Let $q: X\to Y$ and $r: Y\to Z$ be covering maps; let $p=r\circ q$. Show that if $r^{-1}(z)$ is finite for each $z\in Z$, then $p$ is a covering map.
I've almost completed solving this problem, but am stuck at showing that the open set containing a given $z\in Z$ is evenly covered by $p$. I searched on the web and found a solution to this problem, which uses the same method as mine, except that I'm not sure why the reasoning in the last part of this solution makes sense.
Solution:
My goal is to show that $C$ is evenly covered by $p$. But first, I don't understand the statement in the solution that says $C$ is evenly covered by $r$. Also why is $p^{-1}(C)=\bigcup D_{i_\alpha}$? I've been struggling with this last step for a long time, I would greatly appreciate any help.
I think the solution skips or brushes over a couple of steps, which I'll do my best to work through. First of all it's super confusing to have the space be $Z$ and the neighborhood be $Z$ so I'm going to call the neighborhood of $z \in Z$: "$Q$". I agree with, and will assume, the solution up until the construction of $C$.
$$C = \bigcap_{i = 1}^n r(U_i \cap V_i)$$
We need to show that $C$ is evenly covered by $r$, with the slices $U_i \cap V_i$.
Theorem (Munkres, 53.2) Let $p : E \to B$ be a covering map. If $B_0$ is a subspace of $B$, and if $E_0 = p^{-1}(B_0)$, then the map $p_0 : E_0 \to B_0$ obtained by restricting $p$ is a covering map.
Note that $r : Y \to Z$ is a covering map. $C$ is a subspace of $Z$, so the map $r_0 : r^{-1}(C) \to C$ defined by restricting $r$ is a covering map.
The proof from Munkres is informative for our proof so I will include it here, but re-worded for the context of this problem. Given $c \in C$, let $Q$ be our open set in $Z$ containing $c$ that is evenly covered by $r$ from before, so then $\{ V_{\alpha} \}$ is a partition of $r^{-1}(Q)$ into slices. Then $Q \cap C$ is a neighborhood of $c$ in $C$, and the sets: $$V_{\alpha} \cap r^{-1}(C) = V_{\alpha} \cap r^{-1}\Big(\bigcup_{i = 1}^{n} r(U_i \cap V_i)\Big) = V_{\alpha} \cap U_{\alpha}$$ ... are disjoint open sets in $r^{-1}(C)$ whose union is $r^{-1}(Q \cap C)$, and each is mapped homeomorphically onto $Q \cap C$ by $r$. So in the case of our problem, we have that the $r$ evenly covers $C$ with the slices $U_i \cap V_i$.
Proceeding with the proof, as $q$ is a covering map, recall that we set $A_i \subseteq X$ to be disjoint open sets such that: $$q^{-1}(U_i) = \bigcup_{\alpha} A_{i,\alpha}$$
Next, let $$D_{i,\alpha} = q^{-1}(U_i \cap V_i) \cap A_{i,\alpha}$$ ... for each index $i$ and index $\alpha$. These are subsets of the $A_{i,\alpha}$. Since the $q^{-1}(U_i)$ are all disjoint from one another, and each $D_{i,\alpha} \subseteq q^{-1}(U_i)$, each $D_{i,\alpha}$ is disjoint from all other $D_{j,\beta}$. Since the $A_{i,\alpha}$ are disjoint from the other $A_{i,\beta}$ and each $D_{i,\alpha} \subseteq A_{i,\alpha}$, the $D_{i,\alpha}$ are disjoint from the other $D_{i,\beta}$. So, the $\{ D_{i,\alpha} \}$ are all disjoint from one another.
Since $q$ is continuous (covering maps are continuous) and $U_i \cap V_i$ is the intersection of two open sets and thus an open set for each $i$, $q^{-1}(U_i \cap V_i)$ is the pre-image of an open set under a continuous map and is therefore itself open in the co-domain, which is $X$. So $D_{i,\alpha}$ is the intersection of two open sets and thus itself an open set in $X$; we conclude that $\{ D_{i,\alpha} \}$ is a set of disjoint open sets in $X$.
But are the $D_{i,\alpha}$ slices? Actually it suffices to show that they are slices of $r^{-1}(C)$ with respect to $q$, because since the composition of homeomorphisms is a homeomorphism, we immediately have that they are slices of $C$ with respect to $p$ and we are done.
By prior argumentation: $$q^{-1}(r^{-1}(C)) = q^{-1}\big(\bigcup_i (U_i \cap V_i)\big) = \bigcup_i q^{-1}(U_i \cap V_i)$$ By construction of the $A_{i,\alpha}$: $$= \bigcup_{\alpha} \bigcup_{i} q^{-1}(U_i \cap V_i) \cap A_{i,\alpha}$$ $$= \bigcup_{\alpha,i} D_{i,\alpha}$$ We conclude that the $D_{i,\alpha}$ are slices that evenly cover $r^{-1}(C)$ under $q$, and consequentially (as $q, r$ are covering maps, and by our prior logic) are slices that evenly cover $C$ under $p$, and we are done.