perfect map in topology

5. Let $f : X \rightarrow Y$ be a perfect map: a continuous surjective closed map such that $f^{−1}(\{y\})$ is compact for all $y \in Y$ . Prove that if $Y$ is compact, then $X$ is compact.

How to use compactness on $Y$ to show it on $X$.

Thank you for any help.

Solution 1:

I'll give you a start plus proof sketch:

Suppose that $\mathcal{U} = \{U_i : i \in I\}$ is an open cover of $X$.

For every $y$ in $Y$, $F_y := f^{-1}[\{y\}]$ is compact so there is a finite subset $I_y \subset I$ such that $$F_y \subset U_y:=\bigcup_{i \in I_y} U_i$$

For every $U_y$, define $V_y= Y\setminus f[X\setminus U_y]$.

• Check that $y \in V_y$ and all $V_y$ are open in $Y$.
• Now exploit a finite subcover for the $V_y$ in $Y$ to find one for $\mathcal{U}$.