Prove that if X is compact and Y is Hausdorff then a continuous bijection

[madness]

Homework Statement

Prove that if X is compact and Y is Hausdorff then a continuous bijection $$f: X \longrightarrow Y$$ 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).

Homework Equations

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

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.

 

[rasmhop]

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 $x \leftrightarrow \{x\}$. 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 $C \subseteq X$ is closed and show:
C closed => C compact => f(C) compact => f(C) closed.

 

[madness]

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.

 

[madness]

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 $$U_{\lambda}$$. Define an open covering V of C by taking $$V_{\lambda} = f^{-1} \left( U_{\lambda} \right)$$. 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?

 

[rasmhop]

Does that look ok?

Yes except for one minor detail:

and the fact that it covers C relies on the fact that f is bijective.

You don't use the bijectiveness here. If $x \in C$, then $f(x) \in f(C)$ so there exists some $\lambda$ such that $f(x) \in U_\lambda$ because U is a covering of f(C), and then $x \in f^{-1}(U_\lambda) = V_\lambda$. Thus the pre-images $\{V_\lambda\}$ cover C whether f is bijective or not. Apart from that it's good.

 

[madness]

My definition of a covering is that the union of the $$V_{\lambda}$$ 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 $$f^{\left (-1) \right}$$ could give out more elements than are in the set C. Otherwise the bijective property would be used. Now I realized I only used injectivity, I wonder if surjectivity is necessary?

 

[rasmhop]

My definition of a covering is that the union of the $$V_{\lambda}$$ 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 $$f^{\left (-1) \right}$$ could give out more elements than are in the set C. Otherwise the bijective property would be used. Now I realized I only used injectivity, I wonder if surjectivity is necessary?

In that case: Yes you used it, but you don't need it. As you have observed $f^{-1}(U_\lambda)$ is open in X, so $V_\lambda' = f^{-1}(U_\lambda) \cap C$ is open in C given the subspace topology, so C is covered by $\{V_\lambda'\}$ but of course for this problem it doesn't really matter since you know that f is bijective.