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Use the definitions:

  • A continuous map is perfect if it is a closed surjection with compact fibers.
  • A continuous map is proper if the inverse image of each compact set is compact. (In contrast to Bourbaki's more general definition, according to which a map $f : X \rightarrow Y$ is proper when, for every topological space $Z$, the product map $f \times i_{Z}: X \times Z \rightarrow Y \times Z$ is closed.)

The posts (1) Proper map not closed, (2) A continuous surjection is proper if and only if pre-images of compact sets are compact, (3) $f$ proper but not universally closed, and (4) Proper map not closed present examples of continuous surjections that are proper but not perfect (which amounts to saying they are not closed). However, in each of these examples the codomain $Y$ fails to be a Hausdorff space.

Is there such an example where $Y$ is a Hausdorff space? (Of course, for such an example, $Y$ could not be locally compact or, more generally, compactly generated.)

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Let $X$ be an uncountable discrete space and $Y$ be its one-point Lindelöfication: $Y = X \cup \{\infty\}$ and a neighbourhood of $\infty$ is of the form $\{\infty\} \cup X\setminus C$, where $C \subseteq X$ is countable, while any subset of $X$ is still open. An alternative definition is $C$ is closed in $Y$ if $\infty \in C$ or $C \subseteq X$ and $C$ is at most countable.

$Y$ is also called Fortissimo space in counterexamples in topology.

Then as an example: let $Z=Y$ as sets, but in the discrete topology, and $Y$ is the above space, $f:Z \to Y$ is the identity function, which is continuous as the domain is discrete; not closed (as the topology on $Y$ is strictly coarser than the discrete one) and proper as the only compact subsets of $Y$ are the finite ones, and these have compact pre-images. $Y$ is hereditarily normal ($T_5$), so certainly at least Hausdorff.

$Y$ indeed cannot be a $k$-space, as a proper map onto a Hausdorff $k$-space is perfect by a classical theorem.

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  • $\begingroup$ Description of the two spaces needs a bit of revision, as you want $X$ and $Y$ to have the same underlying set. Let $S$ be an uncountable set and let $p \in S$ be a particular point. Let $X$ be $S$ with its discrete topology. Let $Y$ be the space obtained by giving $S$ the "Fortissimo topology": a subset $U$ of $S$ is open for $Y$ if and only if $S \setminus U$ is countable or $p \notin U$. $\endgroup$ – murray Jul 18 '18 at 18:40
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Let $Y$ be any Hausdorff space that is not compactly generated, and let $X$ be the $k$-ification of $Y$ (so $X$ has the same underlying set and a set is closed in $X$ iff its intersection with each compact subset of $Y$ is closed). Then the identity map $X\to Y$ is continuous and surjective, and is proper since the topology on $X$ is the same as the topology on $Y$ when restricted to any compact set. However, it is not a closed map, since if it were then it would be a homeomorphism and $Y$ would be compactly generated.

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  • $\begingroup$ My example is an instance of this construction, I see. I just realised. $\endgroup$ – Henno Brandsma Jul 19 '18 at 16:50

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