# Closed, continuous, surjective map with inverse images compact

Old qual question:

Let $p:X\to Y$ be a closed, continuous, surjective map such that $p^{-1}(y)$ is compact for every $y\in Y$. Let $(U_\alpha)_{\alpha\in A}$ be an open cover of $X$. Show that any $y\in Y$ has an open neighborhood $Y_y$ such that $p^{-1}(Y_y)$ has a finite subcover of $(U_\alpha)_{\alpha\in A}$.

I don't have much of an attempt. So far, let $y\in Y$ be arbitrary. Then there is some $x\in p^{-1}(y)$, which is compact and covered by $(U_\alpha)$ so there is a finite subcover $U_1,\dots,U_n$. That's really all I have.

Let $y$ be in $Y$, since $p^{-1}(y)$ is compact, there exists $U_{i_j},j=1,...,n$ such that $(U_{i_j})\subset (U_{\alpha})$ and $p^{-1}(y)\subset\cup_{j=1}^{j=n}U_{i_j}$. Let $V$ be the complementary space of $\cup_{j=1}^{j=n}U_{i_j}$, $V$ is closed, since $p$ is a closed map, we deduce that $p(V)$ is a closed subset which does not contain $y$ since $p^{-1}(y)\subset\cup_{j=1}^{j=n}U_{i_j}$, this implies that $W=Y-p(V)\subset p(\cup_{j=1}^{j=n}U_{ij})$ is open, contains $y$ and $p^{-1}(W)\subset \cup_{j=1}^{j=n}U_{i_j}$.
• It is just a notation, perhaps $\alpha_j$ was better and I replaced $\alpha$ by $i$ Jan 8 '16 at 1:48