# When is a proper map closed?

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?

• 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? – davik Oct 24 '18 at 22:11

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.)
• $Y$ 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. – Henno Brandsma Oct 24 '18 at 22:08