3
$\begingroup$

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?

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

1 Answer 1

7
$\begingroup$

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.

And a closed continuous bijection is a homeomorphism.

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .