Materials on some construction involving classification of covering spaces

Question

Let $X_u \rightarrow X $ be a universal covering. Let $S $ be any set with a group $\pi(X,x_0) $ acting on it from the right side. Then we get space $S \times X_u $ with a group action $ S \times X_u \times \pi(X,x_0):((s,x),g) \rightarrow (sg,g^{-1}x) \in S \times X_u $ (the action of $\pi(X,x_0)$ comes form the action of deck transformations on $X_u$ - since $X_u$ is a universal covering then they are both isomorphic). Now we define $X_s:=S \times X_u/\pi(X,x_0)$. It is a covering space of $X$. This construction is connected with a clasification of covering spaces, but I can't find anything about it.I will be glad for any sources.

Answer 1

I'll assume that $S$ is given the discrete topology, so that $X_s$ is indeed a covering space.

The space $X_s$ has one component for each orbit of the action of $\pi(X,x_0)$ on $S$, and hence $X_s$ is connected if and only if the action is transitive. In order to connect the question with the classification of connected covering spaces, I'll therefore add to the list of assumptions that the action of $\pi(X,x_0)$ on $S$ is transitive.

Pick one point $s_0 \in S$ and let $\text{Stab}(s_0)$ denote the subgroup of $\pi(X,x_0)$ that stabilizes $s_0$.

The connection with the classification of connected covering spaces can then be stated: $X_s$ is the covering space that is associated to the subgroup $\text{Stab}(s_0)$. This should follow from a close reading of any proof of the classification of connected covering spaces, in any textbook that contains the proof. Look particularly at the portion of the proof that constructs a covering space associated to every subgroup.