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There is a theorem that says that if $f\colon X \rightarrow Y$ is bijective and continuous, then $f$ is an homeomorphism if $X$ is compact and $Y$ is Hausdorff.

What about if $X$ is Hausdorff and $Y$ is compact?

My question arises because I have noted that there a some kind of dual relations between compact and Hausdorff spaces. In this specific problem, I think that it must be false, but I can´t give any counterexample.

Thanks in advance.

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    $\begingroup$ Do you know any examples of continuous bijections that are not homeomorphisms? $\endgroup$ – Eric Wofsey Jun 18 '17 at 0:54
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Let $X=Y=[0,1]$. Let $X$ have the discrete topology and $Y$ have the standard (Euclidean) topology.

We know that $Y$ is compact. Also, $X$ is Hausdorff because every singleton is an open set.

Now consider the identity map, that is, $f:X\to Y, \quad x \mapsto x$.

Clearly, it is bijective. Also it is continuous because the inverse of every open set is open. Actually, every subset of $X$ is open.

It remains to show that this map is not a homeomorphism. $f^{-1}$ is not a continuous map, because $f(\underbrace{\{ 0 \}}_{\text{open in $X$}}) = \{ 0 \}$ is not open in $Y$. So, $f^{-1}$ is not continuous, hence $f$ is not a homeomorphism.

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Hint: Take $f : [0, 2\pi ) \to S$ with $f(t) = e^{it}$ . where $S$ is unit sphere.

$$\lim_{z \to (1,0)} f^{-1} (z) $$

Does not exist!

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