If $q:X\to Y$, $r:Y \to Z$, and $p=r \circ q : X \to Z$ are all covering maps, with $Z$ locally path-connected, and if $p$ is a regular covering then so is $q$.

Note 1. The condition $q$ is a covering map is not necessary, because it is automatically satisfied.

Note 2. regular coverings are sometimes called normal coverings

Note 3.If $X, Y, Z$ are all path-connected, then this is easy, using the following fact:

Proposition. Let $p : X \to Y$ be a covering map with $X$ path-connected and $Y$ path-connected and locally path-connected. Then $p$ is a regular covering iff $p_* \pi_1(X)$ is a normal subgroup of $\pi_1(Y)$.

However, there are no assumptions that $X, Y, Z$ are path-connected. How do I have to proceed?


2 Answers 2


Let $y \in Y$ and $x_1, x_2 \in q^{-1}(x)$. We have to show that there is a deck transformation for $q$ taking $x_1$ to $x_2$.

Since $x_1,x_2 \in p^{-1}(r(y))$, we know that there exists a deck transformation $d$ for $p$ taking $x_1$ to $x_2$.

You assume that $Z$ locally path-connected. Hence also $Y$ and $X$ are locally path-connected. Thus all path-components of these spaces are open, and we conclude that the spaces are the disjoint union of their path-components. Let $P_i$ be the path-components of $X$ containing $x_i$. $d$ is a homeomorphism, hence it maps path components homeomorphically onto path components. Thus $d(P_1) = P_2$. You have either $P_1 = P_2$ or $P_1 \cap P_2 = \emptyset$. In the first case define a new deck transformation $d'$ for $p$ by $d' \mid_{P_1} = d \mid_{P_1}$ and $d' \mid_P = id$ for all path-components $P \ne P_1$. In the second case define a new deck transformation $d'$ for $p$ by $d' \mid_{P_1} = d \mid_{P_1}$, $d' \mid_{P_2} = d^{-1} \mid_{P_2}$ and $d' \mid_P = id$ for all path-components $P \ne P_1, P_2$. Then $d'$ is clearly a bijection. Both $d', (d')^{-1}$ are continuous (recall that $X$ is the disjoint union of its path components). By construction $d'$ takes $x_1$ to $x_2$.

We claim that $d'$ is deck transformation for $q$, that is $q \circ d' = d'$. Obviously $(q \circ d')(x) = q(x)$ for all $x \notin P_1 \cup P_2$. Let $x \in P_1$. There exists a path $u$ in $P_1$ such that $u(0) = x_1$ and $u(1) = x$. Consider the paths $v' = q \circ d' \circ u$ and $v = q \circ u$ in $Y$. They satisfy $v'(0) = y = v(0)$. We have $r \circ v' = r \circ v$, thus $v' = v$ by unique path lifting. Hence $(q \circ d')(x) =(q \circ d' \circ u)(1) = (q \circ u)(1) = q(x)$. Similarly we can show $(q \circ d')(x) = q(x)$ for $x \in P_2$. This proves $q \circ d' = d'$.


We cannot expect that $d$ itself is deck transformation for $q$. As an example take $r : Y = \{-1,1\} \to Z = \{0\}$, $q : X = \{-3,-2,-1,1,2,3\} \to Y, q(x) = \text{sgn}(x)$. Here all spaces have the discrete topology. Let $x_1 = 1, x_2 = 2$. Then any permutation $d$ of $X$ taking such that $d(1) = 2$ is a deck transformation for $p$, but it is not a deck transformation for $q$ unless $d(\{1,2,3\}) = \{1,2,3\}$. Thus we must "adjust" $d$.

  • $\begingroup$ Is $d'$ injective? What happens if $d(x_2) \notin P_1 \cup P_2$? $\endgroup$
    – user302934
    Aug 17, 2019 at 1:42
  • $\begingroup$ How about defining $d'=d$ on $P_1$, $d'=d^{-1}$ on $P_2$, and $d=$id otherwise (in the case $P_1 \neq P_2$) ? $\endgroup$
    – user302934
    Aug 17, 2019 at 1:48
  • $\begingroup$ Ah, you are right: There is no reason why $d(P_2) = P1$ if $P_1 \ne P_2$, thus we can have $d(P_1 \cup P_2) \ne P_1 \cup P_2$. The definition in your second comment will do. I corrected my answer. Thank you for drawing my attention to this point. $\endgroup$
    – Paul Frost
    Aug 17, 2019 at 8:39

The following is essentially 10.6.4 of Topology and Groupoids, using the "algebraic model" of covering maps of spaces by covering morphisms of greoupoids. Here a "loop" in a groupoid $G$ is an element of a vertex group $G(x)= G(x,x)$.

10.6.4 Let $p : H \to G$ be a covering morphism of groupoids. Consider the following conditions:

(a): for all loops $a$ in $G$, either all or no lifts of $a$ are loops;

(b): for all objects $x$ of $H$, the characteristic group $p[H(x)]$ is normal in $G(px)$.

Then (a} $\Rightarrow$ (b), and if $H$ is connected, (b) $\Rightarrow$ (a).

Fig 10.3 on the same page illustrates the idea.


This should help to solve the problem easily, without mentioning paths.

(See also this stackexchange question.)


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