Theorem: Let $X, Y$ be topological spaces where $Y$ is first-countable and Hausdorff and let $f: X\rightarrow Y$ be an injective, continuous map which is proper. Then $f$ is a homeomorphism into it's image.

Proof: The only thing left to show is that $f^{-1}: f(X) \rightarrow X$ is continuous. Since $Y$ is first-countable the subspace $f(X) \subseteq Y$ is first-countable too. In particular $f(X)$ is a compactly generated space. Therefore it is enough to show that for each compact subset $K\subseteq f(X)$ the restriction $f^{-1}|_K: K\rightarrow f^{-1}(K)$ is continuous.

As a compact subset of $f(X)$ the set $K$ is also compact in $Y$. Since $f$ is proper $f^{-1}(K)\subseteq X$ is compact. So $f|_{f^{-1}(K)}: f^{-1}(K) \rightarrow K$ is a continuous and bijective map on a compact space into a Hausdorff space. It is well-known that this already implies that $f|_{f^{-1}(K)}$ is a homeomorphism. In particular the inverse function $f^{-1}|_K: K\rightarrow f^{-1}(K)$ is continuous. This proves the theorem.

Is this proof correct?


Yes, your proof is correct. You should trust yourself!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.