Are these two natural cohomology classes of a manifold constructed from a 1-cochain and a group extension equal?

Question

Let $X$ be a manifold, $G$ and $A$ finite abelian groups and $\epsilon \in H^2(G,A)$ a group cohomology class (for the moment I am assuming there is no action of $G$ on $A$). Given $\alpha \in H^1(X,G)$ there are two natural classes in $H^2(X,A)$ we can construct.

(i) First since $H^2(G,A)$ can be identified with the second cohomology group of $BG$ with $A$ coefficients, and $\alpha$ is associated with a homotopy equivalence class of maps $\widehat{\alpha} :X\rightarrow BG$, we can take the pull-back $\widehat{\alpha}^*\epsilon \in H^2(X,A)$.

(ii) Second, $\epsilon$ identifies uniquely a central extension $$ 1\rightarrow A\rightarrow \widetilde{G}\rightarrow G \rightarrow 1 $$ from which we get the Bockstein map $\beta : H^1(X,G)\rightarrow H^{2}(X,A)$. Then we can construct $\beta(\alpha)\in H^2(X,A)$.

1. First question: is it true in general that $\widehat{\alpha}^*\epsilon=\beta(\alpha)$? If yes, how to prove it? If no in general, when it happen to be true?

2. Then I am interested in if and how this story generalizes when $A$ is a non-trivial $G$ module speficied by a homomorphism $\rho :G \rightarrow \text{Aut}(A)$. In this case the twisted group cohomology $H_{\rho}^2(G,A)$ can be identified with the cohomology of $BG$ with local $A$ coefficients $H^2(BG,\widetilde{A})$, and the first construction similarly leads to a class $\widetilde{\alpha}^*\epsilon \in H^2(X,\widetilde{A})$, however I am not sure whether this cohomology group of $X$ with local coefficients makes sense... As for the second construction we again have an extension but this will be non-central. Nevertheless the first Bockstein map $\beta :H^1(X,G)\rightarrow H^2(X,A)$ can still be constructed in the same way and we can consider $\beta(\alpha)$. However it seems to me that this does not live in the cohomology with local $A$ coefficients, but I am really not sure since there might be subtleties in the construction of the Bockstein map. I would be really greatful to anybody could clarify this situation.

Answer 1

The general problem with this question is the definition of the Bockstein map. If $\widetilde G$ is non-abelian, then cochains with coefficients in $\widetilde G$ doesn't really make sense. The problem is the boundary map. With abelian coefficients, you get some linear combinations, but with non-abelian coefficients, in what order are you going to multiply?

Topologically, the Bockstein map for a short exact sequence $0 \to A \to B \to C \to 0$ of abelian groups comes from the fact that we have a fibre sequence $K(A,n) \to K(B,n) \to K(C,n) \to K(A,n+1)$, and composing with $K(C,n) \to K(A,n+1)$ gives the appropriate map. In your case, if $A \to \widetilde G \to G$ is a central extension, you do at least have a fibre sequence $$K(A,1) \to K(\widetilde G,1) \to K(G,1) \to K(A,2).$$ Then your Bockstein map $H^1(X,G) \to H^2(X,A)$ is defined, and the answer to your first question is yes.

But in your second question, if $A$ is only normal rather than central, you only get $K(A,1) \to K(\widetilde G,1) \to K(G,1)$. Yes, you get a map to cohomology with twisted coefficients, just by pulling back the class for $BG$ to $X$, but you have no Bockstein to compare it with.