# If $q\circ p$ and $q$ are covering maps then $p$ is also a covering map.

I was trying to do this exercise from Hatcher's book:

Let $p:F\to E,q:E\to X$ maps such that $q$ and $q\circ p$ are covering maps. If $X$ is locally path-connected, then show that $p$ is a covering map.

I tried it as follows:

Let $e\in E$. I must find an evenly covered neighborhood (with respect to $p$) of $e$. I think I can't do this just now, but I can take a neighborhood $U$ of $q(e)$ which is path-connected and evenly covered with respect to both $q$ and $q\circ p$. Then $$q^{-1}(U)=\bigsqcup_{i\in I} V_i, (q\circ p)^{-1}(U)=\bigsqcup_{j\in J}W_j$$ such that $q|_{V_i},(q\circ p)|_{W_j}$ are homeomorphisms for each $i,j$. Fix $i$ such that $e\in V:=V_i$.

I want to prove that $p^{-1}(V)$ is the disjoint union of a subfamily of the $W_j$'s, each of them mapped homeomorphically by $p$ onto $V$. But I don't know how to get there.

• Hint: $p\vert_{W_j}$ is a homeomorphism to some $V_i$ – Berci May 14 '18 at 20:03
• It's done. We have $p(W_j)\subseteq V_i$ for some $i$. $q|_{V_i}$ is a homeomorphism, then $p(W_j)=V_i$. But $p|_{W_j}$ is just the composition of $q\circ p|_{W_j}$ with $q|_{V_{i}}^{-1}$, a homeomorphism. Now i think I coud say something like this: If $p^{-1}(V)=\varnothing$ we're done since it's an empty disjount union of copies of $V$. If not, then, since $p^{-1}(V)\subseteq (q\circ p^{-1}(U))$, if for some $j$, $p^{-1}(V)\cap W_j\neq \varnothing$, then since $p$ is now a homeomorphism onto $V$, $p^{-1}(V)\cap W_j=W_j$. If we take all such $W_j$ we get $p^{-1}(V)$. Am I right? – David Molano May 14 '18 at 21:36