Any help please with the following problem?
Let $f:X\rightarrow Y$ be a continuous one-to-one and surjective mapping between
compact metric spaces. Prove that the inverse mapping $f^{-1}:Y\rightarrow X$ is continuous.
I tried the following: I assumed $C$ a subset of $X$, then by compactness of $X$, C is also compact and hence it is closed. $f$ is continuous and $C$ is compact, then $f(C)$ is compact and hence it is closed. Any help how to go from here to prove that $f^{-1}$ is continuous?
Best Answer
Every metric space (for instance, $Y$) is Hausdorff. Then, you can use the fact that every continuous and bijective map $f: X \longrightarrow Y$ is a homeomorphism (that is, $f^{-1}$ is continuous), if $X$ is compact and $Y$ Hausdorff.
To prove that claim, you can do as follows: in order to see that a map $g: Y \longrightarrow X$ is continuous, you can prove that, for every closed subset $C \subset X$, $g^{-1}(C) \subset Y$ is closed. In our case, $g = f^{-1}$, so it suffices to show that, for every closed subset $C \subset X$, $(f^{-1})^{-1}(C) = f(C) \subset Y$ is closed. Right?
But, if $C$ is a closed subset of a compact space $X$, then it's itself compact and the image of a compact set by a continuous map, $f$, is compact. Hence $f(C)$ is a compact subset of a Hausdorff space, $Y$. Every compact subset of a Hausdorff space is closed. Thus, $f(C)$ is closed. qed.