# Proving that a continuous bijective map is a homeomorphism

I believe that the following statement is correct.

Statement. Let $$\varphi: X\to Y$$ be a bijective continuous map between two Hausdorff locally compact spaces $$X$$ and $$Y$$. Suppose that $$\varphi$$ is proper, i.e., for any compact subset $$K\subset Y$$ the set $$\varphi^{-1}(K)$$ is compact in $$X$$. Then $$\varphi$$ is a homeomorphism.

Have you seen this statement (or a more general one that implies this) in some book?

• Most likely, this is in Bourbaki "General Topology". Assuming 1st countability, a proof is here. Commented Feb 3, 2020 at 22:20

Well, $$\phi$$ is a so-called "proper" map, i.e. continuous with inverse images of compact sets compact. It is well-known that if the codomain $$Y$$ is compactly generated (which locally compact Hausdorff spaces are, as well as first countable spaces, e.g.; also known as a $$k$$-space in other texts, like Engelking) then $$\phi$$ is a closed map (i.e. perfect). I believe Bourbaki's Topologie Génerale will cover this general fact.
A direct proof for closedness of $$\phi$$ (which implies that the inverse is continuous) in this case: suppose $$C \subseteq X$$ is closed. Suppose $$y \in \overline{\phi[C]}$$ and let $$K$$ be a compact neighbourhood of $$y$$ in $$Y$$. Then $$\phi^{-1}[K] \cap C$$ is compact in $$X$$ and hence so is $$\phi[\phi^{-1}[K] \cap C]= K \cap \phi[C]$$ and by Hausdorffness of $$Y$$, $$\phi[C] \cap K$$ is closed in $$K$$ which implies $$y \in \phi[C]$$ and so $$\phi[C]$$ is closed.