# 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. [duplicate]

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.

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.