Factorisation of covering maps

Question

It’s a well-known fact that for continuous functions $f:X\to Y$ and $g:Y\to Z$ of locally connected spaces, if $g\circ f$ and $g$ are covering maps, so is $f$. I was wondering if the following statement is also true: If $g\circ f$ and $f$ are covering maps, so is $g$. I can’t find a proper counterexample and am not able to proof it. Any ideas?

Answer 1

If $X$, $Y$ and $Z$ are path connected and locally path connected, under your assumptions, the map $g$ is indeed a covering map. If you only require the spaces to be locally connected, I do not know whether the result still holds.

Below, following Munkres' Topology (Second Edition, Lemma 80.2b, page 485), I give a detailed proof of this result. The main idea of the proof is to take for a given point $z\in Z$ a path connected neighborhood $U$ of $z$ which is evenly covered by $p$ and then write the preimage $g^{-1}(U)$ as a disjoint union of its path components.

Before I present the proof, a quick note on terminology: To me, following Munkres, a covering map is always surjective.

First off, since $p$ is surjective and $p=g\circ f$ holds, the map $g$ is surjective.

Now, let $z\in Z$. Find a neighborhood $U$ of $z$ that is evenly covered by $p$. This exists, since $p$ is a covering map. Without loss of generality we can assume that $U$ is path connected. Namely, if it is not, find a path connected neighborhood $V$ of $z$, which exists since $Z$ is locally path connected. Then $U\cap V$ is a locally path connected neighborhood of $z$. In other words, there exists a path connected neighborhood $U'\subseteq U\cap V$ of $z$. Any such neighboorhood $U'$ is still evenly covered by $p$ and we can consider $U'$ instead of $U$.

Now write $g^{-1}(U)=\coprod V_\beta$, where the $V_\beta$ are the path components of $g^{-1}(U)$. To show that $U$ is evenly covered by $g$, it suffices to show that the restrictions $g_{\restriction{V_\beta}}\colon V_\beta \rightarrow U$ are homeomorphisms.

To see this, first consider the sheets $\{U_\alpha\}$ of $p^{-1}(U)$. They are open and path connected, since the restrictions $p_{\restriction{U_\alpha}}\colon U_\alpha\rightarrow U$ are homeomorphisms. By definition, they are also disjoint. Thus, the sheets are precisely the path components of $p^{-1}(U)$. (If they weren't, we could write a path component of $p^{-1}(U)$ as a disjoint union of non-empty open sets. This would contradict the the connectedness of path components.) Since $f$ is continuous and every $U_\alpha$ is path connected, for every $\alpha$ we have $f(U_\alpha)\subseteq V_\beta$ for some $\beta$. Said differently, for every $\beta$, the set $f^{-1}(V_\beta)$ is equal to the union of a subcollection of the $U_\alpha$. Now, fix $\beta$ and write $f^{-1}(V_\beta)=\bigcup_{i\in I}U_{\alpha_i}$.

At this point, Munkres quotes two results. One result, his Theorem 53.2, gives that the restriction $f_1\colon \bigcup_{i\in I}U_{\alpha_i} \rightarrow V_\beta$ of $f$ is a covering map. We fix $\alpha_0 \in \{\alpha_i \ \vert \ i\in I\}$. The other result, his Lemma 80.1, implies that the restriction $f_0\colon U_{\alpha_0}\rightarrow V_\beta$ is a covering map. Therefore, $f_0$ is continuous, open, and surjective. Next, consider the restrictions $p_0\colon U_{\alpha_0}\rightarrow U$ of $p$ and $g_0\colon V_\beta\rightarrow U$ of $g$. Since $p=g\circ f$, we have $p_0=g_0\circ f_0$. Since $U$ is evenly covered by $p$ and $U_{\alpha_0}$ is a sheet of $p$, we know that $p_0$ is a homeomorphism. Thus, $f_0$ is injective. In other words, $f_0$ is a homeomorphism. With $p_0=g_0\circ f_0$, we conclude that $g_0$ is also a homeomorphism. q.e.d.