# 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.

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.

## marked as duplicate by Paul Frost, Joshua Mundinger, Mike Earnest, Yanior Weg, Hw ChuMay 10 at 0:12

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.