Let $G$ be a discrete topological group that acts freely and properly discontinuously on a space $EG$ which is assumed to be contractible and locally path-connected. We then form the so-called classifying space as the orbit space $BG=EG/G$ and yield a covering projection $p\colon EG\rightarrow BG$, which may also be viewed as a principal $G$-bundle.
I want to prove that if $X$ is a connected $CW$ complex with one 0-cell, and $f,g$ are maps $(X,x_0)\rightarrow (BG,b_0)$, then $f^*EG$ and $g^*EG$ are isomorphic $G$-bundles if and only if the maps $f$ and $g$ are freely homotopic.
I've thought of this: If the bundles are isomorphic, then it suffices to show that $f_\#=g_\#$ since there is a bijection $[X,BG]_*\rightarrow \mathrm{Hom}(\pi_1(X,x_0),G)$ given by $[f]\mapsto \phi\circ f_\#$. Here $\phi$ is any (a-priori given) isomorphism between $\pi_1(BG,b_0)$ and $G$ and $[X,BG]_*$ denotes the family of classes of homotopic maps $X\rightarrow BG$. I got stuck here, so now I'm searching for ideas.
I haven't been able to tackle the other direction either.
Best Answer
Suppose $L:f^*EG\cong g^*EG$ as $G$-bundles over $X$.
They are also covering spaces, and this must be an isomorphism of coverings.
In particular, let $x_0$ be the $0$-cell of $X$ (or in fact, any chosen basepoint), then $L$ induces a bijection on the fibers $p^{-1}(x_0)\to q^{-1}(x_0)$, where $p:f^*EG\to X, q: g^*EG\to X$ are the structure maps.
This bijection is a bijection of $\pi_1(X,x_0)$-sets (this is general covering space theory), and $G$-sets (because $L$ is a morphism of $G$-bundles)
But now, as $G$-sets, both $p^{-1}(x_0)$ and $q^{-1}(x_0)$ are isomorphic to $G$ (which is the fiber of $EG\to BG$) with the regular action, so that this map is necessarily of the form $g\mapsto L(e)g$.
Saying that it's a map of $\pi_1(X,x_0)$-sets then implies the following equation : $$L(e)f_*(\alpha)g= g_*(\alpha) L(e)g$$
for all $\alpha\in \pi_1(X,x_0), g\in G$
Taking $g=e$, we get $L(e)f_*(\alpha)L(e)^{-1} = g_*(\alpha)$, i.e. $f_*$ and $g_*$ are conjugate.
Finally, note that $[X,BG]_*\cong \hom(\pi_1(X,x_0),G)$, but those are pointed-homotopy classes of pointed maps $X\to BG$. If you mod out the RHS by conjugation in $G$, it amounts on the LHS to projecting to unpointed-homotopy classes : $[X,BG]_*\to [X,BG]$
(this is an exercise : if $X,Y$ are nice pointed spaces, then $\pi_1(Y)$ acts on $[X,Y]_*$, and the quotient is exactly $[X,Y]$; you then just need to check that the action of $\pi_1(BG)\cong G$ on $[X,BG]_*$ corresponds to the action of $G$ on $\hom(\pi_1(X),G)$ by conjugation)
It follows that $f,g$ are the same element in $[X,BG]$, that is, they are freely homotopic.
The converse either follows from a very general statement :
Or you can also prove it in this specific case by noting that associating to a covering space its fiber over the basepoint is an equivalence between covering spaces and $\pi_1(X,x_0)$-sets.
If the maps are freely homotopic, they'll induce conjugate morphisms on $\pi_1$, hence (by the above essentially) isomorphic $\pi_1(X,x_0)\times G^{op}$-sets, hence isomorphic $G^{op}$-objects in covering spaces, hence finally, isomorphic $G$-bundles.
So the proof of the converse depends on what you already know regarding covering spaces and bundles.