Characterizing chainable, tree-like, and circle-like continua

In this paper we characterize chainable, tree-like and circle-like continua in the spirit of the following Hemmingsen'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 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 cardinal ord(U ) = sup{|F| : An open cover U of X is called • a chain-like if for U there is an enumeration U = {U 1 , . . . , U n } such that U i ∩ U j = ∅ if and only if |i -j| ≤ 1 for all 1 ≤ i, j ≤ n; • circle-like if there is an enumeration U = {U 1 , . . . , U n } such that U i ∩ U j = ∅ if and only if |i -j| ≤ 1 or {i, j} = {1, n}; • a tree-like if U contains no circle-like subfamily V ⊆ U of cardinality |V| ≥ 3. We recall that a continuum X is called chainable (resp. tree-like, circle-like) if each open cover of X has a chain-like (resp. tree-like, circle-like) open refinement. By a continuum we understand a connected compact Hausdorff 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 of cardinality |U | ≤ 4 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. We recall that a map f : X → Y between two topological spaces is called a U -map, where U is an open cover of X, if there is an open cover V of Y such that the cover f -1 (V) = {f -1 (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 -1 (y) : y ∈ Y } refines U . 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, 1]}, K = {S 1 }). Here S 1 = {z ∈ C : |z| = 1} stands for the circle. It turns out that each 4-K-like continuum is K-like for some extension K of the class K. This extension is defined 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 neighborhood O(x) ⊆ X such that the restriction f ↾ O(x) is injective. For a class of continua K let K be the class of all continua X that admit a locally injective map f : X → Y onto some continuum Y ∈ K. In Proposition 1 we shall prove that each locally injective map f : X → Y from a continuum X onto a tree-like continuum Y is a homeomorphism. This implies that K = K for any class K of tree-like continua. This fact combined with Theorem 3 implies the following characterization: One may ask if the number 4 in this theorem can be lowered to 3 as in the Hemmingsen's characterization of 1-dimensional compacta. It turns out that this cannot be done: the 3-K-likeness is equivalent to being an acyclic It is known that each tree-like continuum is an acyclic curve but there are acyclic curves, which are not tree-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 following chain of equivalences and implications (in which the dotted implication holds under the additional assumption that the continuum X is locally connected): Finally, let us present a factorization theorem that reduces the problem of studying n-K-like continua to the 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 and only if any map f : X → Y to a metrizable compact space Y can be written as the composition f = g • π of a continuous map π : X → Z onto a metrizable n-K-like continuum Z and a continuous map g : Z → Y . 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 only if 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, 1] ∈ 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 shall show that dim X ≤ 1. Let V = {V 1 , V 2 , V 3 } be an open cover of X. Since the space X is 3-K-like, we can find a V-map f : X → T onto a tree-like continuum T . Using the 1-dimensionality of tree-like continua, find an open cover W of T order ≤ 2 such that the cover f -1 (W) = {f -1 (W ) : W ∈ W} is a refinement of V. The continuum 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 1 be a continuous map. Let U = {U 1 , U 2 , U 3 } be a cover of the unit circle S 1 = {z ∈ C : |z| = 1} by three open arcs U 1 , U 2 , U 3 , each of length < π. Such a cover necessarily has ord(U ) = 2. By our assumption there is an open finite cover V of X inscribed in {f -1 (U i ) : i = 1, 2, 3}. 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-open cover of X. It is well known (see e.g. [3]) that there exists a continuous map h : T → S 1 that h • g is homotopic to f . But each map from a tree to the circle is null-homotopic. Hence h • g as well f is null-homotopic too. 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 The number n is called the length of the path (and is equal to the number of edges involved). Each connected graph possesses a natural path metric on the set of vertices V : the distance between two distinct vertices 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 : Moreover, for any distinct vertices v, u ∈ V 3 the balls B(v) and B(u) are disjoint (because d(v, u) ≥ 6 > 2). So we can define a 4-coloring χ on the union v∈V 3 B(v) so that χ is injective on each ball B(u) and χ(v) = χ(w) for each v, w ∈ V 3 . Next, it remains to color the remaining vertices all of order ≤ 2 by four colors so that χ(x) = χ(y) if d(x, y) ≤ 2. It is easy to check that this always can be done. Each graph G = (V, E) can be also thought as a topological object: just embed the set of vertices V as a linearly independent subset into a suitable Euclidean space and consider the union 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 be defined as the degree of the combinatorial graph G (the so-defined degree of Γ does not depend on the choice of a triangulation). It turns out that any graph by a small deformation can be transformed into a graph of degree ≤ 3. This lemma can be easily proved by induction (and we suspect that it is known as a folklore). The following drawing illustrates how to decrease the degree of a selected vertex of a graph. Now we have all tools for the proof of Theorem 3. So, take a class K of 1-dimensional continua and assume that X is a 4-K-like continuum. We should prove that X is K-like. First, we show that X is 1-dimensional. This will follow from Hemmingsen's Theorem 1 as soon as we check that each open cover U of X of cardinality |U | ≤ 3 has an open refinement V of order ≤ 2. Since |U | ≤ 4 and 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 the cover f -1 (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 -1 (W) is an open refinement of U having order ≤ 2. To prove that X is K-like, fix any open cover U of X. Because of the compactness of X, we can additionally assume that the cover U is finite. Being 1-dimensional, the continuum X admits a U -map f : X → Γ onto a topological graph Γ. By Lemma 2, we can assume that deg(Γ) ≤ 3. Adding vertices on edges of Γ, we can find a triangulation (V Γ , E Γ ) of Γ so fine that • the path-distance between any vertices of degree 3 in the graph Γ is ≥ 6; Since Y is a continuum, in particular, a normal Hausdorff space, we may find a partition of unity subordinated to the cover W. This is a family {λ W ∈W λ W (y) = 1 for all y ∈ Y . For every W ∈ W consider the "vertical" family of rectangles 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 most two distinct rectangles. Proof. Assume that besides the rectangle R the set R R,y contains two other distinct rectangles 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 following picture: 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 = R∈R π R : R → R is a well-defined continuous function. It is easy to check that for every rectangle It is easy to check that (g△f )(X) ⊆ R, which implies that the composition h = π • (g△f ) : X → R is well-defined. We claim that h is a U -map onto the continuum L = h(X), which belongs to the class K. Given any rectangle The projection pr Y : L → Y is locally injective because L ⊆ R and for every R ∈ R the restriction pr Y ↾R ∩ L : R ∩ L → Y is injective. Taking into account that Y ∈ K, we conclude that L ∈ K, by the definition of the class K. 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 a homeomorphism. Proof. By the local injectivity of f , there is an open cover U ′ such that for every U ∈ U ′ the restriction f ↾U is injective. Let U be an open cover of X whose second star St 2 (U ) refines the cover For every x ∈ X choose a set U x ∈ U that contains x. Observe that for distinct points x, x ′ ∈ X with Hence for every y ∈ Y the family 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 Now we are able to show that the map f is injective. Assuming the converse, find a point y ∈ Y and two distinct points a, b ∈ f -1 (y). Since X is connected, there is a chain of sets {G 1 , G 2 , . . . , G n } ⊆ U W such that a ∈ G 1 and b ∈ G n . We can assume that the length n of this chain is the smallest possible. In this case all sets G 1 , . . . , G n are pairwise distinct. Let us show that n ≥ 3. In the opposite case a For every i ≤ n consider the point y i = y W i and find sets W i ∈ W and a sequence of elements of the tree-like cover W such that y ∈ W 1 ∩ 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 ≥ 3 there are two numbers 1 ≤ i < j ≤ n such that W i ∩ W j = ∅, |j -i| > 1 and {i, j} = {1, n}. We can assume that the difference k = ji is the smallest possible. In this case k = 2. Otherwise, W i , W i+1 , . . . , W j is a circle-like subfamily of length ≥ 3 in W, which is forbidden. Therefore, j = i+2 and the family {W i , W i+1 , W i+2 } contains at 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. It remains to consider the case the restriction f |U is not injective. This contradiction completes the proof. Proposition 2. If f : X → S 1 is a locally injective map from a continuum X onto the circle S 1 , then X is an arc 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 applying Theorem 1 of [8, §49.VI] (or [9, 5.22(b) and 5.12]), we could find a convergence continuum K ⊆ X. This a non-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 1 is a topological embedding. The intersection U ∩ K, being a non-empty open subset of the continuum K is not zero-dimensional. Consequently, its image f (U ∩ K) ⊆ S 1 also is not zero-dimensional and hence contains a non-empty open subset V of S 1 . 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 , the sequence (f (x n )) n∈ω converges to f (x) ∈ V . So, we can find a number n such that f Therefore, the continuum X is locally connected. By the local injectivity, each point x ∈ X has an open connected neighborhood V homeomorphic to a (connected) subset of S 1 . Now we see that the space X is a compact 1-dimensional manifold (possibly with boundary). So, X is homeomorphic either to the arc or to the circle. In the proof we shall use the technique of inverse spectra described in [5, §2.5] or [4,Ch.1]. Given a continuum X embed it into a Tychonov cube [0, 1] κ of weight κ ≥ ℵ 0 . 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, 1] α of the cube [0, 1] κ and p α : X → X α be the projection map. For any countable subsets α ⊆ β of κ let p β α : X β → X α be the restriction of the natural projection [0, 1] β → [0, 1] α . In such a way we have defined an inverse spectrum S = {X α , p β α : 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 = B and for any increasing sequence {α i } i∈ω ⊂ A with supremum α = 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 to a second countable space Y can be written as the composition f = f α • p α for some index α ∈ A and some continuous 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 that each map f : X → Y factorizes through a metrizable n-K-like continuum. To show that X is n-K-like, fix any open cover U = {U 1 , . . . , U n } of X. By Lemma 5.1.6 of [5], there is a closed cover {F 1 , . . . , F n } of X such that 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 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 assume that 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 find a proper closed-and open subset F Y \ A that contains the point q and conclude that F ∪ A is a proper subcontinuum 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 proper subcontinua C, D such that C ∪ D = A. We can assume that p ∈ D. Then B ∩ D = ∅ (as Y is irreducible between p and q). By Theorem 11. Now we know that Y is either indecomposable or is the union of two indecomposable subcontinua. Applying Theorem 7 of [2], we conclude that the metrizable chainable continuum Y is circle-like. Problem 1. For which families K of connected topological graphs every 4-K-like continuum is K-like? Is it true for the family K = {8} 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?