# Continuous bijection is homeomorphism

(I was able to fix my question while I was writing)

I am asked to show:

Let $X,Y$ be topological spaces with $X$ compact and $Y$ Hausdorff. Let $f:X\to Y$ be continuous and a bijection. Then is $f^{-1}$ continuous. Hence $f$ is a homeomorphism.

To prove this, it is given to use the following statements:

1) Let $(X,\tau)$ be a topological space and $K\subseteq X$ compact. If the topology on $X$ is hausdorff, then is $K$ closed.

2) Let $(X,\tau)$ be a topological space and $K\subseteq X$ compact, $A\subseteq X$ closed. Then is $A$ compact.

3) Let $(X,\tau_X), (Y,\tau_Y)$ be topological spaces. $f: X\to Y$ continuous and $K\subseteq X$ compact. Then is $f(K)\subseteq Y$ compact.

For my prove:

We have to show, that $f^{-1}$ is continuous. Therefor the preimage of open sets is open. I show, that the preimage of closed sets is closed. Therefor if $A\subseteq X$ is closed, then $f(A)\subseteq Y$ is closed.

Let $A\subseteq X$ be closed. $X$ is compact, therefor we have $A$ compact. (Statement 2). Since $f$ is continuous and $A$ compact, it is $f(A)\subseteq Y$ compact. (Statement 3) Since $Y$ is hausdorff it is $f(A)$ closed. (Statement 1) Hence the preimage of closed sets is closed and $f^{-1}$ is continuous.

Thanks in advance.

• Yes, that's it. What's your question? – Berci Aug 12 '18 at 8:24
• @Berci I made a little mistake, which I noticed when I was already done writing. So I fixed that and then posted anyways, because why not. Thanks for your correction. – Cornman Aug 12 '18 at 8:34