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Suppose $G$ is a group and $0 \to A \to B \to C \to 0$ a short exact sequence of abelian groups considered as trivial $G$-modules. There is a corresponding long exact sequence in cohomology. I wonder whether the morphisms in this long sequence are compatible with restrictions to subgroups of $G$.

More precisely, let $H$ be a subgroup of $G$. Then, we have a diagram \begin{array}{c} \dots &\to& H^n(G,A) &\to& H^n(G,B) &\to& H^n(G,C) &\to& H^{n+1}(G,A) &\to& \dots \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ \dots &\to& H^n(H,A) &\to& H^n(H,B) &\to& H^n(H,C) &\to& H^{n+1}(H,A) &\to& \dots \end{array} where the arrows pointing down are cohomological restrictions. Is this diagram commutative?

Pustam Raut
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darko
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  • I'm not an expert, but I think this follows from the "naturality" of the snake lemma. This question (https://math.stackexchange.com/questions/3629887/naturality-of-the-ker-coker-sequence-of-snake-lemma-in-an-abelian-category) might be useful – Sam Ballas May 31 '24 at 13:46
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    Yes. See proposition 1.5.2 of Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg. – Snacc Jun 02 '24 at 02:44
  • Wonderful! @Snacc, please post your comment as an answer so I can mark this question as answered. – darko Jun 03 '24 at 09:14

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This is true. See proposition 1.5.2 of Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg. The proof comes down to working on the level of cocycles and tracking through the definitions of the maps explicitly.

Snacc
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