Composed Covers I have problems solving this seemingly straightforward question.
Let $q : X \rightarrow Z$  be a covering space. Let $p : X \rightarrow Y$ be a covering space. Suppose there is a map $r : Y \rightarrow Z$ such that $q = r \circ p$. Show that $r : Y \rightarrow Z$ is a covering space.
Could someone give me a hint?
Of course I should pick some covering definition and show that $r$ indeed satisfies this.
Thank you

Take a open subset in $U\subseteq Z$ and take the preimage of it along $q^{-1}$. Then you know that $q^{-1}(U)$ is homeomorphic to $U_X\times F_X$, since $q=r\circ p$ you know $U_X\times F_X\cong q^{-1}(U)=(r\circ p)^{-1}(U)=p^{-1}\circ r^{-1}(U)$ is the same (as Zhen Lin pointed out). And now you have to work your way through why $r^{-1}(U)$ is homeomorphic to $U_Y\times F_Y$
As it is homework i didn't want to do it all.
Ps. I find it much more understandable when I draw pictures of commutative diagrams.
