Characterizing chainable, tree-like, and circle-like continua
aa r X i v : . [ m a t h . GN ] J a n CHARACTERIZING CHAINABLE, TREE-LIKE, AND CIRCLE-LIKE CONTINUA
TARAS BANAKH, ZDZIS LAW KOSZTO LOWICZ, S LAWOMIR TUREK
Abstract.
We prove that a continuum X is tree-like (resp. circle-like, chainable) if and only if for each opencover U = { U , U , U , U } of X there is a U -map f : X → Y onto a tree (resp. onto the circle, onto the interval).A continuum X is an acyclic curve if and only if for each open cover U = { U , U , U } of X there is a U -map f : X → Y onto a tree (or the interval [0 , Main results
In this paper we characterize chainable, tree-like and circle-like continua in the spirit of the followingHemmingsen’s characterization of covering dimension [5, 1.6.9].
Theorem 1 (Hemmingsen) . For a compact Hausdorff space X the following conditions are equivalent:(1) dim X ≤ n , which means that any open cover U of X has an open refinement V of order ≤ n + 1 ;(2) each open cover U of X with cardinality |U | ≤ n + 2 has an open refinement V of order ≤ n + 1 ;(3) each open cover { U i } n +2 i =1 of X has an open refinement { V i } n +2 i =1 with T n +2 i =1 V i = ∅ . We say that a cover V of U is a refinement of a cover U if each set V ∈ V lies in some set U ∈ U . The order of a cover U is defined as the cardinalord( U ) = sup {|F | : F ⊆ U with T F 6 = ∅} . An open cover U of X is called • a chain-like if for U there is an enumeration U = { U , . . . , U n } such that U i ∩ U j = ∅ if and only if | i − j | ≤ ≤ i, j ≤ n ; • circle-like if there is an enumeration U = { U , . . . , U n } such that U i ∩ U j = ∅ if and only if | i − j | ≤ { i, j } = { , n } ; • a tree-like if U contains no circle-like subfamily V ⊆ U of cardinality |V| ≥ X is called chainable (resp. tree-like , circle-like ) if each open cover of X hasa chain-like (resp. tree-like, circle-like) open refinement. By a continuum we understand a connected compactHausdorff space.The following characterization of chainable, tree-like and circle-like continua is the main result of this paper.For chainable and tree-like continua this characterization was announced (but not proved) in [1]. Theorem 2.
A continuum X is chainable (resp. tree-like, circle-like) if and only if any open cover U of X ofcardinality |U | ≤ has a chain-like (resp. tree-like, circle-like) open refinement. In fact, this theorem will be derived from a more general theorem treating K -like continua. Definition 1.
Let K be a class of continua and n be a cardinal number. A continuum X is called K -like (resp. n - K - like ) if for any open cover U of X (of cardinality |U | ≤ n ) there is a U -map f : X → K onto some space K ∈ K .We recall that a map f : X → Y between two topological spaces is called a U -map , where U is an opencover of X , if there is an open cover V of Y such that the cover f − ( V ) = { f − ( V ) : V ∈ V} refines the cover U . It worth mentioning that a closed map f : X → Y is a U -map if and only if the family { f − ( y ) : y ∈ Y } refines U . Mathematics Subject Classification.
Primary 54F15, 54F50; Secondary 54D05.
Key words and phrases.
Chainable continuum, tree-like continuum, circle-like continuum.
It is clear that a continuum X is tree-like (resp. chainable, circle-like) if and only if it is K -like for the class K of all trees (resp. for K = { [0 , } , K = { S } ). Here S = { z ∈ C : | z | = 1 } stands for the circle.It turns out that each 4- K -like continuum is b K -like for some extension b K of the class K . This extension isdefined with help of locally injective maps.A map f : X → Y between topological spaces is called locally injective if each point x ∈ X has a neighbor-hood O ( x ) ⊆ X such that the restriction f ↾ O ( x ) is injective. For a class of continua K let b K be the class ofall continua X that admit a locally injective map f : X → Y onto some continuum Y ∈ K . Theorem 3.
Let K be a class of 1-dimensional continua. If a continuum X is - K -like, then X is b K -like. In Proposition 1 we shall prove that each locally injective map f : X → Y from a continuum X onto atree-like continuum Y is a homeomorphism. This implies that b K = K for any class K of tree-like continua. Thisfact combined with Theorem 3 implies the following characterization: Theorem 4.
Let K be a class of tree-like continua. A continuum X is K -like if and only if it is - K -like. One may ask if the number 4 in this theorem can be lowered to 3 as in the Hemmingsen’s characterizationof 1-dimensional compacta. It turns out that this cannot be done: the 3- K -likeness is equivalent to being anacyclic curve. A continuum X is called a curve if dim X ≤
1. It is acyclic if each map f : X → S to the circleis null-homotopic. Theorem 5.
Let K ∋ [0 , be a class of tree-like continua. A continuum X is - K -like if and only if X is anacyclic curve. It is known that each tree-like continuum is an acyclic curve but there are acyclic curves, which are nottree-like [3]. On the other hand, each locally connected acyclic curve is tree-like (moreover, it is a dendrite [9,Chapter X]). Therefore, for any continuum X and a class K ∋ [0 ,
1] of tree-like continua we get the followingchain of equivalences and implications (in which the dotted implication holds under the additional assumptionthat the continuum X is locally connected):4-chainable / / O O (cid:15) (cid:15) K -like O O (cid:15) (cid:15) / / O O (cid:15) (cid:15) / / K -like O O (cid:15) (cid:15) chainable / / K -like / / tree-like / / acyclic curve u u Finally, let us present a factorization theorem that reduces the problem of studying n - K -like continua tothe metrizable case. It will play an important role in the proof of the “circle-like” part of Theorem 2. Theorem 6.
Let n ∈ N ∪ { ω } and K be a family of metrizable continua. A continuum X is n - K -like if andonly if any map f : X → Y to a metrizable compact space Y can be written as the composition f = g ◦ π of acontinuous map π : X → Z onto a metrizable n - K -like continuum Z and a continuous map g : Z → Y . Proof of Theorem 5
Let K ∋ [0 ,
1] be a class of tree-like continua. We need to prove that a continuum X is 3- K -like if and onlyif X is an acyclic curve.To prove the “if” part, assume that X is an acyclic curve. By Theorem 2.1 of [1], X is 3-chainable. Since[0 , ∈ K , the continuum X is 3- K -like and we are done.Now assume conversely, that a continuum X is 3- K -like. First, using Hemmingsen’s Theorem 1, we shallshow that dim X ≤
1. Let V = { V , V , V } be an open cover of X . Since the space X is 3- K -like, we can finda V -map f : X → T onto a tree-like continuum T . Using the 1-dimensionality of tree-like continua, find anopen cover W of T order ≤ f − ( W ) = { f − ( W ) : W ∈ W} is a refinement of V . Thecontinuum X is 1-dimensional by the implication (2) ⇒ (1) of Hemmingsen’s theorem.It remains to prove that X is acyclic. Let f : X → S be a continuous map. Let U = { U , U , U } be a cover ofthe unit circle S = { z ∈ C : | z | = 1 } by three open arcs U , U , U , each of length < π . Such a cover necessarilyhas ord( U ) = 2. By our assumption there is an open finite cover V of X inscribed in { f − ( U i ) : i = 1 , , } . So,there is a tree-like continuum T ∈ K and V -map g : X → T . We can assume that T is a tree and V is a tree-opencover of X . It is well known (see e.g. [3]) that there exists a continuous map h : T → S that h ◦ g is homotopicto f . But each map from a tree to the circle is null-homotopic. Hence h ◦ g as well f is null-homotopic too. HARACTERIZING CHAINABLE, TREE-LIKE, AND CIRCLE-LIKE CONTINUA 3 Proof of Theorem 3
We shall use some terminology from Graph Theory. So at first we recall some definitions.By a ( combinatorial ) graph we understand a pair G = ( V, E ) consisting of a finite set V of vertices and a set E ⊆ (cid:8) { a, b } : a, b ∈ V, a = b (cid:9) of unordered pairs of vertices, called edges . A graph G = ( V, E ) is connected ifany two distinct vertices u, v ∈ V can be linked by a path ( v , v , . . . , v n ) with v = u , v n = v and { v i − , v i } ∈ E for i ≤ n . The number n is called the length of the path (and is equal to the number of edges involved). Eachconnected graph possesses a natural path metric on the set of vertices V : the distance between two distinctvertices equals the smallest length of a path linking these two vertices.Two vertices u, v ∈ V of a graph are adjacent if { u, v } ∈ E is an edge. The degree deg( v ) of a vertex v ∈ V is the number of vertices u ∈ V adjacent to v in the graph. The number deg( G ) = max v ∈ V deg( v ) is called the degree of a graph. By an r -coloring of a graph we understand any map χ : V → r = { , . . . , r − } . In this casefor a vertex v ∈ V the value χ ( v ) is called the color of v . Lemma 1.
Let G = ( V, E ) be a connected graph with deg( G ) ≤ such that d ( u, v ) ≥ for any two vertices u, v ∈ V of order 3. Then there is a 4-coloring χ : V → such that no distinct vertices u, v ∈ V with d ( u, v ) ≤ have the same color.Proof. Let V = { v ∈ V : deg( v ) = 3 } denote the set of vertices of order 3 in G and let ¯ B ( v ) = { v } ∪ { u ∈ V : { u, v } ∈ E } be the unit ball centered at v ∈ V . It follows from deg( G ) ≤ | ¯ B ( v ) | ≤ v ∈ V .Moreover, for any distinct vertices v, u ∈ V the balls ¯ B ( v ) and ¯ B ( u ) are disjoint (because d ( v, u ) ≥ > χ on the union S v ∈ V ¯ B ( v ) so that χ is injective on each ball ¯ B ( u ) and χ ( v ) = χ ( w )for each v, w ∈ V . Next, it remains to color the remaining vertices all of order ≤ χ ( x ) = χ ( y ) if d ( x, y ) ≤
2. It is easy to check that this always can be done. (cid:3)
Each graph G = ( V, E ) can be also thought as a topological object: just embed the set of vertices V asa linearly independent subset into a suitable Euclidean space and consider the union | G | = S { u,v }∈ E [ u, v ] ofintervals corresponding to the edges of G . Assuming that each interval [ u, v ] ⊆ | G | is isometric to the unitinterval [0 , G to the path-metric d on the geometric realization | G | of G .For a point x ∈ | G | by B ( x ) = { y ∈ | G | : d ( x, y ) < } and ¯ B ( x ) = { y ∈ | G | : d ( x, y ) ≤ } denote respectivelythe open and closed unit balls centered at x . More generally, by B r ( x ) = { y ∈ | G | : d ( x, y ) < r } we shall denotethe open ball of radius r with center at x in | G | .By a topological graph we shall understand a topological space Γ homeomorphic to the geometric realization | G | of some combinatorial graph G . In this case G is called the triangulation of Γ. The degree of Γ = | G | will bedefined as the degree of the combinatorial graph G (the so-defined degree of Γ does not depend on the choiceof a triangulation).It turns out that any graph by a small deformation can be transformed into a graph of degree ≤ Lemma 2.
For any open cover U of a topological graph Γ there is a U -map f : Γ → G onto a topological graph G of degree ≤ . This lemma can be easily proved by induction (and we suspect that it is known as a folklore). The followingdrawing illustrates how to decrease the degree of a selected vertex of a graph. r (cid:0)(cid:0)(cid:0)❅❅❅ qqqq Γ G ✲ f r (cid:0)(cid:0)(cid:0)❅❅❅ rr Now we have all tools for the proof of Theorem 3. So, take a class K of 1-dimensional continua and assumethat X is a 4- K -like continuum. We should prove that X is b K -like.First, we show that X is 1-dimensional. This will follow from Hemmingsen’s Theorem 1 as soon as we checkthat each open cover U of X of cardinality |U | ≤ V of order ≤
2. Since |U | ≤ X is 4- K -like, there is a U -map f : X → K onto a continuum K ∈ K . It follows that for some open cover V of K TARAS BANAKH, ZDZIS LAW KOSZTO LOWICZ, S LAWOMIR TUREK the cover f − ( V ) refines the cover U . Since the space K is 1-dimensional, the cover V has an open refinement W of order ≤
2. Then the cover f − ( W ) is an open refinement of U having order ≤ X is b K -like, fix any open cover U of X . Because of the compactness of X , we can additionallyassume that the cover U is finite. Being 1-dimensional, the continuum X admits a U -map f : X → Γ onto atopological graph Γ. By Lemma 2, we can assume that deg(Γ) ≤
3. Adding vertices on edges of Γ, we can finda triangulation ( V Γ , E Γ ) of Γ so fine that • the path-distance between any vertices of degree 3 in the graph Γ is ≥ • the cover { f − ( B ( v )) : v ∈ V Γ } of X is inscribed into U .Lemma 1 yields a 4-coloring χ : V Γ → V Γ such that any two distinct vertices u, v ∈ V Γ with d ( u, v ) ≤ i ∈ U i = S v ∈ χ − ( i ) B ( v ) of themonochrome set χ − ( i ) ⊆ V Γ in Γ. Since open 1-balls centered at vertices v ∈ V Γ cover the graph Γ, the4-element family { U i : i ∈ } is an open cover of Γ. Then for the 4-element cover U = { f − ( U i ) : i ∈ } of the4- K -like continuum X we can find a U -map g : X → Y to a continuum Y ∈ K . Let W be a finite open cover of Y such that the cover g − ( W ) refines the cover U . Since Y is 1-dimensional, we can assume that ord( W ) ≤ W ∈ W find a number ξ ( W ) ∈ g − ( W ) ⊆ f − ( U ξ ( W ) ).Since Y is a continuum, in particular, a normal Hausdorff space, we may find a partition of unity subordi-nated to the cover W . This is a family { λ W : W ∈ W} of continuous functions λ W : Y → [0 ,
1] such that(a) λ W ( y ) = 0 for y ∈ Y \ W ,(b) P W ∈W λ W ( y ) = 1 for all y ∈ Y .For every W ∈ W consider the “vertical” family of rectangles R W = { W × B ( v ) : v ∈ V Γ , χ ( v ) = ξ ( W ) } in Y × Γ and let R = S W ∈W R W . For every rectangle R ∈ R choose a set W R ∈ W and a vertex v R ∈ V Γ suchthat R = W R × B ( v R ). Also let R R = { S ∈ R : R ∩ S = ∅} . Claim 1.
For any rectangle R ∈ R and a point y ∈ W R the set R R,y = { S ∈ R R : y ∈ W S } contains at mosttwo distinct rectangles.Proof. Assume that besides the rectangle R the set R R,y contains two other distinct rectangles S = W S × B ( v S ) and S = W S × B ( v S ). Taking into account that y ∈ W R ∩ W S ∩ W S and ord( W ) ≤
2, we concludethat either W S = W S or W R = W S or W R = W S . If W S = W S , then χ ( v S ) = ξ ( W S ) = ξ ( W S ) = χ ( v S ) . Since B ( v R ) ∩ B ( v S ) = ∅ 6 = B ( v R ) ∩ B ( v S ) the property of 4-coloring χ implies that v S = v S and hence S = S . By analogy we can prove that W R = W S implies R = S and W R = W S implies R = S , whichcontradicts the choice of S , S ∈ R R,y \ { R } . (cid:3) Claim 1 implies that for every rectangle R = W R × B ( v R ) the function λ R : W R → ¯ B ( v R ) ⊆ Γ defined by λ R ( y ) = ( λ W R ( y ) v R + λ W S ( y ) v S , if R R,y = { R, S } for some S = R,v R , if R R,y = { R } is well-defined and continuous. Let π R : R → W R × ¯ B ( v R ) ⊂ ¯ R be the map defined by π R ( y, t ) = ( y, λ R ( y )).The graphs of two functions λ R and λ S for two intersecting rectangles R, S ∈ R are drawn on the followingpicture:
HARACTERIZING CHAINABLE, TREE-LIKE, AND CIRCLE-LIKE CONTINUA 5 ✲ Y ( ) W R ( ) W S ✻ Γ rr v R r v S r R S
It follows that for any rectangles
R, S ∈ R we get π R ↾ R ∩ S = π S ↾ R ∩ S , which implies that the union π = S R ∈R π R : S R → S R is a well-defined continuous function. It is easy to check that for every rectangle R = W × B ( v ) ∈ R we get π − ( W × B ( v )) ⊆ W × B ( v ) . Consider the diagonal product g △ f : X → Y × Γ. It is easy to check that ( g △ f )( X ) ⊆ S R , which impliesthat the composition h = π ◦ ( g △ f ) : X → S R is well-defined. We claim that h is a U -map onto the continuum L = h ( X ), which belongs to the class b K .Given any rectangle R = W × B ( v ) ∈ R , observe that h − ( R ) = ( g △ f ) − ( π − ( W × B ( v ))) ⊆ ( g △ f ) − ( W × B ( v )) == g − ( W ) ∩ f − ( B ( v )) ⊆ f − ( B ( v )) ⊆ U, for some U ∈ U . Hence h is a U -map.The projection pr Y : L → Y is locally injective because L ⊆ S R and for every R ∈ R the restrictionpr Y ↾ R ∩ L : R ∩ L → Y is injective. Taking into account that Y ∈ K , we conclude that L ∈ b K , by the definitionof the class b K . 4. Locally injective maps onto tree-like continua and circle
The following theorem is known for metrizable continua [6].
Proposition 1.
Each locally injective map f : X → Y from a continuum X onto a tree-like continuum Y is ahomeomorphism.Proof. By the local injectivity of f , there is an open cover U ′ such that for every U ∈ U ′ the restriction f ↾ U isinjective. Let U be an open cover of X whose second star S t ( U ) refines the cover U ′ . Here S t ( U, U ) = S { U ′ ∈U : U ∩ U ′ = ∅} , S t ( U ) = {S t ( U, U ) : U ∈ U } and S t ( U ) = {S t ( U, S t ( U )) : U ∈ U } .For every x ∈ X choose a set U x ∈ U that contains x . Observe that for distinct points x, x ′ ∈ X with f ( x ) = f ( x ′ ) the sets U x , U x ′ are disjoint. In the opposite case x, x ′ ∈ U x ∪ U x ′ ⊆ S t ( U x , U ) ⊆ U for some set U ∈ U ′ , which is not possible as f ↾ U is injective.Hence for every y ∈ Y the family U y = { U x : x ∈ f − ( y ) } is disjoint. Since f is closed and surjective, theset V y = Y \ f ( X \ S U y ) is an open neighborhood of y in Y such that f − ( V y ) ⊆ S U y .Since the continuum Y is a tree-like, the cover V = { V y : y ∈ Y } has a finite tree-like refinement W . For every W ∈ W find a point y W ∈ Y with W ⊆ V y W and consider the disjoint family U W = { U ∩ f − ( W ) : U ∈ U y W } .It follows that f − ( W ) = S U W and hence U W = S W ∈W U W is an open cover of X .Now we are able to show that the map f is injective. Assuming the converse, find a point y ∈ Y and twodistinct points a, b ∈ f − ( y ). Since X is connected, there is a chain of sets { G , G , . . . , G n } ⊆ U W such that a ∈ G and b ∈ G n . We can assume that the length n of this chain is the smallest possible. In this case all sets G , . . . , G n are pairwise distinct.Let us show that n ≥
3. In the opposite case a ∈ G = U ∩ f − ( W ) ∈ U W , b ∈ G = U ∩ f − ( W ) ∈ U W and G ∩ G = ∅ . So, a, b ∈ U ∪ U ⊆ S t ( U , U ) ⊆ U for some U ∈ U ′ and then the restriction f ↾ U is notinjective. So n ≥ TARAS BANAKH, ZDZIS LAW KOSZTO LOWICZ, S LAWOMIR TUREK
For every i ≤ n consider the point y i = y W i and find sets W i ∈ W and U i ∈ U y i such that G i = U i ∩ f − ( W i ) ∈U W i . Then ( W , . . . , W n ) is a sequence of elements of the tree-like cover W such that y ∈ W ∩ W n and W i ∩ W i +1 = ∅ for all i < n . Since the tree-like cover W does not contain circle-like subfamilies of length ≥ ≤ i < j ≤ n such that W i ∩ W j = ∅ , | j − i | > { i, j } 6 = { , n } . We can assume thatthe difference k = j − i is the smallest possible. In this case k = 2. Otherwise, W i , W i +1 , . . . , W j is a circle-likesubfamily of length ≥ W , which is forbidden. Therefore, j = i +2 and the family { W i , W i +1 , W i +2 } containsat most two distinct sets (in the opposite this family is circle-like, which is forbidden). If W i = W i +1 , then U i = U i +1 as the family U W i is disjoint. The assumption W i +1 = W i +2 leads to a similar contradiction. Itremains to consider the case W i = W i +2 = W i +1 . Since the sets U i , U i +2 ∈ U y i are distinct, there are distinctpoints x i , x i +2 ∈ f − ( y i ) such that x i ∈ U i and x i +2 ∈ U i +2 . Since x i , x i +2 ∈ U i ∪ U i +2 ⊂ S t ( U i , U ) ⊂ U forsome U ∈ U ′ , the restriction f | U is not injective. This contradiction completes the proof. (cid:3) Proposition 2. If f : X → S is a locally injective map from a continuum X onto the circle S , then X is anarc or a circle.Proof. The compact space X has a finite cover by compact subsets that embed into the circle. Consequently, X is metrizable and 1-dimensional. We claim that X is locally connected. Assuming the converse and applyingTheorem 1 of [8, § K ⊆ X . This anon-trivial continuum K , which is the limit of a sequence of continua ( K n ) n ∈ ω that lie in X \ K .By the local injectivity of f , the continuum K meets some open set U ⊆ X such that f ↾ U : U → S is a topological embedding. The intersection U ∩ K , being a non-empty open subset of the continuum K isnot zero-dimensional. Consequently, its image f ( U ∩ K ) ⊆ S also is not zero-dimensional and hence containsa non-empty open subset V of S . Choose any point x ∈ U ∩ K with f ( x ) ∈ V . The convergence K n → K ,implies the existence of a sequence of points x n ∈ K n , n ∈ ω , that converge to x . By the continuity of f , thesequence ( f ( x n )) n ∈ ω converges to f ( x ) ∈ V . So, we can find a number n such that f ( x n ) ∈ V ⊆ f ( U ∩ K ) and x n ∈ U . The injectivity of f ↾ U guarantees that x n ∈ U ∩ K which is not possible as x n ∈ K n ⊂ X \ K .Therefore, the continuum X is locally connected. By the local injectivity, each point x ∈ X has an openconnected neighborhood V homeomorphic to a (connected) subset of S . Now we see that the space X is acompact 1-dimensional manifold (possibly with boundary). So, X is homeomorphic either to the arc or to thecircle. (cid:3) Proof of Theorem 6
In the proof we shall use the technique of inverse spectra described in [5, § X embed it into a Tychonov cube [0 , κ of weight κ ≥ ℵ .Let A be the set of all countable subsets of κ , partially ordered by the inclusion relation: α ≤ β iff α ⊆ β .For a countable subset α ⊆ κ let X α = pr α ( X ) be the projection of X onto the face [0 , α of the cube [0 , κ and p α : X → X α be the projection map. For any countable subsets α ⊆ β of κ let p βα : X β → X α bethe restriction of the natural projection [0 , β → [0 , α . In such a way we have defined an inverse spectrum S = { X α , p βα : α, β ∈ A } over the index set A , which is ω -complete in the sense that any countable subset B ⊂ A has the smallest upper bound sup B = S B and for any increasing sequence { α i } i ∈ ω ⊂ A withsupremum α = S i ∈ ω α i the space X α is the limit of the inverse sequence { X α i , p α i +1 α i , ω } . The spectrum S consists of metrizable compacta X α , α ∈ A , and its inverse limit lim ←− S can be identified with the space X .By Corollary 1.3.2 of [4], the spectrum S is factorizing in the sense that any continuous map f : X → Y toa second countable space Y can be written as the composition f = f α ◦ p α for some index α ∈ A and somecontinuous map f α : X α → Y .Now we are able to prove the “if” and “only if” parts of Theorem 6. To prove the “if” part, assume thateach map f : X → Y factorizes through a metrizable n - K -like continuum. To show that X is n - K -like, fixany open cover U = { U , . . . , U n } of X . By Lemma 5.1.6 of [5], there is a closed cover { F , . . . , F n } of X suchthat F i ⊂ U i for all i ≤ n . Since F i and X \ U i are disjoint closed subsets of the compact space X = lim ←− S ,there is an index α ∈ A such that for every i ≤ n the images p α ( X \ U i ) and p α ( F i ) are disjoint and hence W i = X α \ p α ( X \ U α ) is an open neighborhood of p α ( F i ). Then { W , . . . , W n } is an open cover of X α suchthat p − α ( W i ) ⊆ U i for all i ≤ n . HARACTERIZING CHAINABLE, TREE-LIKE, AND CIRCLE-LIKE CONTINUA 7
By our assumption the projection p α : X → X α can be written as the composition p α = g ◦ π of a map π : X → Z onto a metrizable n - K -like continuum Z and a map g : Z → X α . For every i ≤ n consider theopen subset V i = g − ( W i ) of Z . Since Z is n - K -like, for the open cover V = { V , . . . , V n } of Z there is a V -map h : Z → K onto a space K ∈ K . Then the composition h ◦ π : X → K is a U -map of X onto the space K ∈ K witnessing that X is an n - K -like continuum.Now we shall prove the “only if” part of the theorem. Assume that the continuum X is n - K -like. We shallneed the following lemma. Lemma 3.
For any index α ∈ A there is an index β ≥ α in A such that for any open cover V = { V , . . . , V n } of X α there is a map f : X β → K onto a space K ∈ K such that f ◦ p β : X → K is a p − α ( V ) -map.Proof. Let B be a countable base of the topology of the compact metrizable space X α such that B is closedunder unions. Denote by U the family of all possible n -set covers { B , . . . , B n } ⊆ B of X α . It is clear that thefamily U is countable.Each cover U = { B , . . . , B n } ∈ U induces the open cover p − α ( U ) = { p − α ( B i ) : 1 ≤ i ≤ n } of X . Since thecontinuum X is n - K -like, there is a p − α ( U )-map f U : X → K U onto a space K U ∈ K . By the metrizability of K U and the factorizing property of the spectrum S , for some index α U ≥ α in A there is a map f α U : X α U → K U such that f U = f α U ◦ p α U . Consider the countable set β = S U∈ U α U , which is the smallest lower bound of theset { α U : U ∈ U } in A . We claim that this index β has the required property.Let V = { V , . . . , V n } be any open cover of X α . By Lemma 5.1.6 of [5], there is a closed cover { F , . . . , F n } of X α such that F i ⊆ V i for all i ≤ n . Since B is the base of the topology of X α and B is closed under finiteunions, for every i ≤ n there is a basic set B i ∈ B such that F i ⊆ B i ⊆ V i . Then the cover U = { B , . . . , B n } belongs to the family U and refines the cover V . Consider the map f = f α U ◦ p βα U : X β → K and observe that f ◦ p β = f α U ◦ p α U is a p − α ( U )-map and a p − α ( V )-map. (cid:3) Now let us return back to the proof of the theorem. Given a map f : X → Y to a second countable space,we need to find a map π : X → Z onto a metrizable n - K -like continuum Z and a map g : Z → Y such that f = g ◦ π . Since the spectrum S is factorizing, there are an index α ∈ A and a map f : X α → Y such that f = f ◦ p α . Using Lemma 3, by induction construct an increasing sequence ( α n ) n ∈ ω in A such that for every i ∈ ω and any open cover V = { V , . . . , V n } of X α i there is a map f : X α i +1 → K onto a space K ∈ K such that f ◦ p α i +1 is a p − α i ( V )-map.Let α = sup i ∈ ω α i = S i ∈ ω α i . We claim that the metrizable continuum X α is n - K -like. Given any open cover U = { U , . . . , U n } of X α = lim ←− X α i , we can find i ∈ ω such that the sets W i = X α i \ p αα i ( X α \ U i ), i ≤ n , form anopen cover W = { W , . . . , W n } of X α i such that the cover ( p αα i ) − ( W ) refines the cover U . By the choice of theindex α i +1 , there is a map g : X α i +1 → K onto a space K ∈ K such that g ◦ p α i +1 : X → K is a p − α i ( W )-map.It follows that g ◦ p αα i +1 : X α → K is a ( p αα i ) − ( W )-map and hence a U -map, witnessing that the continuum X α is n - K -like.Now we see that the metrizable n - K -like continuum X α and the maps π = p α : X → X α and g = f ◦ p αα : X α → Y satisfy our requirements. 6. Proof of Theorem 2
The “chainable and tree-like” parts of Theorem 2 follow immediately from the characterization Theorem 4.So, it remains to prove the “circle-like” part. Let K = { S } . We need to prove that each 4- K -like continuum X is K -like. Given an open cover U of X we need to construct a U -map of X onto the circle. By Theorem 6,there is a U -map onto a metrizable 4- K -like continuum Y . It follows that for some open cover V of Y the cover f − ( V ) refines U . The proof will be complete as soon as we prove that the continuum Y is circle-like. In thiscase there is a V -map g : Y → S and the composition g ◦ f : X → S is a required U -map witnessing that X is circle-like.By Theorem 3, the metrizable continuum Y is b K -like. By Proposition 2, each continuum K ∈ b K is homeo-morphic to S or [0 , Y is circle-like or chainable. In the first case we are done.So, we assume that Y is chainable.By [9, Theorem 12.5], the continuum Y is irreducible between some points p, q ∈ Y . The latter means thateach subcontinuum of X that contains the points p, q coincides with Y . We claim that Y is either indecomposable TARAS BANAKH, ZDZIS LAW KOSZTO LOWICZ, S LAWOMIR TUREK or Y is the union of two indecomposable subcontinua. For the proof of this fact we will use the argument of [9,Exercise 12.50] (cf. also [7, Theorem 3.3]).Suppose that Y is not indecomposable. It means that there are two proper subcontinua A, B of Y such that Y = A ∪ B . By the choice of the points p, q , they cannot simultaneously lie in A or in B . So, we can assumethat p ∈ A and q ∈ B .We claim that the closure of the set Y \ A is connected. Assuming that Y \ A is disconnected, we can finda proper closed-and open subset F ( Y \ A that contains the point q and conclude that F ∪ A is a propersubcontinuum of Y that contains both points p, q , which is not possible. Replacing B by the closure of Y \ A ,we can assume that Y \ A is dense in B . Then Y \ B is dense in A .We claim that the sets A and B are indecomposable. Assuming that A is decomposable, find two propersubcontinua C, D such that C ∪ D = A . We can assume that p ∈ D . Then B ∩ D = ∅ (as Y is irreduciblebetween p and q ). By Theorem 11.8 of [9], the set Y \ ( B ∪ D ) is connected. Let Z and Z be open disjointsubsets of X such that B ⊆ Z and D ⊆ Z . Since Y is 4- { S } -like, for the open cover Z = { Z , Z , Y \ ( B ∪ D ) } of Y there exists a Z -map h : Y → S . Thus h ( B ) ∩ h ( D ) = ∅ and S \ ( h ( B ) ∪ h ( D )) is the union of twodisjoint open intervals W , W . Since h is a Z -map, Y \ ( B ∪ D ) = h − ( W ) ∪ h − ( W ) which contradicts theconnectedness of the set Y \ ( B ∪ D ).Now we know that Y is either indecomposable or is the union of two indecomposable subcontinua. ApplyingTheorem 7 of [2], we conclude that the metrizable chainable continuum Y is circle-like.7. Open Problems
Problem 1.
For which families K of connected topological graphs every 4- K -like continuum is K -like? Is it truefor the family K = { } that contains 8, the bouquet of two circles?Also we do not know if Theorem 4 can be generalized to classes of higher-dimensional continua. Problem 2.
Let k ∈ N and K be a class of k -dimensional (contractible) continua. Is there a finite number n such that a continuum X is K -like if and only if it is n - K -like? Acknowledgment
The authors express their sincere thanks to the anonymous referee whose valuable remarks and suggestionshelped the authors to improve substantially the results of this paper.
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Institute of Mathematics, Uniwersytet Humanistyczno-Przyrodniczy JanaKochanowskiego, ul. ´Swietokrzyska 15, 25-406 Kielce, Poland
E-mail address : [email protected], [email protected] (T.Banakh) Department of Mathematics, Ivan Franko National University of Lviv, Ukraine
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