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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

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    Dear Jake, I think a complete, non hand-waving proof, would be rather messy to write-up in complete detail. Could you tell us who gave you this homework ? And post the teacher's solution in due time: I wonder how long it will be! I have written a complete solution but I have used a non-trivial theorem in Spanier's classic *Algebraic Topology*.2011-12-15

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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 $Y$.

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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.