Proper in one argument implies total properness?

Question

Let $X$, $Y$, $Z$ be locally compact Hausdorff spaces. Furthermore, $X$ is compact. We have a continuous map $$f : X \times Y \to Z$$ satisfying that for every fixed $x \in X$ the map $$f_x\colon Y \to Z , \quad y \mapsto f(x,y)$$ is proper. Does this imply that $f$ itself is proper?

Answer 1

No. It need not be that $f$ is proper.

Consider the map $f:\mathbb{R}\times [0,1]\rightarrow \mathbb{R}$ given by $$f(x,t)=tx^2+x.$$ This a homotopy of proper maps $\mathbb{R}\rightarrow\mathbb{R}$. However $f$ is not a proper map. To see this observe that the sequence $(-n,1/n)$ is unbounded in the domain but satisfies $f(-n,1/n)=0$. Thus $f^{-1}(0)$ is not compact.

I learned of this counterexample from the introduction of Thomas Rot's paper Homotopy classes of proper maps out of vector bundles (arxiv).

Answer 2

This problem is a pretty hard nut. I came so close to a solution with some theorems I've never heard before, but something is still missing. The proof works for finite $X$ though. Here is what I have so far:

Continuous maps into locally compact spaces are proper iff closed (See nLab, Proposition 4.1.). It is therefore sufficient to prove, that $f$ is closed. Let $x\in X$. Since $X$ is a Hausdorff space, the singleton subset $\{x\}\subseteq X$ is closed. Because of the definition of the product topology, $\{x\}\times Y=\pi_X^{-1}(\{x\})$ is closed, where $\pi_X\colon X\times Y\rightarrow X,(x,y)\mapsto x$ is the canonical projection.

Let $A\subseteq X\times Y$ be closed, then $A\cap(\{x\}\times Y)$ is closed. Let $A_x=\pi_Y(A\cap(\{x\}\times Y))\subseteq Y$ be the slice of $A$ through $\{x\}$. Since $X$ is compact, the projection map $\pi_Y\colon X\times Y\rightarrow Y$ is closed (See here, but with $X$ and $Y$ switched.) and therefore $A_x$ is closed (and locally compact since the superset $Y$ is). We have: \begin{equation} A=\bigcup_{x\in X}A\cap(\{x\}\times Y) =\bigcup_{x\in X}\{x\}\times A_x. \end{equation} Since $f_x$ is proper, it is closed. Therefore $f_x(A_x)$ is closed (and locally compact since the superset $Y$ is) and we have: \begin{equation} f(A)=\bigcup_{x\in X}f(\{x\}\times A_x) =\bigcup_{x\in X}f_x(A_x). \end{equation} If $X$ is finite, $f(A)$ would be closed as the finite union of closed sets and we would already be finished, so assume it's not. $f_x(A_x)^\complement$ is open and also dense in $Y$, since: \begin{equation} \overline{f_x(A_x)^\complement} =(f_x(A_x)^\circ)^\complement =(f(\{x\}\times A_x)^\circ)^\complement \supseteq f((\{x\}\times A_x)^\circ)^\complement =f(\{x\}^\circ\times A_x^\circ)^\complement =\emptyset^\complement =Y. \end{equation} Since $Y$ is a locally compact Hausdorff space, the open and dense subset $f_x(A_x)^\complement$ is locally compact (See here, Corollary 1.10.). Since $f$ is continuous, $f_x(A_x)$ is locally compact (See here).

(Arbitrary) intersections of locally compact subsets in Hausdorff spaces are locally compact (using the definition that every open neighborhood is the superset of a compact neighborhood), since the compact neighborhoods are all closed (due to the superset being Hausdorff), their (arbitrary) intersection being closed and therefore compact (since being included in all compact neighborhoods).

Therefore $f(A)=\bigcap_{x\in X}f_x(A_x)^\complement$ is locally compact and can be written as $f(A)=B\cap C$ with $B,C\subseteq Z$, where $B$ is closed and $C$ is open (See here, Proposition 1.9.). Well, the $C$ ruins it.

Answer 3

I present here a proof I came up with which works in case closedness can be described using converging sequences. I guess it won't be a problem to generalize this to the general case using nets instead of sequences.

By what Samuel Adrian Antz pointed out we have to show that $f$ maps closed subsets to closed subsets. Let $C \subset X \times Y$ be closed and let $((x_n,y_n))_n \in C$ be a sequence such that $(f(x_n,y_n))_n$ converges to a point $z \in Z$. We have to show $z \in f(C)$.

Since $X$ is compact, we find a subsequence of $(x_n)_n$ which converges in $X$, lets say to a point $x$. Without loss of generality $(x_n)_n$ converges to $x$.

Because $f$ is continuous the sequence $(f(x,y_n))_n$ converges to $z$. Let $U$ be a compact neighborhood of $z$. Then $f_x^{-1}(U)$ is compact and contains almost all $y_n$. In particular the limit $y \in Y$ of a subsequence of $(y_n)_n$ is contained in $f_x^{-1}(U)$. Without loss of generality $(y_n)_n$ converges to $y$.

Hence, we have shown that $((x_n,y_n))_n$ converges to $(x,y)$. Because $C$ is closed we have $(x,y)\in C$ which proves $z \in f(C)$.

I hope this proof is correct. Any comments?