# Characterization of proper maps using filters

A continuous function $f \colon X \to Y$ is called proper if it is closed and if $f^{-1}(\{y\})$ is compact for every $y \in Y$. I want to prove that a continuous function $f \colon X \to Y$ is proper if and only if for every ultrafilter $\mathcal U$ on $X$ and every point $y \in Y$ such that $f[\mathcal U]$ converges to $y$ (here, $f[\mathcal U]$ is the filter generated by sets of the form $f(U)$, $U \in \mathcal U$), there is an $x \in X$ such that $\mathcal U$ converges to $x$ and $f(x) = y$.

Here is what I have been able to argue so far: Suppose $f$ is proper. Pick an ultrafilter $\mathcal U$ on $X$ and a $y \in Y$ such that $f[\mathcal U]$ converges to $y$. Since $f$ is proper, $A := f^{-1}(\{y\})$ is compact. Since $\mathcal U$ is an ultrafilter, it either contains $A$ or it doesn't. In the form case, $\mathcal U$ converges to some $x \in A$ and we're done. In the latter case, $\mathcal U$ contains some set $B$ such that $A \cap B$ is empty. I do not know how to proceed from here. Any tips? I have not attempted the reverse implication.

Edit: In the argument above, I meant to say that either $A \in \mathcal U$ or $A' := X \setminus A$ is in $\mathcal U$. In this latter case, $f(A')$ is in $f[\mathcal U]$. Since $f[\mathcal U]$ converges to $y$, the closure of every set in $f[\mathcal U]$ contains $y$, which means $y \in \overline{f(A')}$. Since $f$ is closed, $\overline{f(A')}=f(\overline{A'})$. Thus, there is an $x$ in the closure of $A'$ such that $f(x) = y$. Now I just need to argue that $\mathcal U$ converges to $x$. Suppose it does not. Then $\mathcal U$ contains the compliment $U'$ of some open set $U$ containing $x$, which means that $f(U')$ is contained in $f[\mathcal U]$, and since $f(U)$ is in $f[\mathcal U]$, $f(U) \cap f(U') \in f[\mathcal U]$. I'm not sure where to go from here...

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"Since $\mathcal U$ is an ultrafilter, it either contains $A$ or it doesn't." Uhh, what? You don't need to be an ultrafilter for that. :P Either $A \in \mathcal U$ or $A \notin \mathcal U$ whether $\mathcal U$ is an ultrafilter or not. What did you want to say exactly? – Patrick Da Silva May 23 '13 at 20:40
@PatrickDaSilva He probably meant $X \setminus A \in \mathcal{U}$? – Henno Brandsma May 23 '13 at 20:47
We should use that the image filter of $\mathcal{U}$ converges to $y$... – Henno Brandsma May 23 '13 at 20:47
@Henno : Probably, yes... – Patrick Da Silva May 23 '13 at 20:50
A useful characterisation of closed maps, which is in this flavour: $f$ is closed iff for every $y \in Y$ and every open set $O$ that contains $f^{-1}[\{y\}]$, there is an open subset $U$ that contains $y$ and such that $f^{-1}[U] \subset O$; I learned this as closed maps are "reversely" continuous... – Henno Brandsma May 23 '13 at 20:52

Suppose that $\mathscr{U}$ does not converge to any point of $A$. Then each $x\in A$ has an open nbhd $V_x$ such that $V_x\notin\mathscr{U}$. $A$ is compact, so there is a finite $F\subseteq A$ such that $\{V_x:x\in F\}$ covers $A$. Let $V=\bigcup_{x\in F}V_x$; then $V$ is an open nbhd of $A$, and $V\notin\mathscr{U}$.

Since $f$ is closed, $f[X\setminus V]$ is closed in $Y$; let $W=Y\setminus f[X\setminus V]$. Then $W$ is an open nbhd of $y$. And $f[\mathscr{U}]\to y$, so $W\in f[\mathscr{U}]$. Thus, there is some $U\in\mathscr{U}$ such that $f[U]\subseteq W$. And this leads immediately to a contradiction; how?

For the converse, let $f$ be a continuous map with the stated property. Suppose that there is a $y\in Y$ such that $A=f^{-1}[\{y\}]$ is not compact; then there is a non-convergent ultrafilter $\mathscr{U}$ on $X$ such that $A\in\mathscr{U}$. Clearly $y\in f[U]$ for each $U\in\mathscr{U}$, so $f[\mathscr{U}]\to y$ and by hypothesis there is an $x\in A$ such that $\mathscr{U}\to x$, contradicting the choice of $\mathscr{U}$. Thus, $f$ has compact fibres, and it only remains to show that $f$ is closed. $\hspace{10 in}$ Suppose that $F\subseteq X$ is closed, but $A=f[F]$ is not closed in $Y$. Fix $y\in\operatorname{cl}A\setminus A$, and let $\mathscr{A}$ be an ultrafilter on $A$ converging to $y$. Let $\mathscr{F}=\{f^{-1}[S]:S\in\mathscr{A}\}$; $\mathscr{F}$ is a filter base on $F$. Extend $\mathscr{F}$ to an ultrafilter $\mathscr{U}$ on $X$. Let $N$ be a nbhd of $y$ in $Y$. Then $N\cap A\in\mathscr{A}$, $f^{-1}[N]\in\mathscr{F}\subseteq\mathscr{U}$, and $N\supseteq f\big[f^{-1}[N]\big]\in f[\mathscr{U}]$, so $N\in f[\mathscr{U}]$. Thus, $f[\mathscr{U}]\to y$. But $F\in\mathscr{U}$, and $F$ is closed, so any limit of $\mathscr{U}$ must be in $F$ and therefore is not sent to $y$ by $f$. This contradiction shows that $A$ must be closed and hence that $f$ is a closed map.

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I don't see the contradiction. $f(U) \subseteq W$ implies that $U \subseteq f^{-1}(W)$ and I know that $f^{-1}(W)$ is open and contains $A$. – echoone May 25 '13 at 0:09
@echoone: Look carefully at the definition of $W$: $U\subseteq f^{-1}[W]\subseteq V$. But $U\in\mathscr{U}$, so $V\in\mathscr{U}$, a contradiction. – Brian M. Scott May 25 '13 at 0:18
Of course! Thanks. – echoone May 25 '13 at 17:31
@echoone: You’re welcome. – Brian M. Scott May 25 '13 at 18:04

$\Rightarrow)$ Suppose $y\in Y$ is arbitrary and $C=f^{-1}(\{y\})$. Let $\mathcal U$ be an ultrafilter on $X$ with $f(\mathcal U)\to y$. then $$y\in \bigcap_{U\in \mathcal U }\overline{f(U)}\subseteq \bigcap_{U\in \mathcal U }\overline{f(\overline U)}\subseteq \bigcap_{U\in \mathcal U }{f(\overline U)}$$ So $$(\forall U\in \mathcal U)(\overline{U}\cap C\ne \emptyset)$$ Let $\mathcal F$ be the filter generated by $\{\overline{U}\cap C \mid U\in \mathcal U\}$ on $C$. Since $C$ is compact $\mathcal F$ has some limit(=cluster) point $c\in C$; that is $$c\in \bigcap_{F\in \mathcal F}\overline{F}\cap C\subseteq \bigcap_{U\in \mathcal U}\overline{\overline{U}\cap C}\subseteq \bigcap_{U\in \mathcal U}\overline{U}$$ and so $c$ is a limit point of $\mathcal U$. Since $\mathcal U$ is an ultrafilter: $$\mathcal U\to c$$ and so $$f(\mathcal U)\to f(c)=y$$

$\Leftarrow)$ Suppose $y\in Y$ is arbitrary and $C=f^{-1}(\{y\})$. Suppose $\mathcal F$ is a filter on $C$. It is enough to prove $\mathcal F$ has limit point in space $C$. $\mathcal F$ is a base for a filter $\mathcal G$ on $X$. $\mathcal G$ is contained in some ultrafilter $\mathcal U$. We have

$$f(\mathcal F) \to y$$

So

$$f(\mathcal U) \to y$$ By assumption there's some $c\in C$ with $$\mathcal U\to c$$ which implies $c$ is a limit point of $\mathcal F$ in space $X$ and so in space $C$.

closedness of $f$ is not proved in $\Leftarrow$. If anything is wrong let me know.

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Why would $C$ intersect all members of $\mathcal{U}$? Where do you use that $f[\mathcal{U}] \rightarrow y$, which is the assumption we start with? – Henno Brandsma May 24 '13 at 3:39
In the beginning, you use $A$ when I think you mean $C$. Also, it is not clear to me what $\mathcal U_C$ is. – echoone May 24 '13 at 3:43
Nevermind. I see that $\mathcal U_C$ is the trace of $\mathcal U$ on $C$. But it is not clear that the trace would exist to begin with. – echoone May 24 '13 at 3:55