# A closed map is inveritible?

Let $(X,d_X)$ and $(Y,d_y)$ metric spaces, in which X is compact. Let $f:X\to Y$ be a continuous map. So

If $f$ is injective the $\underline{f}:X\to f(X),y\to f(y)$ is a homeomorphism and $f(X)$ is compact metric space.

Proof: If $f$ is injective, $\underline{f}$ is a homeomorphism, and if $f$ is bijective we have $f=\underline{f}$,therefore it is an homeomorphism.

Question:

How do we know $f$ is invertible? That is also a condition for $f$ to be a homeomorphism. Is it due to the fact $f$ is a closed map?

• $\underline{f}$ is invertible because it is injective and surjective onto $f(X)$. – Michael Lee Oct 20 '17 at 15:10
Yes, as the comment says, by the given conditions, the map is a bijection. We have to show that $f^{-1}|_{f(C)}$ is continuous, for which it is enough to show that closed sets in $X$ are sent to closed sets in $f(X)$. This is true because any closed set $C\subset X$ is compact (because $X$ is compact), hence $f|_{C}$ being continuous, $f(C)\subset f(X)$ is compact, hence closed because any metric space is Hausdorff, and any compact subset of a Hausdorff space is closed.