Proper map

Let ${\displaystyle f\colon X\to Y}$ be a closed map, such that ${\displaystyle f^{-1}(y)}$ is compact (in X) for all ${\displaystyle y\in Y}$. Let ${\displaystyle K}$ be a compact subset of ${\displaystyle Y}$. We will show that ${\displaystyle f^{-1}(K)}$ is compact.

Let ${\displaystyle \{U_{\lambda }\vert \lambda \ \in \ \Lambda \}}$ be an open cover of ${\displaystyle f^{-1}(K)}$. Then for all ${\displaystyle k\ \in K}$ this is also an open cover of ${\displaystyle f^{-1}(k)}$. Since the latter is assumed to be compact, it has a finite subcover. In other words, for all ${\displaystyle k\ \in K}$ there is a finite set ${\displaystyle \gamma _{k}\subset \Lambda }$ such that ${\displaystyle f^{-1}(k)\subset \cup _{\lambda \in \gamma _{k}}U_{\lambda }}$. The set ${\displaystyle X\setminus \cup _{\lambda \in \gamma _{k}}U_{\lambda }}$ is closed. Its image is closed in Y, because f is a closed map. Hence the set

${\displaystyle V_{k}=Y\setminus f(X\setminus \cup _{\lambda \in \gamma _{k}}U_{\lambda })}$ is open in Y. It is easy to check that ${\displaystyle V_{k}}$ contains the point ${\displaystyle k}$. Now ${\displaystyle K\subset \cup _{k\in K}V_{k}}$ and because K is assumed to be compact, there are finitely many points ${\displaystyle k_{1},\dots ,k_{s}}$ such that ${\displaystyle K\subset \cup _{i=1}^{s}V_{k_{i}}}$. Furthermore the set ${\displaystyle \Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}}$ is a finite union of finite sets, thus ${\displaystyle \Gamma }$ is finite.

Now it follows that ${\displaystyle f^{-1}(K)\subset f^{-1}(\cup _{i=1}^{s}V_{k_{i}})\subset \cup _{\lambda \in \Gamma }U_{\lambda }}$ and we have found a finite subcover of ${\displaystyle f^{-1}(K)}$, which completes the proof.