Suppose that the above diagram is commutative and the three maps are continuous. Moreover, suppose that $(X,p)$ is a covering space of $Y$ and $(X,q)$ is a covering space of $Z$. I want to prove that $(Y,r)$ is a covering space of $Z$ too.
In order to do that I suppose I have to prove that if $U \subset Z$ is an elementary neighborhood for the covering space $(X,q)$, and call $V$ an arc component of $r^{-1}(U)$, then $V$ is mapped homeomorphically onto $U$ by $r$. I know also that if $\tilde V \subset X$ is an arc component of $p^{-1}(V)$ then the restricted projection to $\tilde V$ togheter with $\tilde V$ is a covering space of $V$ (i.e. $(\tilde V, p_{| \tilde V})$ is a covering space of $V$). So, by using this information, how can I prove the claim?
Let $z$ be an element of $Z$, write $q^{-1}(z)=\{x_1,..x_n\}$ and $r^{-1}(z)=\{y_1,..,y_m\}$. Since $q, r$ are covering. For every $j$, there exists $x_{i_j} $ such that $p(x_{i_j})=y_j$. There exists open subsets $U_i, V_{i} $ of $x_{i_j} $ such that the restriction of $p$ to $U_i$ and $q$ to $V_i$ is a homeomorphism onto their images. Let $W_i=q_{\mid U_i} ^{-1}(q(U_i) \cap q(V_i) )$. It is a subset of $U_i\cap V_i$. $q_{\mid W_i}:W_i\rightarrow q(W_i) $ and $p:W_i\rightarrow p(W_i)$ is an homeomorphism. We deduce that :$r_{\mid p(W_i)}=q_{\mid W_i}\circ p_{\mid W_i} ^{-1}:p(W_i) \rightarrow q) W_i) $ is a homeomorphism.