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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...

user850424
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  • Yes they form a commutative diagram – user850424 Apr 10 '21 at 11:59
  • If these were actual homeomorphisms we would have that $q\circ f = p$ giving that $f^{-1} \circ q^{-1} = p^{-1}$ so $f^{-1} = p^{-1} \circ q$. Is there something we can do similar to this in the case of a local homeomorphism? – memerson Apr 10 '21 at 12:12

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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 $f(U^*)$.

Watanabe
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    Your definition of a local homeomorphism is almost right, but you should add the requirement that $f(U)$ is open in $Y$ (see here). For instance, the inclusion map of a line into the plane satisfies your condition but is not a local homeomorphism. – Moishe Kohan Apr 10 '21 at 14:26
  • Okay thanks. I changed the answer accordingly. – Watanabe Apr 10 '21 at 15:55