I wonder if there is a concrete formula to express the isomorphism in the well known Shapiro's Lemma that $H^i(G, \text{CoInd}_{H}^{G}(M)) \simeq H^i(H, M)$, where $H \subset G$ is a subgroup of $G$, $M$ is a $\mathbb{Z}[H]$-module, and $\text{CoInd}_{H}^{G}(M)$ is the co-induced $\mathbb{Z}[G]$-module. Here by 'concrete' I mean, for instance, given an $i$-cocycle $\phi \in Z^i(H, M)$ considered as a map $\phi: H \times \cdots \times H \longrightarrow M$, I wish to find a corresponding cocycle in $Z^i(G, \text{CoInd}_{H}^{G}(M))$, or vice versa.
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2$\begingroup$ Let $\pi: \text{Coind}^G_H(M) = Hom_H(\mathbb{Z}G,M)\to M,\, x \mapsto x(1)$. The isomorphism $\varphi: H^\ast(G,\text{Coind}^G_H(M)) \to H^\ast(H,M)$ from Shapiro's lemma can be described in the following way: Let $f: G \times \cdots \times G \to Hom_H(\mathbb{Z}G,M)$ be a cocycle. Then $\varphi([f])$ is represented by the cocycle $\pi \circ f: H \times \cdots \times H \to M$ (see Brown, Cohomology of Groups, III.8, exercise 2). $\endgroup$– tj_Commented Jan 9, 2017 at 2:15
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$\begingroup$ It's more interesting to define the inverse and to show that they compose on cocycle level to the identity up to a coboundary. $\endgroup$– tj_Commented Jan 9, 2017 at 2:17
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1$\begingroup$ A related question: mathoverflow.net/a/256208/7709. $\endgroup$– Mark WildonCommented Jan 9, 2017 at 11:36
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2$\begingroup$ @tj_: Could you explain what the inverse is? $\endgroup$– Rene RecktenwaldCommented May 8, 2018 at 14:15
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1 Answer
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D. Naidu constructs an explicit inverse map $H^n(H,A) \rightarrow H^n(G,\operatorname{CoInd}_G^H(A))$ for the case $n=1,2$. For the construction and proof, check out lemma 2.1, 2.2 in his paper Categorical Morita equivalence for group-theoretical categories.
