# $p:X\to Y$ covering space if $q\circ p:X\to Z$ and $q:Y\to Z$ covering spaces and $Z$ locally path-connected.

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?

• 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 $q(z,i) = z$. – 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 $$y = v(1) = p(u(1)) \in p(X)$$.

Update:

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 $$p(X) = Y$$.