In this question I was told to prove the following: let $X,Y,Z$ be topological spaces, $g:Y\to Z$ and $h:X\to Z$ local homeomorphisms and assume that $f:X\to Y$ continuous satisfies $h=g\circ f$. Then $f$ is a local homeomorphism. So I set out to prove the following: for all $x_{0}\in X$ there exists an open neighbourhood $U\subseteq X$ of $x_{0}$ such that $f(U)\subseteq Y$ is open and $f\big|_{U}:U\to f(U)$ is a homeomorphism.
It turned out that my direct approach got a bit confusing with choices of subsets at many stages. I would be very grateful if you could give me your opinion on this. Thank you.
Proof: Fix $x_{0}\in X$. Let $U_{0}$ be a neighbourhood of $x_{0}$ such that $h\big|_{U_{0}}$ is a homeomorphism onto the image $h(U_{0})$, where $h(U_{0})$ is open in $Z$. By assumption we can choose an open neighbourhood $V\subseteq Y$ of $y_{0}:=f(x_{0})$ such that $g\big|_{V}$ is a homeomorphism onto the image and $W:=g(V)\subseteq h(U_{0})$ and also $W$ is open in $Z$. Finally we can use continuity of $f$ in order to choose an open neighbourhood $U\subseteq U_{0}$ of $x_{0}$ such that $f(U)\subseteq V_{0}$. We claim that $f(U)$ is open in $Y$ and $f\big|_{U}$ is a homeomorphism onto the image.
Let $O:=h(U)$ and define $\tilde{h}:U\to O$, $\tilde{f}:U\to V$ and $\tilde{g}:V\to W$ by restriction, i.e. $\tilde{h}(x):=h(x)$, $\tilde{f}(x)=f(x)$ and $\tilde{g}(y):=g(y)$ for all $x\in U$ and $y\in V$ respectively. Note that $\tilde{h}$ is a homeomorphism. As $h=g\circ f$, we have $O\subseteq W$. Hence $\tilde{g}^{-1}:O\to V$ is well defined and indeed $f(U)=\tilde{f}(U)=\tilde{g}^{-1}\circ\tilde{h}(U)$. As $h(U)\subseteq Z$ is open and $W\subseteq Z$ is open, $O\subseteq Z$ is open and hence $f(U)\subseteq V$ is open, implying that $f(U)$ is open in $Y$. Clearly $\tilde{g}^{-1}\circ\tilde{h}:U\to f(U)$ is invertible with continuous inverse by construction, thus indeed $f\big|_{U}$ is a homeomorphism onto the image.
