Let $G$ be a group, $H \leq G$ a subgroup of finite index. The transfer map $\mathrm{tr} \colon H^*(H,-) \Rightarrow H^*(G,-)$ is uniquely determined as a morphism of $\delta$-functors by its behavior on the 0th cohomology groups: $\mathrm{tr}_A^0 \colon A^H \to A^G, a \mapsto \sum_{\overline{g} \in H \setminus G} ag$ ($A$ is any right $G$-module). Now there is an obvious approach to find explicit formulas for $\mathrm{tr}_A^n \colon H^n(H,A) \to H^n(G,A)$ for higher $n$, namely:
- Pick any exact sequence $0 \to A \to \hat{A} \to A^* \to A$ of $G$-modules such that $H^k(H,\hat{A}) = 0$ for all $k$. E.g. take $\hat{A} = \mathrm{CoInd}^G(A) = \mathrm{Hom}_\mathbb{Z}(G,A)$.
- Derive a formula for $\mathrm{tr}^n$ inductively by looking at the commutative diagram: $\begin{array}{ccc} H^{n-1}(H,A^*) & \xrightarrow{\delta'} & H^n(H,A)\\ \downarrow{\mathrm{tr}_{A^*}^{n-1}} & & \downarrow{\mathrm{tr}_A^n}\\ H^{n-1}(G,A^*) & \xrightarrow{\delta} & H^n(G,A) \end{array}$
However, this approach becomes very complicated. Are there more elegant ways to find these formulas? I would also be happy about a reference, where those formulas are given.
PS: I consider $H^n(G,A)$ as the cohomology of the standard complex here, i.e. co-cycles modulo co-boundaries.