# Compactness and continuity

I am having some trouble with this:

Let $f : X \to Y$ be continuous and bijective with $X$ compact then $f^{-1} : Y \to X$ is continuous? I am not quite sure if this is true, if is not I cant find any counter example, but if it is true I am stuck with the proof. I need some help. Thank you

-
oh that´s true, my fault – user62758 Feb 18 '13 at 5:19

Take $Y$ to be a set with the same cardinality as $X$ and give it the topology $\tau=\{ \emptyset, Y\}$. Consider a (set theoretic) bijection $f: X\to Y$. This is continuous but it's inverse is not, unless $X$ had the same topology that $Y$ does.

-
Oh that works, thank you – user62758 Feb 18 '13 at 5:46

To show $f^{-1}$ is continuous, you just need to show that $f = (f^{-1})^{-1}$ is an open (or closed) mapping. Since we're working with $X$ compact, it is probably easier to show $f$ is a closed mapping. First of all, let $U \subseteq X$ be closed (hence compact because $X$ is compact). Since $f$ is continuous, $f(U)$ is compact, so it's closed and you're done.

I forgot to mention that I'm assuming $Y$ is a Hausdorff space.

-
yes i know that if Y is hausdorff then it is true, but when y is not then \$f ^{-1} is not continuous – user62758 Feb 18 '13 at 5:29
sure, but could you clarify that in your problem then? – cats Feb 18 '13 at 5:31
and regardless, if you know why it works in a Hausdorff space, then it is not too hard to come up with a counter example – cats Feb 18 '13 at 5:34
sorry, I guess the problem is that f is not closed so neither an homeomorphism – user62758 Feb 18 '13 at 5:35