Pullback of a local homeomorphism is a local homeomorphism 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...
 A: 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.
A: I filled in the details in CSLeek answer 
Let
$$\require{AMScd}
\begin{CD}
P @>>^{p_{2}}> Y\\
@VV^{p_{1}}V @VV^gV \\
X @>>^f> Z\\
\end{CD}
$$
be a pullback square and suppose $g:Y\to Z$ is a local homeomorphism
we want to show that $p_{1}$ is a local homeomorphism. Let $x\in P$
and $y=p_{2}\left(x\right)$. Because $g$ is a local homeomorphism
there exists an open neighborhood $\ U_{y}$ s.t. $y\in U_{y}$ , $g\left(U_{y}\right)$ is open and $g|_{U_{y}}:U_{y}\to g\left(U_{y}\right)$
is a homeomorphism. We apply the pullback lemma on the diagram 
$$\require{AMScd}
\begin{CD}
p^{-1}_{2}(U_{y})@>>>U_{y}\\
@VVV @VVV \\
P @>>> Y\\
@VVV @VVV \\
X @>>> Z\\
\end{CD}
$$
and conclude that 
$$\require{AMScd}
\begin{CD}
p^{-1}_{2}(U_{y})@>>>U_{y}\\
@VVV @VVV \\
X @>>> Z\\
\end{CD}
$$
 is a pullback square. We notice that $p_{2}^{-1}\left(U_{y}\right)\subset p_{2}^{-1}g^{-1}\left(g\left(U_{y}\right)\right)=p_{1}^{-1}f^{-1}\left(g\left(U_{y}\right)\right)$ 
so $p_{1}|_{p_{2}^{-1}\left(U_{y}\right)}:p_{2}^{-1}\left(U_{y}\right)\to f^{-1}g\left(U_{y}\right)$.
Now we apply the pullback lemma again on the diagram
$$\require{AMScd}
\begin{CD}
p^{-1}_{2}(U_{y})@>>>U_{y}\\
@VVV @VVV \\
f^{-1}g(U_{y}) @>>> g(U_{y})\\
@VVV @VVV \\
X @>>> Z\\
\end{CD}
$$
 and conclude that diagram
$$\require{AMScd}
\begin{CD}
p^{-1}_{2}(U_{y})@>>>U_{y}\\
@VVV @VVV \\
f^{-1}g(U_{y}) @>>> g(U_{y})\\
\end{CD}
$$
 is pullback square. Because isomorphisms are stable under pullbacks
we get that $p_{1}|_{p_{2}^{-1}\left(U_{y}\right)}:p_{2}^{-1}\left(U_{y}\right)\to f^{-1}g\left(U_{y}\right)$
is a homeomorphism this ends the proof. 
 Note as to the method suggested by Hamed. The map $p_{2}$ may not be an open map as in the pullback square 
$$\require{AMScd}
\begin{CD}
\bigcup_{k\in \mathbb{Z}} \left[2\pi k, 2\pi k+\frac{\pi}{2}\right]  @>>> \mathbb{R}\\
@VVV @VV^{e^{it}}V \\
\left[0,\frac{\pi}{2}\right] @>>{e^{it}}> S^{1}\\
\end{CD}
$$
