Continuous map between topological spaces, with two local homeomorphisms is a local homeomorphism Let $X,Y,Z$ be topological spaces. Let $f:X\rightarrow Y$ be a contionuous map. I also have two local homeomorphisms $p:X\rightarrow Z$ and $q:Y\rightarrow Z$. Such that these functions form a commutative diagram.
I need help to show that $f$ is also a local homeomorphism...
I tried to show it by taking an elemet $y\in Y$ and an open neighbourhood of $y$ named $U$ such that $U$ is homeomorphic to $q(U)$ and taking the inverse image of  $U$ by $f$..... I have no idea what to do with this...

Assuming that the definition of local homeomorphism is: for all $x\in X$ there exists an open neighbourhood $U$ of $x$ such that $f:U\to f(U)$ is an homeomorphism and $f(U)$ is open.
Let $x\in X$ be a point and $U\subset X$ be the open neighbourhood of $x$ such that $p:U\to p(U)$ is an homeomorphism.
Analogously, for $f(x)\in Y$, there exists a neighbourhood $V\subset Y$ of $f(x)$ such that $q:V\to q(V)$ is an homeomorphism.
Restrict $U$ to a open subset $U^*\subset U$ such that $f(U^*)\subset V$, namely $U^*:=U\cap f^{-1}(V)$ and note that this is still a neighbourhood of $x$ inside $U$. Then
$$q:f(U^*)\to q(f(U^*))\qquad \& \qquad p:U^*\to p(U^*)$$ are still homeomorphisms. By the commutativity, $q(f(U^*))=p(U^*)$ and $f:U^*\to f(U^*)$ is exactly the composition of two homeomorphisms $f=q^{-1}\circ p$. By the fact that an homeomorphism is an open map, then $p(U^*)$ is open and so is $f(U^*)$.
