If $q \circ p$ and $q$ are coverings with $q \circ p$ normal, then $p$ is normal

Question

It has been asked here and here for an explanation of Hatcher 1.3.16. Here is the question:

Given maps $X \to Y\ \to Z$ such that both $Y \to Z$ and the composition $X \to Z$ are covering spaces, show $X \to Y$ is a covering space if $Z$ is locally path-connected, and show that this covering space is normal if $X \to Z$ is a normal covering space.

Let $p:X\to Y,q: Y \to Z$ be the maps in question. I can prove the first part in a fairly straightforward way, and my solution turns out to be essentially the same one given in the second link.

But the linked questions do not actually mention the second part of this problem. Luckily, it is partially addressed here. But even that answer by Mathemagician is not a full solution--in fact I think the crux of the exercise is the part missing, namely, showing we have a deck transformation that factors through $p$.

So, starting with:

Let $y \in Y$ and $x,x' \in p^{-1}(y)$ be distinct. We are given $q \circ p$ is normal, i.e. $\exists h \in \mathrm{Homeo}(X)$ such that $h(x)=x'$ and $q \circ p \circ h = q \circ p$.

We follow Mathemagician's argument:

$h$ is a deck transformation for $q \circ p$, not for $q$. It need not be the case that $(p \circ h)(x')=y$ in order for $(q \circ p)(x') = (q \circ p)(x) = q(y)$. It may be the case that $(p \circ h)(x')$ instead equals an element of the fibre of $q(y)$ not equal to $y$. But we can recover a deck transformation by carefully amending this problem. We consider two cases.

Case 1: Suppose $x,x'$ lie in the same connected component. Then define $g \in \mathrm{Homeo}(X)$ to be equal to $h$ on the connected component of $x$, and the identity elsewhere.

Case 2: Suppose $x,x'$ lie in separate connected components. Call them $C,C'$. Then define $g$ to equal $h$ on $C$ and $h^{-1}$ on $C'$, and the identity elsewhere.

It remains to be shown that $g$ is a deck transformation. If it is, then we are done, since $g(x) = x'$. But I'm not sure how to do this. I hoped to use Proposition 1.37 of Hatcher:

If $X$ is path-connected and locally path-connected, then two path-connected covering spaces $p_1:X_1 \to X$ and $p_2:X_2 \to X$ are isomorphism via an isomorphism $f:X_1 \to X_2$ taking basepoint $x_1 \in p_1^{-1}(x_0)$ to basepoint $x_2 \in p_2^{-1}(x_0)$ iff $p_{1*}(\pi_1(X_1,x_1)) = p_{2*}(\pi_1(X_2,x_2))$.

However, I'm not sure the path-connected portion of the hypothesis is fulfilled. Other than that, I'm not sure how to proceed.

Answer 1

We can relax the condition "$Z$ locally path-connected" to "$Z$ locally connected". See my answer to Exercise 1.3.16 in Hatcher.

Observe that $X$ is locally connected because $Z$ is locally connected and $q \circ p : X \to Z$ is a local homeomorphisms. This implies that all components of $X$ are open in $X$.

Let $x, x' \in X$ such that $p(x) = p(x') = y$. We have to find a deck transformation $h$ for $p$ such that $h(x) = x'$.

Let $C, C'$ be the components of $x, x'$ and let $g : X \to X$ be a deck transformation for $q \circ p$ such that $g(x) = x'$. It maps components of $X$ homeomorphically onto components of $X$ and we have $g(C) = C'$ because $g(x) = x'$. Then we have $$(p\circ g)(\xi) = p(\xi) \text{ for all } \xi \in C . \tag{1}$$ Note that this implies that also $(p \circ g^{-1})(\xi) = p(\xi)$ for all $\xi \in C'$.

To prove $(1)$, observe that $(q \circ p) \circ g = q \circ p$ implies that $p \mid_C, (p \circ g) \mid_C : C \to Y$ are both lifts (with respect to $q$) of the map $(q \circ p) \mid_C : C \to Z$. Since $p(x) = p(x') = p(g(x)) = (p \circ g)(x)$ and $C$ is connected, we conclude that $p \mid_C = (p \circ g) \mid_C$. See Hatcher's Proposition 1.34.

Then we get the desired $h$ as follows: If $C = C'$, then take $$h(\xi) = \begin{cases} g(\xi) & \xi \in C \\ \xi & \xi \in X \setminus C \end{cases}$$ and if $C \ne C'$, then take $$h(\xi) = \begin{cases} g(\xi) & \xi \in C \\ g^{-1}(\xi) & \xi \in C' \\ \xi & \xi \in X \setminus (C \cup C') \end{cases}$$

These definitions produce continuous maps because $X \setminus C$ and $X \setminus (C \cup C')$ are open.