Let $p \colon X \rightarrow Y$ be a perfect map, and let $Y$ be compact. Show that $X$ is compact.

Question

Let $p \colon X \to Y$ be a closed continuous map such that $p^{-1} ( \{y \} )$ is compact, for each $y \in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact. [Hint: If $U$ is an open set containing $p^{-1} ( \{ y \} )$, there is a neighborhood $W$ of $y$ such that $p^{-1} (W)$ is contained in $U$.

I know that this exact question has been asked on this site, but I would like some guidance and hints to approach to solve it on my own. Here's my approach so far.

Let $X = \cup_\alpha U_\alpha$, with $U_\alpha$ open. Now, for $p^{-1}(\{y\}) \subseteq U_\alpha$, we may find $W_y$ such that $p^{-1}(W_y)$ is contained in $U_\alpha$. Now, $Y = \cup_{y \in Y} W_y$. So we get a finite subcover $W_{y_1} \ldots W_{y_n}$. So get $U_{\alpha_i}$ that contains $W_{y_i}$. As $\cup_i p^{-1}(W_{y_i}) = \cup_i p^{-1}(\cup W_{y_i})$, $\cup_i U_{\alpha_i} = X$ as desired.

Now, this proof doesn't seem to use many of the assumptions, so it can't be right.

Any help would be appreciated, thanks!

Answer 1

First, if the map is not surjective, how can you be sure to be able to even say $p^{-1}(\{y\}) \subseteq U_{\alpha}$? You are considering as a trivial step the fact that we can find $W_y$ such that $p^{-1}(W_y)$ is contained in $U_{\alpha}$, while here you are using all the needed assumptions.

Look at this: Perfect map: If $Y$ is compact, then $X$ is compact?