# 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)?

## 1 Answer

It it useful to have an algebraic model of covering maps and this is well provided for by the notion of covering morphism of groupoids. This is given in the book Topology and Groupoids (T&G), as it was in the 1968, 1988, differently titled, editions. For more see this stackexchange answer. (The earliest use of this notion I have found is in a 1951 Annals paper of PA Smith, and called a "regular morphism" of groupoids.)

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 $a$.

This explains why covering maps do not satisfy all of the "2 out of three rule".

However the "semicoverings " of J Brazas (see the above stack exchange link) do satisfy this rule.