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
 A: We will suppose that our spaces are locally connected, so that connected components are open and closed.
The space  $Z$ can be covered by open connected subsets  over which $q$ is trivial, and since the restriction of $res(p):p^{-1} (r^{-1}(U) = q^{-1}(U) \to r^{-1}(U)$ is still a covering , we may and will henceforth assume that $q$ is a trivial covering and that $Z$ is connected.   
The core of the proof
Take a connected component $V\subset X$ of $X$ ( a sheet of the trivial covering  $q$) .
 Its image $p(V)$ will be a connected component of $Y$,  according to Spanier's Algebraic Topology, Chap.3, Theorem 14, page 64.
But then $res(r):p(V)\to Z$   is a homeomorphism and since, by surjectivity of $p$, the space $Y$ is a disjoint union of such $p(V)$, the map $r:Y\to Z$ is a trivial covering whose sheets are exactly the connected components of $Y$.
A: 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.
