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Let f be a one-to-one function from a compact Hausdorff space X onto a compact Hausdorff space Y. Show that if f[K] is compact for every compact K subset of X, then f is continuous.

I know that f restricted to K must be continuous, because the continuous image of a compact set is compact. I'm not sure how to prove that f is continuous, other than the fact that every b in Y has a distinct f(b) in X because f is one-to-one and onto.

I'm not sure how to set this problem up.

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"I know that f restricted to K must be continuous, because the continuous image of a compact set is compact." You're mixing up your implication; consider any surjective map from [0,1] to {0,1}. –  Gaffney Apr 26 at 19:51

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Hint: show that $f$ is a closed map. Then $f^{-1}$ is continuous, so $f^{-1}(K)$ is compact when $K$ is. Use this to show that $f$ is continuous.

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I know that every compact space is closed. So each K is closed in X and is mapped to f[K], which is closed in Y (because f is one-to-one and onto each f[K] is distinct). This shows that f is a closed map? Am I missing anything? –  user114634 Apr 26 at 21:38
    
@user114634: To show $f$ is a closed map, you have to start with an arbitrary closed subset $A\subseteq X$ and show that $f(A)$ is closed in $Y$. –  Jack Lee Apr 26 at 22:05
    
Well, If I take an arbitrary closed subset A of X. Then A is compact because closed subsets of compact spaces are compact. Thus, every closed space in X is compact and since f is one-to-one and surjective, there exists an f[A] in Y such that f[A] is compact (by our assumption). Thus, f is a closed map. –  user114634 Apr 26 at 22:36
    
Not sure how to prove that f is continuous... –  user114634 Apr 26 at 23:27
    
@user114634 A map is continuous iff the inverse image of every closed set is closed. Since $f$ is a bijective closed map, $f^{-1}$ is continuous. Now use the fact that continuous maps preserve compact sets ($f^{-1}$ in particular) to show that $f$ is also a closed map. –  Seth Apr 26 at 23:58

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