an isomorphism: $H^{q-1}(G,C) \xrightarrow{\delta} H^q(G,A)$.

Question

This is from P$570$ of Rotman's 'Intro to Homological Algebra'.

$A^* = Hom_\mathbb{Z}(\mathbb{Z}G, A)$ where A is a $G$-module and $G$ is a group. There is an imbedding $i: A\to Hom_\mathbb{Z}(\mathbb{Z}G, A)$and an exact sequence of $G$-modules:

$0\to A \to A^* \to C \to 0$

We have an isomorphism as follows: $H^{q-1}(G,C) \xrightarrow{\delta} H^q(G,A)$. Here $\delta$ is the connecting homomorphism of the long exact sequence.

Rotman states that the isomorphism is because $A^*$ is $G$-coinduced. However, there's no theorems ahead of this point that supports this argument. Is it because of Theorem 1.5.3 from here: https://www.math.ucla.edu/~sharifi/groupcoh.pdf

Any help would be appreciated!

Answer 1

Co-induced modules $A^*$ are acyclic for group cohomology: $H^q(G,A^*)=0$ for $q\ge1$.

Part of the long exact sequence of cohomology here is $$H^{q-1}(G,A^*)\to H^{q-1}(G,C)\stackrel\delta\to H^{q}(G,A)\to H^{q}(G,A^*).$$ For $q\ge2$ this reduces to $$0\to H^{q-1}(G,C)\stackrel\delta\to H^{q}(G,A)\to0$$ which means that $\delta$ is an isomorphism.