I know this question has been asked before, but I couldn't find a satisfying answer so I hope to find it now.
Let $p:X\to Y$ and $q:Y\to Z$ be maps such that $q$ and $q\circ p$ are covering map and $Z$ is locally path-connected. Then $p$ is a covering map, too. So, I choose $y\in Y$ and choose $U_y$ a neighbourhood of $y$ such that $q|_{U_y}:U_y\to q(U_y)$ is a homeomorphism. Then, the set $\{q(U_y)\colon y\in Y\}$ covers $Z$. We see that $(q\circ p)^{-1}(q(U_y))=p^{-1}(U_y)$. Now, I that I only have to show that $p|_{p^{-1}(U_y)}$ is a homeomorphism. How can I do this, and where do I need locally path-connectedness?
Best Answer
Let us first observe that Hatcher's definition of a covering map is somewhat unusual because he does not require surjectivity. Compare this with the definition in Wikipedia. We come back to this point later.
What you have to show is that
(1) $p^{-1}(U_y) = \bigcup_{\alpha \in A} V_\alpha$ with pairwise disjoint open $V_\alpha$ (which are called sheets over $U_y$) such that each $p_\alpha : V_\alpha \stackrel{p}{\rightarrow} U_y$ is a homeomorphism.
This has nothing to do with showing that $p|_{p^{-1}(U_y)}$ is a homeomorphism. In general the latter will be false. Moreover, your approach cannot work because you do not use fact that $q$ is a covering map; you only invoke that $q$ is a local homeomorphism.
To prove (1), let $y \in Y$ and $W$ be an open neighborhood of $q(y)$ which is evenly covered by $q$ and by $q \circ p$. That is, $q^{-1}(W) = \bigcup_{\beta \in B} U_\beta$ with pairwise disjoint open $U_\beta$ such that each $q_\beta : U_\beta \stackrel{q}{\rightarrow} W$ is a homeomorphism, and $(q \circ p)^{-1}(W) = \bigcup_{\gamma \in C} V_\gamma$ with pairwise disjoint open $V_\gamma$ such that each $(q \circ p)_\gamma : V_\gamma \stackrel{q \circ p}{\rightarrow} W$ is a homeomorphism.
Let $\beta$ be the unique index with $y \in U_\beta$. Then take $U_y = U_\beta$. We get $p^{-1}(U_y) \subset (q \circ p)^{-1}(W)$, but it is not at all clear that $p^{-1}(U_y) = \bigcup_{\gamma \in C'} V_\gamma$ for some $C' \subset C$. Only if that is the case, we can conclude that for all $\gamma \in C'$ the map $p_\gamma : V_\gamma \to U_y$ is a homeomomorphism (because then $(q \circ p)_\gamma = q_\beta \circ p_\gamma$, i.e. $p_\gamma = q^{-1}_\beta \circ (q \circ p)_\gamma$).
So what can assure that for all $\beta$ there exists $C(\beta) \subset C$ such that $p^{-1}(U_\beta) = \bigcup_{\gamma \in C(\beta)} V_\gamma$? We shall see that this is true if $W$ is connected.
In fact, in this case the $V_\gamma$ are connected, that is, they are the components of $(q \circ p)^{-1}(W)$. This is a distinguishing feature of connected evenly covered $W$. No $V_\gamma$ can intersect more than one $p^{-1}(U_\beta)$ because otherwise we would get a partition of $V_\gamma$ into pairwise disjoint open sets. Therefore each $V_\gamma$ is contained in a unique $p^{-1}(U_{\beta(\gamma)})$. Since each $p^{-1}(U_\beta)$ is covered by the collection of all $V_\gamma$, we see that each $p^{-1}(U_\beta)$ is the union of suitable subcollection $\{ V_\gamma \mid \gamma \in C(\beta) \}$.
Which assumption assures that each $z \in Z$ has a connected open neighborhhod which is evenly covered by $q$ and by $q \circ p$? This is that each $z \in Z$ has a neighborhood base consisting of connected open neighborhoods, i.e. the local connectedness of $Z$. This is a little weaker than local path-connectedness. So we can explicitly state
If we adopt the standard definition of covering maps (including surjectivity), then $p$ is not necessarily a covering map because it may lack surjectivity. An example is this. Let $Y = Z \times \{ 0 , 1 \}$ and $q$ be the projection onto the first factor, let $X = Z$ and $q = id$, $p(z) = (z,0)$. Then $q$ and $q \circ p = id$ are (surjective) covering maps, but $p^{-1}(Z \times \{ 1 \})$ is empty.
This shows that we need an additional assumption assuring that $p$ is surjective. A trivial solution is to simply amend the requirement that $p$ is surjective. A less trivial alternative is to assume that $Y$ is connected. See my answer to $p:X\to Y$ covering space if $q\circ p:X\to Z$ and $q:Y\to Z$ covering spaces and $Z$ locally path-connected..