Classification of $G$-principal Bundle and classification of $G$-coverings: a bridge between the two?

Question

I encountered the following sentence in an exercise (the context is irrelevant)

Let $G\cong\langle s_1,s_2,\dots , s_g \mid R \rangle$ be a discrete one relator group. Consider the $G$-principal covering associated to the projection $p\colon \ast_{i=1}^g\mathbb{Z} \to G $, $q\colon Y \to B( \ast_{i=1}^g\mathbb{Z})$

I think there are two reasonable interpretations of this sentence:

1) let $K:=\ker(p)$, we have an associated (up to iso) covering to $K$. We know that the Deck transformation group is isomorphic to $N(K)/K\cong G$ and hence it's a $G$-covering.

and

2) let $\tilde{p} \colon B(\ast_{i=1}^g\mathbb{Z})\to BG$ be the map associated to $p \in \hom_{Grp}(\ast_{i=1}^g\mathbb{Z}, G)$ (both spaces are E-M spaces) and do the pullback of $\xi \colon EG\to BG$ via $\tilde{p}$ so we obtain a $G$-principal bundle, but being $G$ discrete, it's a $G$-principal cover.

I was curios about these two ways to interpret the sentence and so I asked myself wether was possible to draw some lines between the two:

I explain clearly what I mean by that. Under the assumption of $G$ being a discrete group, the $G$ bundle builded in $2)$ is a covering of $B(\ast_{i=1}^g\mathbb{Z})$, ad hence it's classified by a subgroup of $\ast_{i=1}^g\mathbb{Z}$ (up to conjugation). With the l.e.s for a Serre Fibration one can prove (according to me) that the cardinality of such subgroup has to be the same of the cardinality of $\ker(p)$, but I really don't know wether it is possible to show that this covering is isomorphic to the one classified by $\ker(p)$ in general.

My question therefore is if one can say something more between these two constructions or if the exercise should have been more specific

Answer 1

Okay, I was using terminology a little carelessly. There are two things you might mean by "$G$-cover" and they're not the same:

• A Galois cover with automorphism group $G$.
• A principal $G$-bundle.

For example, consider the trivial principal $G$-bundle $X \times G$ on any reasonable space $X$. This is not a Galois cover because $X \times G$ is not connected, even if $X$ is, and it does not have automorphism group $G$: its automorphism group, as a cover, is one copy of the symmetric group $S_{|G|}$ for every connected component of $X$.

In any case, I think the second interpretation was the intended one.