If $q\circ p$ and $p$ are covering maps, under what conditions is $q$ a covering map?

Let $B,C,D$ be topological spaces and $p,q,r$ be continuous maps such that the following commutes: $$\newcommand{\ra}{\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!} \newcommand{\da}{\hphantom{#1}\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\sum}\right.} \newcommand{\sea}{\searrow{\scriptstyle#1}\!\!\!\!\!\!} % \begin{array}{ccc} C & \ra{p} & D\\ &\sea{r} & \da{q}\\ & & B\\ \end{array}$$

We know:

• If $p,q$ are covering maps, $r$ need not be. This is discussed e.g. in composition of covering maps
• If $D$ is locally connected and path-connected and $r,q$ are covering maps, then so is $p$. (More generally, if $B,C,D$ are locally connected and $p$ is surjective.)

Question: Under what conditions does $p$ and $q\circ p$ being covering maps imply that $q$ is one?

I got the following, if I'm not mistaken:

Proposition. If $p$ and $r$ are covering maps with $p$ Galois, then $q$ is a covering map.

Proof. $\DeclareMathOperator{Aut}{Aut}$ $q$ is open because $r$ and $p$ are local homeomorphisms. Let $U$ be an open neighborhood of $b\in B$ trivializing $r$, with $r^{-1}(U)=\sqcup U_i$. Then $q^{-1}(U)=p(\sqcup U_i)$ because $p$ is surjective. Each $p(U_i)$ is homeomorphic to $U$ because $p$ is injective on the $U_i$'s. It remains to prove that $p(U_i)$ and $p(U_j)$ are either equal or disjoint. If $p(x)\in p(U_i)\cap p(U_j)$, then $U_j\cap U_i^\alpha\neq\varnothing$ for some $\alpha\in\Aut(p)$. But $U_j$ and $U_i^\alpha$ are both sheets above $U$ for $r$, so they're equal. $\blacksquare$

Can the condition that $p$ be Galois be weakened (without too much effort)?

• Feb 19 '17 at 23:14

Now covering morphisms of groupoids satisfy the "2 out of 3 " rule: if $p,q,r$ are morphisms of groupoids such that the composition $p=qr$ is defined, then if 2 out of 3 of $p,q,r$ are covering morphism, so is the third. (For one part of this one needs $Ob(r)$ is surjective.) See 10.2.3 of T&G.
If $a: Y \to X$ is a covering map of spaces and $\pi_1$ is the fundamental groupoid functor, then $\pi_1a: \pi_1 Y \to \pi_1 Y$ is a covering morphism of groupoids. (10.2.1 of T&G.) There is also a local condition on $X$ which is expressed in T&G Section 10.5 by saying that $X$ is "semilocally $\chi_a$ connected". The point is that each $x \in X$ has a neighbourhood $U$ such that for each $y \in a^{-1}(x)$ the inclusion $U \to X$ lifts to a map $U \to Y$ taking $x$ to $y$ and so the image of $\pi_1(U,x)$ in $\pi_1(X,x)$ is contained in the image of $\pi_1(Y,y)$ under $a$.