# Characterization of proper and closed functions on Hausdorff spaces

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.

• 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. May 15, 2020 at 19:29
• 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. May 15, 2020 at 19:31
• 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 $f^{-1}(V) \subset U_1 \cup \ldots \cup U_k$. May 15, 2020 at 19:37
• 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. May 15, 2020 at 19:39

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.

• I see. But did you actually use the Hausdorff property? May 15, 2020 at 19:50
• @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. May 15, 2020 at 19:56
• @mathemagician99 No, it's well-known that for this part Hausdorffness is not used on domain nor codomain. May 15, 2020 at 21:27

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).

• 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. May 16, 2020 at 8:47
• @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 $X$. May 16, 2020 at 8:48