Suppose we have pullback diagram of topological spaces:
I want to prove:
If $g:Y\to Z$ is a local homeomorphism (etale map), then $p_1:P\to X$ is a local homeomorphism.
My idea: First of all $P=\{(x,y)\in X\times Y\mid f(x)=g(y)\}$. For $(x,y)\in P$ there exists an open set $y\in V\subset Y$ such that $g(V)$ is open and $g|_V: V\to g(V)$ is a homeomorphism. Consider the open set $(x,y)\in p_2^{-1}(V)\subset P$, which we define as $W:=p_2^{-1}(V)$.
One checks that $p_1(W)=f^{-1}(g(V))$ so $p_1(W)$ is open (using the properties of local homeomorphism $g$). The map $p_1|_{W}: W\to p_1(W)$ is clearly surjective. Injectivity is also easily verified using the commutativity of the diagram.
My problem is how to show $p_1|W: W\to p_1(W)$ is open. Actually I managed to show that $p_1|W$ is open if $p_2|W$ is open. But showing $p_2|W$ is open has proven to be difficult. I have a feeling I'm missing something obvious...
So far you have a commutative diagram \begin{array}{ccc} W & \rightarrow & V \\ \downarrow & & \downarrow \\ X & \rightarrow & Z \\ \end{array} where $V \to Z$ is the inclusion of an open set. Now, it's easy to verify that this is actually a pullback square, i.e. $W$ is the fiber product $V \times_X Z$. On the other hand, since $V \to Z$ is an inclusion, we can use the universal property the fiber product to show $V \times_X Z$ is $f^{-1}(V)$ and the map to $X$ is the inclusion. It follows easily now that $W \to X$ is open (in fact a homeomorphism onto its image) since $V$ is open.