Cup-products and Transgression maps.

Question

This question is related to Lyndon-Hochschild-Serre spectral sequence and cup products.

I have the followin result by J.S Milne in his book Arithmetic duality theorems pg 105.

Let $$0 \rightarrow C \rightarrow W \rightarrow G \rightarrow 0$$ is an exact sequence of groups corresponding to the canonical class $u \in H^{2}(G,C)$, and $C$ is of finite index in $W$

Let $M$ be a $W$ module on which $C$ acts trivially. Then by Hochschild-serre gives a spectral sequence $$0 \rightarrow H^{1}(G,M) \rightarrow H^{1}(W,M) \rightarrow H^{1}(C,M)^{G} \overset{\tau}\rightarrow H^{2}(G,M) $$ where $\tau$ is the transgression map. Now he goes on to prove that in this case $\tau(\alpha) = - \alpha \cup u$ (cup-product) for some $\alpha \in H^{0}(G,Hom(C,M))$.

Now consider the LHS for group homology, with $C,W,G,M$ as above ($C$ acting trivially on $M$)

$$ H_{2}(G,M) \overset{\tau'}\longrightarrow H_{1}(C,M)_{G} \overset{cor}\longrightarrow H_1(W,M) \overset{coinf}\longrightarrow H_{1}(G,M) \longrightarrow 0$$

Does it follow that $\tau'$ is induced by a cup product?

Thank you

Answer 1

The homology analogue of the result stated by Milne is that the transgression in the Lyndon-Hochschild-Serre spectral sequence for group homology $$ H_2(G;M)\to H_0(G;H_1(C;M)) = H_1(C;M)_G $$ is given (up to a sign) by cap product with the extension class $H^2(G;C)$. This makes sense since the cap product is a map $$ H^2(G;C)\otimes H_2(G;M)\to H_0(G;C\otimes M)=H_0(G;H_1(C; M)), $$ the final equality since $C$ is abelian.

This is supposedly well-known, see page 3 of this paper. I don't know a reference to an explicit proof, though.