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

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!

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?