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Let $G$ be a group, and let $H$ be a subgroup of index $m$. Let $A$ be a $G$-module. we have restriction $$\mathrm{Res}: H^n(G,A)\to H^n(H,A)$$ and co-restriction $$\mathrm{Cor}: H^n(H,A)\to H^n(G,A).$$ It is known that $$\mathrm{Cor}\circ \mathrm{Res}(c)=mc.$$ I don't see why the converse holds:

Why $$\mathrm{Res}\circ \mathrm{Cor}(c)=mc$$ for every $c\in H^n(H,A)$?

In this direction applying Res doesn't do anything to the function, and we stay with some sum $\sum_{i=1}^m g_i f(g_i^{-1}p)$ for every $p\in P_n$ for a projective resolution $P_n$ of $\mathbb Z$ as a $G$-module. The different terms in the sum seem like distrinct functions, and we cannot put the $g_i^{-1}$ outside because $f$ is only a $\mathbb Z H$-module homomorphism.

Here's the claim in Dummit & Foote: enter image description here

Emolga
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1 Answers1

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You say that $f$ is only an $H$-module homomorphism, but in fact it's a $G$-equivariant: we started with a $G$-invariant homomorphism, "forgot" it was $G$-invariant, and then applied the formula with the sums. Since it was $G$-invariant, each term in the formula with the sum is just $f(p)$.

EDIT: as discussed in the comments, the other claim about corestricting and then restricting is false. Reference Neukirch Schmidt and Wingberg Corollary 1.5.7 for the correct formulation.

hunter
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  • Thank you! I now edited the question to exchange the order of restriction and co-restriction. Do you maybe see why this other direction holds? – Emolga Jul 09 '20 at 16:25
  • @Emolga the proof from Dummit-Foote you have copy/pasted is for the other direction, though, no? – hunter Jul 09 '20 at 17:31
  • From $CR = m$ you get formally $RCRC = mRC$; this implies $RC = m$ whenever $RC$ is surjective. However it is not generally true. For example, suppose $H$ is the trivial group and $G$ is the cyclic group of order $2$ with the non-trivial element acting by negation on $M = \mathbb{Z}$. Then $M^G = 0$, $M^H = \mathbb{Z}$, so corestricting and then restricting must act by $0$ on $H^0$ and not multiplication by $2$. – hunter Jul 09 '20 at 18:01
  • So it is a mistake in the book? – Emolga Jul 09 '20 at 19:16
  • sorry, just saw the last line. It's definitely false on H^0. Do they say anything afterward? – hunter Jul 09 '20 at 20:31
  • They don't. (Strange) – Emolga Jul 09 '20 at 20:41
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    I think in Dummit and Foote it's just a typo for "it follows that Cor \circ Res is mult. by m on the cohomology groups", i.e. they are just saying this descends to cycles mod boundaries. They are not trying to say that this works in the other order (which is false.) – hunter Jul 09 '20 at 22:16