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

My solution: Let $p:X\to Y$ and $q:Y\to Z$ be such that $q$ and $q\circ p$ are covering spaces. Let $y\in Y$ and let $W$ be a path-connected neighborhood of $z=q(y)\in Z$ which is evenly covered by $q$ and $q\circ p.$ 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 $q\circ p.$ Since $q:V\to W$ is a homeomorphism, we conclude that $p:U\to V$ 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?

  • $\begingroup$ You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z \times \{0,1 \}$ with $p(z) = (z,0)$ and $q(z,i) = z$. $\endgroup$ – Paul Frost Dec 29 '18 at 14:56

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 $y = v(1) = p(u(1)) \in p(X)$.


Given $p,q$ as in your question (i.e. $q \circ p$ and $q$ are assumed to be covering projections), let us first observe that if if one of $X, Y, Z$ is locally connected, then so are the other two because the three spaces are locally homeomorphic. Let us verify the following more general results:

(1) If $Z$ is locally connected, then each $y \in Y$ has an open neighborhood $U$ which is evenly covered by $p$. Note that this is trivially true if $p^{-1}(U) = \emptyset$. See my answer to Exercise 1.3.16 in Hatcher. However, $p$ is in general not a covering projection because surjectivity is not guaranteed.

(2) In $Z$ is locally connected, then $p(X)$ is open and closed in $Y$ and $p : X \to p(X)$ is a covering projection.

In fact, for $y \in Y$ let $U$ be an open evenly covered neighborhood of $y$. If $y \in p(E)$, then $p^{-1}(U) \ne \emptyset$ since it contains $p^{-1}(y)$. Hence $U = p(p^{-1}(U)) \subset p(X)$, i.e. $p(X)$ is open. If $y \notin p(X)$, then $p^{-1}(U)$ must be empty because otherwise we would get $y \in U = p(p^{-1}(U)) \subset p(X)$. Hence $U \cap p(E) = \emptyset$, i.e. $Y \setminus p(X)$ is open. That $p : X \to p(X)$ is a covering projection is now clear.

(3) If $Y$ is connected and locally connected, then $p$ is a covering projection.

$p(X)$ is open and closed, thus $p(X) = Y$.


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