Let $X,Y,Z$ be arc-connected and locally arc-connected spaces. Suppose that $q: X\to Z$ and that $p: X\to Y$ are covering spaces. Suppose there is a continuous function $r: Y\to Z$ such that $r\circ p=q$. Prove that $r$ is also a covering space.

We know that $r$ is continuous, so we can only verify that $r$ is surjective and that if $c\in Z$ then there is an open $U$ of $Z$ such that $r^{-1}(U)=\sqcup_{\alpha\in A}V_{\alpha}$ where $V_{\alpha}$ are open in $Y$ for all $\alpha\in A$ and $r|_{V_{\alpha}}: V_{\alpha}\to U$.

To see the overjection, let's take $c\in Z$, therefore there is a $a\in X$ such that $q(a)=c$ since $q$ is surjective it because it is a covering function. Then $r(p(a))=c$ and as $p(a)\in Y$, then $p(a)$ is a preimage of $c$ and so $r$ is onto.

Let $c\in Z$, since $q$ is a covering application, there is an open $U$ of $Z$ such that $c\in U$ and $q^{-1}(U)=\sqcup_{\alpha\in A}V_{\alpha}$ and $q|_{V_{\alpha}}: V_{\alpha}\to U$ is a homeomorphism. Consider the family $\{p(V_{\alpha})\}_{\alpha\in A}$, let's see that $r^{-1}(U)=\sqcup_{\alpha\in A}p(V_{\alpha})$ and that $r|_{p(V_{\alpha})}: p(V_{\alpha})\to U$ is a homeomorphism. In effect, as $r\circ q$, then $p(q^{-1}(U))=p(\sqcup_{\alpha\in A}V_{\alpha})=\sqcup_{\alpha\in A}p(V_{\alpha})$ but $r^{-1}(U)=p(q^{-1}(U))=\sqcup_{\alpha\in A}p(V_{\alpha})$.

How do I prove that $r|_{p(V_{\alpha})}: p(V_{\alpha})\to U$ is a homeomorphism? Thank you very much.


We shall use some elementary fact about functions.

(a) Let $f : A \to B, g : B \to C$ be two functions. If $g \circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).

(b) $f : A \to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M \subset B$.

Let us show that $r$ is a covering map.

(1) $r$ is a surjective open map.

We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M \subset Y$. This shows that $r$ is an open map.

Let $z \in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = \bigcup_\alpha U_\alpha$, where the $U_\alpha$ are open in $X$ and the $q_\alpha = q : U_\alpha \to W$ are homeomorphisms.

Let $p_\alpha = p : U_\alpha \to p(U_\alpha)$ and $r_\alpha = r : p(U_\alpha) \to W$. Each $p(U_\alpha)$ is open in $Y$ and the $p_\alpha, r_\alpha$ are continuous open maps.

(2) The $p_\alpha$ and the $r_\alpha$ are homeomorphisms.

It remains to show that they are bijective. We have $r_\alpha p_\alpha = q_\alpha$. Hence $p_\alpha$ is bijectve (use (a)). But this implies that also $r_\alpha$ is bijectve.

Let $V = p(q^{-1}(W)) = \bigcup_\alpha p(U_\alpha)$ which is open in $Y$.

(3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.

We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.

Her comes the crucial point which uses the fact that $W$ is path connected.

(4) Either $p(U_{\alpha_1}) \cap p(U_{\alpha_2}) = \emptyset$ or $p(U_{\alpha_1}) = p(U_{\alpha_2})$.

So let $p(U_{\alpha_1}) \cap p(U_{\alpha_2}) \ne \emptyset$. Since the $p(U_{\alpha_i})$ are path connected, so is their union. Consider $y_i \in p(U_{\alpha_i})$ and let $x_i \in U_{\alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{\alpha_1}) \cup p(U_{\alpha_2}) \subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_\alpha$, we conclude that $u_i$ is contained in $U_{\alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{\alpha_1}), p(U_{\alpha_2})$. This means $y_1,y_2 \in p(U_{\alpha_1}) \cap p(U_{\alpha_2})$. We conclude $p(U_{\alpha_1}), p(U_{\alpha_2}) \subset p(U_{\alpha_1}) \cap p(U_{\alpha_2})$ which implies $p(U_{\alpha_1}) = p(U_{\alpha_2})$.

This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{\alpha})$. Now (2) implies that $W$ is evenly covered by $r$.


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