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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.

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    $\begingroup$ Even if $A$ is central, $\tilde G$ is usually not abelian. $\endgroup$ Jun 3, 2023 at 12:03
  • $\begingroup$ What is the simplest example? $\endgroup$ Jun 3, 2023 at 12:04
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    $\begingroup$ $1\to \mathbb{Z}/2 \to D_8 \to \mathbb{Z}/2\times\mathbb{Z}/2 \to 1$ ? $\endgroup$ Jun 3, 2023 at 12:06
  • $\begingroup$ Right, I'll edit the question. Do you agree that the first question doesn't change much since it is enough to have $\beta :H^1(X,G)\rightarrow H^2(X,A)$? $\endgroup$ Jun 3, 2023 at 12:25
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    $\begingroup$ 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 $n=1$. $\endgroup$ Jun 3, 2023 at 13:27

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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.

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  • $\begingroup$ In this case it seems to me that the pull-back of the extension class, taking values in cohomology with twisted coefficients is a sort of generalization of the Bockstein: it coincide with the Bockstein for trivial twisting but holds more generally. I am a bit surprised that this does not have an "algebraic" definition. $\endgroup$ Jun 5, 2023 at 8:43
  • $\begingroup$ 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 $K(G,1)\rightarrow K(A,2)$? $\endgroup$ Jun 5, 2023 at 8:45
  • $\begingroup$ Let me say it again. If $\widetilde G$ is not abelian then there is no algebraic definition of a Bockstein. $\endgroup$ Jun 5, 2023 at 8:53
  • $\begingroup$ Yes sorry, I meant a reference for the case in which the three groups are abelian. $\endgroup$ Jun 5, 2023 at 9:04
  • $\begingroup$ 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. $\endgroup$ Jun 5, 2023 at 9:17

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