I am interested in whether the transgression maps for group cohomology and group homology are related via a version of the universal coefficient theorem.
Let $G$ be a group, $H$ a normal subgroup of $G$ and let $A$ and $B$ finite rank free $\mathbb{Z}$-modules equipped with actions of $G$ and a $G$-equivariant perfect pairing
$$A \times B \rightarrow \mathbb{Z}.$$
The Lyndon-Hochschild-Serre spectral sequences in cohomology and homology give transgression maps:
$$d^2 : \operatorname{H}^1(H,B \otimes \mathbb{C}^{\times})^{G/H} \rightarrow \operatorname{H}^2(G/H, (B \otimes \mathbb{C}^{\times})^H),$$
$$d_2 : \operatorname{H}_2(G/H, A_H) \rightarrow \operatorname{H}_1(H, A)_{G/H}.$$
Assume that $H$ acts trivially on $A$ and $B$, and suppose that the order of $G/H$ is $m < \infty$.
Since $H$ acts trivially on $B$, the universal coefficients theorem gives a map
$$\operatorname{H}^1(H,B \otimes \mathbb{C}^{\times}) \rightarrow \operatorname{Hom}(\operatorname{H}_1(H, \mathbb{Z}), B \otimes \mathbb{C}^{\times}).$$
Since $A$ is torsion-free, $B \otimes \mathbb{C}^{\times} = \operatorname{Hom}(A, \mathbb{C}^{\times})$, so $$\operatorname{Hom}(\operatorname{H}_1(H,\mathbb{Z}),B \otimes \mathbb{C}^{\times})^{G/H} = \operatorname{Hom}(\operatorname{H}_1(H,\mathbb{Z}) \otimes A, \mathbb{C}^{\times})^{G/H} $$ $$= \operatorname{Hom}(\operatorname{H}_1(H, A), \mathbb{C}^{\times})^{G/H} = \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H}, \mathbb{C}^{\times}).$$
We thus get an isomorphism between the domain of $d^2$ and dual of the codomain of $d_2$. We can't use the universal coefficient theorem for the other pair since $G/H$ may not act trivially, but we can replace it as follows. Tate-Nakayama duality says that cup product induces a perfect pairing $$\hat{\operatorname{H}}^n(G/H, A) \times \hat{\operatorname{H}}^{-n}(G/H, B) \rightarrow \mathbb{Z} / m \mathbb{Z},$$ where $\hat{\operatorname{H}}$ denotes Tate cohomology.
Via the exponential sequence $$0 \rightarrow B \rightarrow B \otimes \mathbb{C} \rightarrow B \otimes \mathbb{C}^{\times} \rightarrow 0$$ one obtains an isomorphism
$$\operatorname{H}^2(G/H, (B \otimes \mathbb{C}^{\times})^H) = \operatorname{H}^2(G/H, B \otimes \mathbb{C}^{\times}) = \operatorname{H}^3(G/H, B).$$
Following this isomorphism by Tate-Nakayama duality, we finally get a partial diagram
\begin{array}{cccc} \operatorname{H}^1(H,B \otimes \mathbb{C}^{\times})^{G/H} &\xrightarrow{d^2}&\operatorname{H}^2(G/H, B \otimes \mathbb{C}^{\times}) \\ \downarrow & & \downarrow \\ \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H},\mathbb{C}^{\times}) & & \operatorname{Hom}(\operatorname{H}_2(G/H,A),\mathbb{C}^{\times}) \end{array}
Does adding $$d_2^{\vee} : \operatorname{Hom}(\operatorname{H}_1(H,A)_{G/H}, \mathbb{C}^{\times}) \rightarrow \operatorname{Hom}(\operatorname{H}_2(G/H, A_H), \mathbb{C}^{\times})$$ make the diagram commute? Does anyone know a reference for such a statement?
