Proper map

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

Let ${\displaystyle \left\{U_{a}:a\in A\right\}}$  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 every ${\displaystyle k\in K,}$  there exists a finite subset ${\displaystyle \gamma _{k}\subseteq A}$  such that ${\displaystyle f^{-1}(k)\subseteq \cup _{a\in \gamma _{k}}U_{a}.}$  The set ${\displaystyle X\setminus \cup _{a\in \gamma _{k}}U_{a}}$  is closed in ${\displaystyle X}$  and its image under ${\displaystyle f}$  is closed in ${\displaystyle Y}$  because ${\displaystyle f}$  is a closed map. Hence the set $${\displaystyle V_{k}=Y\setminus f\left(X\setminus \cup _{a\in \gamma _{k}}U_{a}\right)}$$  is open in ${\displaystyle Y.}$  It follows that ${\displaystyle V_{k}}$  contains the point ${\displaystyle k.}$  Now ${\displaystyle K\subseteq \cup _{k\in K}V_{k}}$  and because ${\displaystyle K}$  is assumed to be compact, there are finitely many points ${\displaystyle k_{1},\dots ,k_{s}}$  such that ${\displaystyle K\subseteq \cup _{i=1}^{s}V_{k_{i}}.}$  Furthermore, the set ${\displaystyle \Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}}$  is a finite union of finite sets, which makes ${\displaystyle \Gamma }$  a finite set.

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