In Page 53 of "Lie Groups" by Duistermaat and Kolk, we find the following:
A map $f:X\rightarrow Y$ between topological spaces is proper if $f^{-1}(K)$ is compact for each compact $K\subseteq Y$.
It is then claimed that if $X$ and $Y$ are Hausdorff then $f$ is a closed map (that is, if $C\subseteq X$ is closed then $f(C)$ is closed in $Y$).
I have not been able to prove this. In various locations, you can find a proof that if $Y$ is Hausdorff and locally compact then proper implies closed (e.g. this answer on MSE).
Is this result true just under the Hausdorff condition? If not, what is a counter-example?
Best Answer
This is not true if you merely assume $Y$ is Hausdorff (or even if you assume both $X$ and $Y$ are Hausdorff). For instance, let $S$ be an uncountable set and fix an element $a\in S$. Let $Y$ be $S$ with the topology such that a set is open iff it is either cocountable or does not contain $a$. Let $X$ be $S$ with the discrete topology. Then $X$ and $Y$ are Hausdorff, and the identity map $f:X\to Y$ is proper since every compact subset of $Y$ is finite. However, $f$ is not closed.
(More generally, you could take $Y$ to be any Hausdorff space which is not compactly generated and let $f:X\to Y$ be its $k$-ification. So, combining this with Stefan Hamcke's answer at the linked question, a necessary and sufficient condition for this theorem to be valid for a Hausdorff space $Y$ is that $Y$ is compactly generated.)