This problem is bugging me for some time.

$f:(X,\mathcal{T}) \to (Y,\mathcal{T}')$ is a continuous bijection where $(X,\mathcal{T})$ is compact and $(Y,\mathcal{T}')$ is $T_2$ (i.e. Hausdorff and $T_1$). Then show that $f$ must be a homeomorphism.

So far I have failed to understand any relation between the compactness of domain and $T_2$-ness of the range. I tried in vain to prove that $f(U)$ is open in $(Y,\mathcal{T}')$ whenever $U$ is so in $(X,\mathcal{T})$. I know $f$ makes $X$ and $Y$ respectively $T_2$ and compact and hence makes both of them $T_4$ spaces. But that is not coming to any use. I would be glad if someone helps me with a solution. Thank you.


1 Answer 1


Hint: You only have to prove that it's a closed map.

Now, if you take a closed subspace $A\subseteq X$ you know it is (BLANK) and so its image is (BLANK). But, by Hausdorffness, (BLANK) subspaces are closed.

  • 1
    $\begingroup$ Those blanks are all "compact", aren't they? Thanks a lot for your cool hint. And I feel like an idiot, as usual. :-( $\endgroup$
    – Sayantan
    Dec 12, 2012 at 7:52
  • 6
    $\begingroup$ Haha, there is no reason to feel like an idiot--there are just tricks you pick up as you go along. Chin up sailor! $\endgroup$ Dec 12, 2012 at 7:56

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