# If $g \circ f$ and $f$ are covering maps, then so is $g$

Apologies for the long winded question. Recently I had a topology midterm which stated the following: If $g \circ f: X \to Z$ is a cover and $f: X \to Y$ is a cover, then prove that $g: Y \to Z$ is a cover. There were no assumptions on the spaces $X,Y,Z$, we were supposed to find the loosest conditions possible so that the above statement was true. A few weeks after the midterm, and much discussion with my classmates, we have not been able to find a solution yet. My idea for the proof is as follows: I will assume all spaces are locally connected. Let $p=g \circ f$ and fix $z \in Z$. Let $U$ be a connected fundamental neighborhood with respect to $p$, i.e. we may write $p^{-1}(U)$ as a collection of disjoint open sets each homeomorphic to $U$. Then, let $\{U_\alpha \}$ be the collection of connected components of $g^{-1}(U)$ these are open by the locally connected hypothesis. Let $p^{-1}(U)= \{W_\beta\}$. By definition, we know that $f(W_\beta) \subset U_\alpha$ for one such $\alpha$. This is because each $W_\beta$ is connected, being locally homeomorphic to $U$ and thus can map into at most one $U_\alpha$. Now, $f^{-1}(U_\alpha)$ must be a subcollection of $W_\beta$, call this collection $\{V_{\beta,\alpha}\}$. I claimed that $$f: \cup_\alpha V_{\beta,\alpha} \to U_\alpha$$ is a covering map. However, my instructor pointed out that this map may not be surjective in general. If it is then I believe we can conclude that $f: V_{\beta,\alpha} \to U_\alpha$ is a homeomorphism, which would complete the proof because $p: V_{\beta,\alpha} \to U$ is a homeomorphism. Hence $g$ must be too. My question is, does anyone know if we can prove that $f: V_{\beta,\alpha} \to U_\alpha$ is surjective here? Or is my approach wrong? Any solutions would be appreciated because so far, no solutions have been posted to the midterm and I would like to figure this out.

This is not true in general, but it is true if you assume that $f$ is surjective.

Consider $Y=X\coprod U$, the disjoint union of $X$ and $U$ and $Z=X$, take $f:X\rightarrow Y$ the canonical imbedding $h:Y\rightarrow X$ any map and $g_{\mid X}=Id_X, g_{\mid Y}=h$. $g\circ f=Id_X$ and $f$ are coverings, but $g$ is not always a covering.

• Okay, if we take in our definition of cover that a covering map is surjective (i.e. we may assume f is surjective) then how does the proof follow? – Sheel Stueber May 25 '18 at 21:15