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For a seminar on group cohomology I want to avoid having the students talk about group rings and resolutions for the time being and do everything with explicit inhomogenous cochains.

Now let $A$ be a $G$ module where $G$ is a (finite) group. Let $H$ be a subgroup (not necessarily normal). Let $CoInd_H^G(A)=Map(G/H,A)$ be the coinduced module with the $G$ action given by $(g.f)(x)=g.(f(g^{-1}x))$ for $x\in G/H$.

Evaluation at $H$ gives a map $p:Map(G/H,A)\rightarrow A$.

Given a cochain $f:G^n\to CoInd_H^G(A)$ we obtain a map $p\circ f|_H:H^n\to A$. This induces $H^n(G,CoInd_H^G(A))\rightarrow H^n(H,A)$. I want to see that this is an isomorphism, but I cannot find the inverse.

Rene Recktenwald
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  • For $n=0$, the composite map $$A^H/N_H(A)\hookrightarrow (CoInd_G^H(A))^H/N_H(CoInd^H_G(A))\stackrel{cor}{\to}(CoInd_G^H(A))^G/N_G(CoInd^H_G(A)),$$ gives the inverse map for Shapiro's isomorphism. Since, corestriction and inclusions can be viewed as arising from dimension 0, I think $H^n(H,A)\hookrightarrow H^n(H,CoInd_G^H(A))\stackrel{cor}{\to}H^n(G,CoInd_G^H(A))$ must be the inverse to Shapiro's isomorphism in dimension $n$. – Himanshu Shukla Sep 22 '21 at 12:49

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D. Naidu constructs an explicit inverse map $H^n(H,A) \rightarrow H^n(G,CoInd_G^H(A))$ for the case $n=1,2$. For the construction and proof, check out lemma $2.1, 2.2$ in his paper https://arxiv.org/pdf/math/0605530.pdf.

Harshit Yadav
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