Let \(\mathcal{B}(X; G)\) be the set of isomorphism classes of bundles of groups \(E \to X\) with fiber \(G\), and let \(E_0 \to B\operatorname{Aut}(G)\) be the bundle corresponding to the “identity” action \(\rho: \operatorname{Aut}(G) \to \operatorname{Aut}(G)\). Show that the map \([X, B\operatorname{Aut}(G)] \to \mathcal{B}(X; G)\), \([f] \mapsto f^*(E_0)\), is a bijection if \(X\) is a CW complex, where \([X, Y]\) denotes the set of homotopy classes of maps \(X \to Y\).
Solution
Regard \(\operatorname{Aut}(G)\) as a discrete group. Then \[ B\operatorname{Aut}(G)=K(\operatorname{Aut}(G),1), \] and the bundle \(E_0\to B\operatorname{Aut}(G)\) associated to the identity action is the universal bundle of groups with fiber \(G\).
Assume first that \(X\) is connected and choose a basepoint. A bundle of groups \(E\to X\) with fiber \(G\) has monodromy \[ \rho_E:\pi_1(X)\longrightarrow\operatorname{Aut}(G), \] well-defined up to conjugation by an automorphism of the chosen fiber. Conversely, a homomorphism \[ \rho:\pi_1(X)\to\operatorname{Aut}(G) \] defines the bundle \[ \widetilde X\times_{\pi_1(X)}G\longrightarrow X, \] where the deck group acts on \(G\) through \(\rho\). These two constructions are inverse up to bundle isomorphism. Thus \[ \mathcal B(X;G) \cong \operatorname{Hom}(\pi_1X,\operatorname{Aut}(G))/\text{conjugacy}. \]
Since \(B\operatorname{Aut}(G)\) is a \(K(\operatorname{Aut}(G),1)\) and \(X\) is a CW complex, the standard classification of maps into a \(K(\pi,1)\) gives \[ [X,B\operatorname{Aut}(G)] \cong \operatorname{Hom}(\pi_1X,\operatorname{Aut}(G))/\text{conjugacy}. \] Under this correspondence, a map \(f:X\to B\operatorname{Aut}(G)\) corresponds to the monodromy homomorphism \(f_*\). The pullback \(f^*E_0\) has exactly this monodromy, so the displayed identification is precisely the map \[ [f]\longmapsto f^*(E_0). \] Hence it is a bijection. For disconnected \(X\), apply the same argument independently on each component.