Suppose we have pullback diagram of topological spaces:

pullback diagram

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.


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 $$\require{AMScd} \begin{CD} \bigcup_{k\in \mathbb{Z}} \left[2\pi k, 2\pi k+\frac{\pi}{2}\right] @>>> \mathbb{R}\\ @VVV @VV^{e^{it}}V \\ \left[0,\frac{\pi}{2}\right] @>>{e^{it}}> S^{1}\\ \end{CD} $$


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