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?

  • $\begingroup$ just a note, in two of my books proper maps are by definition closed, so it's interesting/weird that your book doesn't define it that way. Anyone know which definition is more natural? $\endgroup$
    – davik
    Oct 24, 2018 at 22:11

1 Answer 1


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.)

  • 1
    $\begingroup$ $Y$ and $X$ are even hereditarily normal, so separation axioms won't make a better result possible. And $Y$ is even only not locally compact at $a$, and $X$ is locally compact, so it's pretty close to optimal. $\endgroup$ Oct 24, 2018 at 22:08

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