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...
 A: 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?
Added: You didn’t ask about it, but I’ve added a spoiler-protected proof of the converse.

 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.

A: $\Rightarrow)$ 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.
