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Let $p:Y\to X$ and $q:Z\to X$ be covering maps (of course $X,Y,Z$ are all Hausdorff, arcwise connected and locally arcwise connected) and $g:Y\to Z$ a morphism such that $q\circ g=p$. Then, $g$ is a covering map.

First: I know this question was asked and answered here, but I am having trouble with one final piece that was not addressed in any of the answers.

I'll sketch the proof as I have it thus far:

Let $z\in Z$ and $U$ a neighborhood of $q(z)$ in $X$ that is evenly covered by $p$ and $q$, say $p^{-1}(U)=U\times F$ and $q^{-1}(U)=U\times E$ (where $E$ and $F$ are discrete spaces). Let $e_0\in E$ such that $z\in U\times\{e_0\}$. Then, $g^{-1}(U\times\{e_0\})=p^{-1}q(U\times\{e_0\})=p^{-1}(U)=U\times F$. Further, for each $f\in F$, $(q|_{U\times\{e_0\}})^{-1}\circ p|_{U\times\{f\}}$ is a homeomorphism from $U\times\{f\}$ to $U\times\{e_0\}$. The problem is that $g|_{U\times\{f\}}(U\times\{f\})=q^{-1}\circ p|_{U\times\{f\}}(U\times \{f\})=q^{-1}(U)=U\times E$.

I don't see how $g|_{U\times\{f\}}$ is a homeomorphism, i.e. how $g|_{U\times\{f\}}=(q|_{U\times\{e_0\}})^{-1}\circ p|_{U\times\{f\}}$.

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1 Answer 1

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Let $z \in Z$ and $U$ a neighborhood of $q(z)$ in $X$ such that $p^{-1}(U) = U \times F \subseteq Y$ and $q^{-1}(U) = U \times E \subseteq Z.$ Since $z \in q^{-1}(U)$ there exists $e_0$ such that $z \in U \times \{e_0\}$. Let $V \subset Z$ denote this open subset. You want to prove $V$ is the neighborhood evenly covered by $g$.

You wrote: $g^{-1}(U \times \{e_0\}) = U \times F$ which is not always true. What you want to do is find a set $J \subseteq F$ such that $g^{-1}(U \times \{e_0\}) = U \times J$.

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