Dtyjlui

I have done the problem, but I'm confused about why the local path-connectedness of $Z$ is necessary.

My solution: Let $p:Xto Y$ and $q:Yto Z$ be such that $q$ and $qcirc p$ are covering spaces. Let $yin Y$ and let $W$ be a path-connected neighborhood of $z=q(y)in Z$ which is evenly covered by $q$ and $qcirc p.$ Let $V$ be the path component of $q^{-1}(W)$ which contains $y$ (observe that $V$ is then a sheet of $W$ by $q$). Any sheet $U$ of $p^{-1}(V)$ is mapped homeomorphically to $W$ by $qcirc p.$ Since $q:Vto W$ is a homeomorphism, we conclude that $p:Uto V$ is a homeomorphism, so $p$ is a covering space.

It seems like I used the local path-connectedness of $Z$ in my solution, letting $W$ be a path-connected neighborhood of $q(y),$ but this use seems artificial. It isn't clear to me why $W$ must be path-connected, other than when one tries to isolate an evenly covered neighborhood $V$ of $y.$ However, it seems like local connectedness is enough to do this.

So my question is this: is local path-connectedness of $Z$ necessary (and if so, is there some instructive counterexample when $Z$ is not locally path-connected) or can this hypothesis be weakened?

It can be weakened. We shall use the following well-known fact about covering projections:

Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.

In this result no further connectedness assumptions on $X,Y,Z$ are needed.

Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.

Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove

If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.

This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.

Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.