Locally Sierpinski Quotients

Given any non-trivial, connected topological space X, it is possible to define an equivalence relation ∼ on it such that the topological quotient space X/ ∼ is the Sierpinski space. Locally Sierpinski spaces are generalizations of the Sierpinski space and here we address the following question. Does a statement like the one above hold if Sierpinski is replaced by (proper) locally Sierpinski ? The answer is no and we will give below a few counterexamples. The situation where a homeomorphism group acts on a topological n-manifold will also be analysed, the conclusion being that the cases n = 1, n > 1 are radically different. MSC 2000: 54F65 1. Locally Sierpinski spaces A topological space X is said to be a locally Sierpinski space, l. S. space in short, if every point x ∈ X has an open neighbourhood Ux homeomorphic to the Sierpinski space ({0, 1}, {∅, {0}, {0, 1}}). If in an l. S. space X a unitary set {x} is open we say that x is a centre. The remaning points will be called satellites. The centres and the open sets homeomorphic to the Sierpinski space form a base for the topology of X. Therefore X is locally (path-)connected and its (path-)connected components are open. Moreover there is a bijection between the set of centres and the set of components of X. Such a bijection associates to each centre the component which contains it. Examples of connected l. S. spaces are often referred to in topology textbooks [3]. They are obtained as follows. Let X be a set with, 0138-4821/93 $ 2.50 c © 2005 Heldermann Verlag 358 S. Carter, F. J. Craveiro de Carvalho: Locally Sierpinski Quotients at least, two points. Fix p ∈ X and define a set to be open if it is either the empty set or contains p. L. S. spaces can be characterized as locally (path-)connected spaces whose (path-)components are subspaces of the same type as X. 2. Existence of quotient maps Proposition 1. Let X be a connected topological space. If X can be partitioned as X = A ∪ ( ⋃ i∈I Ai), I 6= ∅, where A is open and, for each i ∈ I, Ai is closed and A ∪ Ai is open, then there is a quotient map p : X → S, where S is an l. S. space with #I as the number of satellites. Conversely, if there is a quotient map p : X → S, where S is l. S., then X can be partitioned as X = A ∪ ( ⋃ i∈I Ai), with A open, for each i ∈ I, Ai closed, A ∪Ai open and #I the number of satellites. Proof. Assume that there is a quotient map p : X → S, c is the centre of S and xi, i ∈ I, are the satellites. Let then A = p−1(c) and Ai = p −1({c, xi}), i ∈ I. Assume now that X = A∪ ( ⋃ i∈I Ai), I 6= ∅, where A is open and, for each i ∈ I, Ai is closed and A∪Ai is open. Add an extra point c to I to obtain a new set J and make this new set into an l. S. space by taking as a topological basis the subsets {c}, {c, i}, i ∈ I. Then define p : X → S by p(A) = c, p(Ai) = i, i ∈ I. This map is obviously surjective and continuous. It is also a quotient map. Suppose that U ⊂ J , nonempty and different from J , is such that p−1(U) is open. Then c ∈ U and therefore U is open in J . If c were not in U then U would be closed and p−1(U) would also be closed. This contradicts the connectedness of X. By assuming that X is locally connected, we can actually formulate the first part of the proposition, but not the second, in terms of the components Xλ, λ ∈ Λ, of X. The proof would follow as above. For each λ ∈ Λ, we would have a map p : Xλ → Sλ and would use these maps to obtain p : X → S, where S is the disjoint union of the Sλ’s. As to a counterexample for the converse consider the disjoint union of a Sierpinski space and a trivial two point space. That is, consider X = {a, b, c, d}, with topology τX = {∅, X, {a, b, c}, {a, b}, {c}, {c, d}}, and the Sierpinski space {c, d}, with {c} open. The map p : X → S given by p({a, b, d}) = {d}, p(c) = c is a quotient map. 3. Examples 1. Let X be a T1 connected space and r be an integer such that 1 ≤ r < #X. Then a connected l. S. space with r satellites is obtainable from X. One fixes r points in X, one takes for A their complement and each Ai is formed by just one of the fixed points. 2. Let X be a set with at least 2 points and let us fix p ∈ X. Define U ⊂ X to be open if U = X or p / ∈ U . This space is connected and the only l. S. space which is obtainable from it is the Sierpinski space itself. 3. LetX now be the real numbers and define U to be open if it is empty or an open interval of the form (−∞, a) or (−∞,+∞). Again only the Sierpinski space is obtainable. S. Carter, F. J. Craveiro de Carvalho: Locally Sierpinski Quotients 359 4. S and other spheres As mentioned earlier any finite, connected l. S. space can be obtained from the circle S. However if the quotient space is a space of orbits originated by the action of a subgroup G of the homeomorphism group of S the possibilities are severely restricted. For R this problem was solved in [1]. From now on we will assume that a group of homeomorphisms G acts on S. Proposition 1. No free action on S gives rise to an l. S. space. Proof. Assume that S/G is l. S., p : X → S/G is the projection and c is the centre of S/G. If p−1(c) is S with one point removed then that point would be a fixed point for every element of G. Suppose then that p−1(c) is not S with one point removed. Take a component _ x0y0 of p −1(c) and choose distinct x1, y1 in it. There is f ∈ G which maps x1 to y1. Since f maps the closed arc _ x0y0 to itself, it must have a fixed point. Proposition 2. If S/G is an l. S. space then it has one or two satellites. Both cases can occur. Proof. Let c be the centre of S/G, p the projection which, since we are dealing with a group action, is open, and p−1(c) = ⋃ i∈I Ci, where the Ci’s are the connected components. We will show that the inverse image of the set of satellites is precisely the set of end points of the open arcs Ci. Obviously the end points of the Ci’s project to satellites. If x ∈ p−1(s), where s is a satellite, consider the open set {c, s} and its inverse image. Let C denote the connected component in that inverse image which contains x. If no point other than x lies in C∩p−1(s), x is an end point of one of the Ci’s. On the other hand, if there is another x ′ in C ∩ p−1(s) then there must exist a component Ci such that Ci ⊂ _ xx ′ . The end points of Ci are in the orbit of x and, consequently, x must be an end point of a component Cj. If the inverse image of the set of satellites is a singleton then S/G is the Sierpinski space. If not the Ci’s are arcs with distinct end points. These end points give rise to one or two orbits and therefore the quotient is the Sierpinski space or has two satellites. The Sierpinski space can be obtained by letting G be the group of S homeomorphisms which fix, for instance, the north pole or, alternatively, fix the subset formed by the north and south pole. An l. S. space with two satellites can be obtained by letting G be the group of S homeomorphisms which fix, for instance, the north pole and the south pole. Combining the corresponding result for R [1] and the next lemma gives us another way of recovering Proposition 2. Lemma 1. Let X be a connected, locally path-connected, semilocally simply connected topological space and let (X̃, π) be its universal covering. Assume there is a homeomorphism group G such that X/G is l. S. Then there is also a group G̃ such that X̃/G̃ and X/G are homeomorphic. Proof. Let G̃ = {f̃ : X̃ → X̃ | π ◦ f̃ = f ◦ π, for some f ∈ G} and let it act on X̃. Then x1, x2 ∈ X̃ are in the same orbit iff the same happens to π(x1), π(x2). Consider now the 360 S. Carter, F. J. Craveiro de Carvalho: Locally Sierpinski Quotients projection p : X → X/G and compose it with π. This composition is a quotient map which induces exactly the same equivalence relation on X̃ as the G̃-action. Therefore the spaces X̃/G̃ and X/G are homeomorphic. Let now X be a topological space and G be a homeomorphism group. If Y is a homogeneous topological space and H is its homeomorphism group then the quotient spaces X/G and X × Y/G×H are homeomorphic. Hence, for instance, the torus S × S can be acted upon by a group of homeomorphisms such that the quotient is an l. S. space with one or two satellites. In the case of S, n ≥ 2, we have Proposition 3. Given a positive integer r, there is a group G of S homeomorphisms, n ≥ 2, such that S/G is an l. S. space with r satellites. Proof. The result follows from the following facts: S is the 1-point compactification of R and any homeomorphism of R extends naturally to a homeomorphism of S. It is known [2] that given two n-sequences (x1, . . . , xr), (y1, . . . , yr) of distinct points in R , n ≥ 2, there is a homeomorphism f : R → R such that f(xi) = yi, i = 1, . . . , r. We can therefore take the group G formed by the S homeomorphisms which fix r points. 5. Topological n-manifolds In view of the results of the previous section it is only natural to ask what sort of l. S. spaces can be obtained if one starts with a topological n-manifold, n ≥ 2. We start with an independent proof of a particular case of Proposition 5 in [2]. In what follows each time we refer to a chart for an n-manifold it will be understood that the codomain is R. Proposition 1. Let F be a finite subset of R, n ≥ 2, and let x, y ∈ R \ F . Then there is a homeomorphism h : R → R which is pointwise fixed on F , is the identity outside a compact set and such that h(x) = y. Proof. We will write xy for the line segment determined by x and y. Case 1: xy ∩ F = ∅ There is no loss of generality in supposing x = (0,−1), y = (0, 1) ∈ Rn−1×R. Then there exist r, s > 0 such that (Dn−1(r)× [−1−s, 1+s])∩F = ∅, where Dn−1(r) = {p ∈ Rn−1 | ‖ p ‖≤ r}. Let f : [−1 − s, 1 + s] → [−1 − s, 1 + s] be a homeomorphism which fixes the end-points and maps −1 to 1 and define h : Dn−1(r) × [−1 − s, 1 + s] → Dn−1(r) × [−1 − s, 1 + s] by h(p, t) = (p, (1 − ‖p‖ r )f(t) + ‖p‖ r t). This h is a continuous bijection and, due to compactness and Hausdorfness, it is a homeomorphism. On the boundary of Dn−1(r) × [−1 − s, 1 + s] the homeomorphism h is the identity and we get the required homeomorphism for R if we extend it by taking it to be the identity outside Dn−1(r)× [−1− s, 1 + s]. Case 2: xy ∩ F 6= ∅ We can find z ∈ R such that xz ∩ F = ∅ and zy ∩ F = ∅. By Case 1, there are Rhomeomorphisms h1 and h2, both of which are pointwise fixed on F , are the identity outside S. Carter, F. J. Craveiro de Carvalho: Locally Sierpinski Quotients 361 compact subsets K1, K2, respectively, and such that h1(x) = z, h2(z) = y. If we take h = h2 ◦ h1 we have a homeomorphism as required.