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Let $X$ and $Y$ be Hausdorff spaces and $f:X\to Y$. Show that if $f$ is a closed function (i.e. it maps closed sets to closed sets) and $f^{-1}(y)$ is compact in $X$ for every $y\in Y$ then $f$ is a proper map. (i.e. the preimage of compact sets in $Y$ is compact in $X$) Show furthermore that if $f$ is proper and $Y$ is a locally compact Hausdorff space then $f$ is closed.

My ideas:

For the first part, I considered an arbitrary open cover $\bigcup_{i\in I} U_i$ of $f^{-1}(K)$ in $X$ where $K$ is compact in $Y$. I tried mapping the complement of that cover, which is closed, with $f$ and then going again to the complement, so $Y\setminus f(X\setminus\bigcup_{i\in I} U_i)$, with the intention that it may be an open cover of $K$ which I could reduce to a finite open cover but that did not work out.

Moreover I looked at the preimages $f^{-1}(y_i)$ where $y_i\in U_i$ and $K\subseteq \bigcup_{i=1}^n U_i$ (a finite open cover) knowing that those preimages will be compact, but this is also not enough to lead me to the right idea. I don't know how to use the Hausdorff property. I may miss a crucial point, but in general properness is a property for preimages and closedness is a property for images.

For the second part, I sadly don't have any ideas.

Could someone please help me out?

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  • $\begingroup$ Are you familiar with filters? The characterisation of quasicompact spaces as those where every filter has an adherent point suggests a way to prove the first part, if you are. $\endgroup$ May 15, 2020 at 19:29
  • $\begingroup$ Unfortunately, I am not. But I have certainly heard about them and may study them in the future. However, I think this problem has to have a solution without using filters. $\endgroup$ May 15, 2020 at 19:31
  • $\begingroup$ Yes, of course. It's just easier (IMO) to see a proof via filters. So, take an open cover of $f^{-1}(K)$. For each $y \in K$, you can select finitely many elements $U_1, \ldots, U_k$ of the cover with $f^{-1}(y) \subset U_1 \cup \ldots \cup U_k$. Use these sets and the closedness of $f$ to find a neighbourhood $V$ of $y$ such that $f^{-1}(V) \subset U_1 \cup \ldots \cup U_k$. $\endgroup$ May 15, 2020 at 19:37
  • $\begingroup$ I’m pressed for time at the moment and will have to come back to the second part, but I’ve written up an answer to the first part; you really were on the right track. $\endgroup$ May 15, 2020 at 19:39

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You were on the right track for the first part; you just didn’t realize that you have to use the fact that the fibres of $f$ are compact.

Let $K\subseteq Y$ be compact, and let $\mathscr{U}$ be an open cover of $f^{-1}[K]$ in $X$. Let

$$\mathscr{V}=\left\{\bigcup\mathscr{F}:\mathscr{F}\subseteq\mathscr{U}\text{ is finite}\right\}\;;$$

$\mathscr{V}$ is also an open cover of $f^{-1}[K]$. For each $V\in\mathscr{V}$ let $W_V=Y\setminus f[X\setminus V]$; since $f$ is closed, $W_V$ is open in $Y$. Let $y\in K$; $f$ is perfect, so $f^{-1}[\{y\}]$ is compact, some finite $\mathscr{F}\subseteq\mathscr{U}$ covers $f^{-1}[\{y\}]$, and and therefore there is a $V\in\mathscr{V}$ such that $f^{-1}[\{y\}]\subseteq V$. Then $f^{-1}[\{y\}]\cap(X\setminus V)=\varnothing$, so $y\in W_V$. Thus, $\mathscr{W}=\{W_V:V\in\mathscr{V}\}$ is an open cover of $K$. $K$ is compact, so $\mathscr{W}$ has a finite subcover; let $\mathscr{V}_0$ be a finite subset of $\mathscr{V}$ such that $\{W_V:V\in\mathscr{V}_0\}$ covers $K$.

Clearly $\mathscr{V}_0$ covers $f^{-1}[K]$. Moreover, for each $V\in\mathscr{V}_0$ there is a finite $\mathscr{F}_V\subseteq\mathscr{U}$ such that $V=\bigcup\mathscr{F}_V$. Let $\mathscr{U}_0=\bigcup_{V\in\mathscr{V}_0}\mathscr{F}_V$; then $\mathscr{U}_0$ is a finite subset of $\mathscr{U}$ that covers $f^{-1}[K]$, and $f$ is proper.

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  • $\begingroup$ I see. But did you actually use the Hausdorff property? $\endgroup$ May 15, 2020 at 19:50
  • $\begingroup$ @mathemagician99: Not that I can see; it really doesn’t seem to be needed for this argument. At no point, for instance, do we need to argue that a set is closed because it’s compact. $\endgroup$ May 15, 2020 at 19:56
  • $\begingroup$ @mathemagician99 No, it's well-known that for this part Hausdorffness is not used on domain nor codomain. $\endgroup$ May 15, 2020 at 21:27
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For the second part (Brian already handled part 1): we need that $Y$ is compactly generated (or $k$-space, definitions are subtle here), not exactly local compactness:

$Y$ is called compactly generated iff $C \subseteq Y$ is closed iff $C \cap K$ is closed in $K$ (in the subspace topology) for all compact $K \subseteq Y$. Fact: all locally compact Hausdorff spaces are compactly generated, (and BTW also all first countable spaces (a large class, that includes all metric spaces, e.g.)).

Now let $f$ be proper, and $C \subseteq X$ be closed. To see $f[C]$ is closed in $Y$ we let $K \subseteq Y$ be an arbritary compact subspace of $Y$. Then note that $$f[C] \cap K = f[f^{-1}[K] \cap C]$$ and $f^{-1}[K] \cap C$ is compact in $X$ (as $f$ is proper and $C$ is closed and a closed subspace of a compact subset remains compact, and so its image under $f$ is also compact (and hence closed in $K$, as $Y$ is Hausdorff (!)). So $Y$ being compactly generated (and $K$ being arbitrary) implies that indeed $f[C]$ is closed and $f$ is a closed map.

That a locally compact Hausdorff space is indeed compactly generated is not too hard to see: if $C$ is not closed, then let $p \in \overline{C}\setminus C$. Then $p$ has a compact neighbourhood $K_p$ and then $K_p \cap C$ is also not closed in $K_p$, as $p$ is still in its closure, but not in it. So that proves the right to left implication in the definition (and the left to right holds in all spaces).

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  • $\begingroup$ Could you please explain again why $f^{-1}(K)\cap C$ has to be closed and why $f(C)\cap K$ being compact implies that it is closed. You mentioned that it has to do with Hausdorff spaces. $\endgroup$ May 16, 2020 at 8:47
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    $\begingroup$ @mathemagician99 $f^{-1}[K]$ is compact, and $C \cap f^{-1}[K]$ is a closed subset of a compact set so compact (always true). So $f[C]\cap K$ is compact and thus closed (as $Y$ is Hausdorff ). Thm: if $C \subseteq X$ is compact and $X$ is Hausdorff then $C$ is closed in $X$. $\endgroup$ May 16, 2020 at 8:48

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