# compactness and continuity in metric spaces

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?

• Warning: it is not true that every subset of a compact space $X$ is itself compact! For example, the open interval $(0,1)$ is not compact, though it is a subspace of the compact interval $[0,1]$. However, every closed subset of a compact space is itself compact. – bradhd Dec 12 '11 at 3:01
• As for the rest of the argument: recall that a map $g: Y\to X$ is continuous if and only if, for each closed subset $C\subset X$, the preimage $g^{-1}(C)$ is a closed subset of $Y$. What is this preimage when $g =f^{-1}$? – bradhd Dec 12 '11 at 3:01
• @Brad in this case, the preimage is $f(C)$, right? – M.Krov Dec 12 '11 at 3:05

## 3 Answers

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.

It is a standard fact that a function $g : A \to B$ is continuous if and only if the inverse image of every closed $C \subseteq B$ is closed in $A$. [It is good to compare this with the definition of continuity, which says that the inverse image of open sets is open.] If you have not done so already, I recommend that you show this fact as an exercise. Thus to prove $f^{-1} : Y \to X$ is continuous, it suffices to show that $f$ is a closed map; i.e., the (forward) image of every closed set in $X$ is closed in $Y$.

You have almost showed this final statement. However your proof has a small, but important, error. It is not true that if $C \subseteq X$, then $C$ is compact. However, if $C$ is closed in $X$ (which itself is compact), then it is true that $C$ is compact. The rest of your proof then carries over as written.

You went a right way. Because $f$ is a bijection, it has the inverse function $f^{-1}$, which has to be shown to be continuous as well as $f$ is. A function between topological spaces is continuous if a preimage of every closed set is closed. So take $C$ closed in $X$. Because of the compactness of $X$ it is also compact. Continuous functions preserve compactness, so $f(C)$ is compact, and hence closed. Since $(f^{-1})^{-1}(C)=f(C)$, we get that the preimage of $C$ by $f^{-1}$ is closed. So $f^{-1}$ is continuous.

$Y$ does not need to be compact, but it has to be Hausdorff. However, since $f$ is a homeomorphism, $Y$ appears to be compact, too. And of course every metric space is Hausdorff.