former case we may puncture S at a point p in the complement of the union of simple closed geodesics, producing anon-surjective, superinjective map C ( S ) → C ( S (cid:114) { p } ) ∼ = C ( S ) , where the isomorphism comes from fact that S and S (cid:114) { p } are homeomorphic. In thelatter case, we may again remove a point p missing all simple closed geodesics to get anon-surjective, superinjective map C ( S ) → C ( S (cid:114) { p } ) ∼ = C ( S (cid:48) ), where S (cid:48) is obtained from S by removing an open disk, then glue in a disk minus a Cantor set to S (cid:48) produce a surface S (cid:48)(cid:48) ∼ = S . This gives a superinjective map C ( S (cid:48) ) → C ( S (cid:48)(cid:48) ) ∼ = C ( S ), and composing with themaps above gives the required non-surjective, superinjective map in this case. (cid:4) We are now going to prove part (2) of Theorem 4. The proof is divided into two lemmas.In what follows, we denote by M the Loch Ness Monster surface, which we again recallis the unique, up to homeomorphism, infinite-genus surface with exactly one end. Lemma 6.1.
Let S be an arbitrary infinite-type surface. Then there exists a superinjectivemap C ( S ) → C ( M ) . Figure 3.
One of the steps towards obtaining the surface S Proof.
We construct the desired map via a series of intermediate steps. First, let S bethe surface obtained from S by first replacing each puncture by a boundary component,and then gluing, to each of these new boundary components, a torus with one boundarycomponent. Observe that if S has any planar ends, then they must belong to a Cantorset. We have a superinjective map(3) C ( S ) → C ( S ) . Fix a principal exhaustion P ⊂ P ⊂ . . . of S , namely an exhaustion of S by compactsubsurfaces such that every component of ∂P i is separating in S . Let X n , X n , . . . , X nk n be the connected components of P n \ P n − . We puncture S along y ni ∈ X ni , and thenreplace each of these punctures by a boundary component b ni , obtaining a new surface S and a superinjective map(4) C ( S ) → C ( S )The surface S is naturally equipped with a principal exhaustion Q ⊂ Q ⊂ . . . com-ing from the obvious subsurface embedding S → S ; abusing notation, we denote theconnected components of P n \ P n − in S by Z n , Z n , . . . , Z nk n .Now, we glue a sphere with k n boundary components to the union of the Z ni , thusobtaining a connected surface Y n which contains Q n , see Figure 3. Moreover, Y n ⊂ Y n +1 ,so we may consider S = (cid:83) Y n . Since S = (cid:83) Q n , we have that S contains a subsurfacehomeomorphic to S , and in particular there is a superinjective map(5) C ( S ) → C ( S )It remains to show that S is homeomorphic to M . By construction, S has infinite genusand no planar ends. Since the complement of any finite-type subsurface of S containsonly one component of infinite-type, we have that S has a single end which is not planar.Therefore, S is homeomorphic to M , as desired. At this point, the composition of themaps in (3),(4) and (5) gives the superinjective map C ( S ) → C ( M ) . This finishes the proof of the lemma. (cid:4)
Next, we prove that every surface S (cid:48) of infinite genus contains a subsurface homeomor-phic to M . Lemma 6.2. If S is a surface with infinite genus then it contains a subsurface S ∼ = M .In particular, there is a superinjective map C ( M ) → C ( S ) .Proof. Let N be a neighborhood of a non-planar end e . Let c , c , · · · ⊂ N be a set ofpairwise disjoint separating curves, each of which bound bounding a one-holed torus. Let a i be an arc connecting c i − and c i such that a i ∩ a j = ∅ . IG MAPPING CLASS GROUPS AND THE CO-HOPFIAN PROPERTY 19
Now, the boundary of a neighborhood of (cid:83) a i ∪ c i contains a separating arc α . We defineΣ to be the component of S \ α containing every c i (here, we take the complement S \ α to be without boundary). It follows that Σ has infinite genus. Moreover, the complementof any compact subsurface contains a single non-compact component, so Σ has a singleend and is therefore homeomorphic to M . (cid:4) Finally, we put together the above pieces in order to prove Theorem 4:
Proof of Theorem 4.
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Instituto de Ciencias Matem´aticas, ICMAT (CSIC-UAM-UC3M-UCM)
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