# 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 $r\circ p=q$.

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 $$r$$.