$X,Y$ are compact Hausdorff. $f$ is bijective continuous. Is $f$ a homeomorphism?

Question

Let $X,Y$ be compact Hausdorff spaces. Let $f:X\to Y$ be one-to-one, onto and continuous. Show that $f$ is a homeomorphism.

I came up with this "proof" but I am very sure it is wrong. In order to show that $f$ is a homeomorphism, we only need to show that $f^{-1}$ is continuous. So I considered an open set $U\subset X$, and wanted to show that $f^{-1}(U)$ is open in $Y$. Suppose that's not the case, then $f^{-1}(U)$ is closed in $Y$. Then its preimage $f^{-1}f^{-1}(U)$ is closed in $X$ since $f$ is continuous. Then that implies that $U$ is closed in $X$, a contradiction.

I feel like I messed up with the last part. Also, how do I use "compact" and "Hausdorff" here? I know that a compact set in a Hausdorff space has to be closed.

Answer 1

So, let's consider an open $K \subset X$, then its complement is closed. Closed subspace of a compact space is also compact. Let $f: X \rightarrow Y$ - continous map. Then, since the image of a compact set is compact (the map is continous), $Y - f(K)$ is compact. Since the compact subspace of a Hausdorff space is closed, $Y - f(K)$ is closed and then $f(K)$ is open.

So, $f$ maps open spaces to open spaces and this also is true for $f^{-1}$.

Answer 2

The proof is wrong because if a set is not open that doesn't mean it is closed. A set may be neither open nor closed. Instead work with the complement of $U$, that is closed.

The point is that a closed subset of a compact set is compact and a compact subset of a hausdorff space is closed. So show that $f(\text{closed})=\text{closed}$ and that will show $f^{-1}$ is continuous.