# question about disconnected normal covering map

I have stuck on the problem in Hatcher, Algebraic topology, which claim that if the covering map $q\circ p:X\rightarrow Y \rightarrow Z$ is normal, then the covering $p:X\rightarrow Y$ also is normal.

(Problem 16 in Section 1.3)

The messy part of this problem (I think) is that he only gives the condition that each space is locally path-connected.

This opens the case that X is not connected, so normality of the projection of the fundamental group of the covering space is not enough to show that $p$ is normal.

I think the direct proof should be needed, but have no idea what should I do.

The definition of a normal (regular) cover is that the deck transformations act transitively on each fibre. Hence suppose $y \in Y$ and $x_1,x_2 \in p^{-1}(\{y\})$. Then since $p(x_1)=p(x_2)$ we get $q\circ p(x_1)=q \circ p(x_2)$ and by assumption $q\circ p$ is regular and so we have a deck transformation $g:X \to X$ such that $g(x_1)=x_2$. But note this is a deck transformation for $q \circ p$ and not $p$. So we have $q \circ p \circ g=q\circ p$. Now we want $p \circ g=p$ but this may not actually occur, so we need to modify $g$ on each connected component. If $x_1$ and $x_2$ lie in the same connected component of $X$ then we define $h:X \to X$ to be equal to $g$ on that connected component and equal to the identity otherwise. If they lie on different connected components, say $x_i \in C_i$ for $i=1,2$, then define $h:X \to X$ to be equal to $g$ on $C_1$ and $g^{-1}$ on $C_2$ and the identity on the other connected components.
Need to check that $p \circ h=p$ in both cases.