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Topology Proof

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that if X is compact and Y is Hausdorff then a continuous bijection [tex] f: X \longrightarrow Y [/tex] is a homeomorphism. (You may assume that a closed subspace of a compact space is compact, and that an identification space of a compact space is compact).


    2. Relevant equations

    A space X is compact if every open cover {[tex] \left{ U_{\lambda} | \lambda \in \Lambda \right}[/tex]} , [tex] \cup U_{\lambda} = X [/tex].
    A space is Hausdorff if for any pair of distinct points x,y in Y, there exist open sets separating them.


    3. The attempt at a solution

    Going on the hint of what I can assume, I define the equivalence class x~x' if f(x) = f(x'), with projection p(x) = [x]. Then f = gp: X -> X/~ -> Y. It may be easier to prove p and g are homeomorphisms than f. The identification space is compact, and so is any closed subset. We are given that f is a continuous bijection, so only need to show that f inverse is continuous, ie f(U) is open for any U open, or perhaps more easily that f(C) is closed for C closed in this case. A compact subset of a Hausdorff space is closed, so if I can prove that p maps a compact space to a compact space then I think I've proved p is a homeomorphism.
    Then I would need to prove g.
     
    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 15, 2009 #2
    I must admit that I don't really see why you would use the identification space. I don't think that your identification will help much because f is injective so f(x)=f(x') imply x=x' which means that ~ is just equality and therefore X is naturally homeomorphic to X/~ by the correspondence [itex]x \leftrightarrow \{x\}[/itex]. Thus proving g a homeomorphism is pretty much the same as proving f a homeomorphism.

    Anyway the way I would do it without identification spaces is to realize that a continuous function maps compact subspaces to compact subspaces in general (either refer to your book or show it directly by considering a covering of f(C) where C compact, taking pre-images of the covering, using compactness to find a finite subcover of the pre-images and then finally using this to get a finite subcover of f(C)) and then simply assume [itex]C \subseteq X[/itex] is closed and show:
    C closed => C compact => f(C) compact => f(C) closed.
     
  4. Nov 15, 2009 #3
    Yeah you're right about my identification space, I didn't think about that. I think I was already going in the direction of showing compact spaces map to compact spaces, I'll try you're suggestion.
     
  5. Nov 15, 2009 #4
    Ok I copied the structure of an earlier proof and it seems straight forward:
    Take a compact subspace C in X. Let U be an open covering of f(C) with sets [tex] U_{\lambda} [/tex]. Define an open covering V of C by taking [tex] V_{\lambda} = f^{-1} \left( U_{\lambda} \right) [/tex]. Note that the openness of V relies on the continuity of f, and the fact that it covers C relies on the fact that f is bijective. Now taking a finite refinement of V gives a finite refinement of U, so f(C) is compact.

    If C is closed, then C is compact in X. If f(C) is compact then f(C) is closed in Y, since Y is Hausdorff. So f(C) is closed for every C closed. Hence f inverse is continuous.

    Does that look ok?
     
  6. Nov 15, 2009 #5
    Yes except for one minor detail:
    You don't use the bijectiveness here. If [itex]x \in C[/itex], then [itex]f(x) \in f(C)[/itex] so there exists some [itex]\lambda[/itex] such that [itex]f(x) \in U_\lambda[/itex] because U is a covering of f(C), and then [itex]x \in f^{-1}(U_\lambda) = V_\lambda[/itex]. Thus the pre-images [itex]\{V_\lambda\}[/itex] cover C whether f is bijective or not. Apart from that it's good.
     
  7. Nov 15, 2009 #6
    My definition of a covering is that the union of the [tex] V_{\lambda} [/tex] is strictly equal to C, not a subset. It may be that all the x in C are covered, but if the map is not injective then [tex] f^{\left (-1) \right} [/tex] could give out more elements than are in the set C. Otherwise the bijective property would be used. Now I realised I only used injectivity, I wonder if surjectivity is necessary?
     
  8. Nov 15, 2009 #7
    In that case: Yes you used it, but you don't need it. As you have observed [itex]f^{-1}(U_\lambda)[/itex] is open in X, so [itex]V_\lambda' = f^{-1}(U_\lambda) \cap C[/itex] is open in C given the subspace topology, so C is covered by [itex]\{V_\lambda'\}[/itex] but of course for this problem it doesn't really matter since you know that f is bijective.
     
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