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.