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

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.

• Even if $A$ is central, $\tilde G$ is usually not abelian. Jun 3, 2023 at 12:03
• What is the simplest example? Jun 3, 2023 at 12:04
• $1\to \mathbb{Z}/2 \to D_8 \to \mathbb{Z}/2\times\mathbb{Z}/2 \to 1$ ? Jun 3, 2023 at 12:06
• Right, I'll edit the question. Do you agree that the first question doesn't change much since it is enough to have $\beta :H^1(X,G)\rightarrow H^2(X,A)$? Jun 3, 2023 at 12:25
• In general, the Bockstein for a short exact sequence $0\to A \to B \to C \to 0$ of abelian groups is induced by composition with $K(C,n) \to K(A,n+1)$. When $B$ is non-abelian, this only makes sense when $n=1$. Jun 3, 2023 at 13:27

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.
• Could you suggest a reference for the equivalence, in the central extension case, between the algebraic definition of the Bockstein and the geometric one composing with $K(G,1)\rightarrow K(A,2)$? Jun 5, 2023 at 8:45
• Let me say it again. If $\widetilde G$ is not abelian then there is no algebraic definition of a Bockstein. Jun 5, 2023 at 8:53
• This is really just Yoneda's lemma. Every natural transformation between representable functors is representable. This is where the map $K(G,n) \to K(A,n+1)$ comes from. Jun 5, 2023 at 9:17