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I'm looking for a detailed proof of the Hochschild-Serre spectral sequence for Galois Cohomology:

If we have $H$ a normal subgroup of $G$ (profinite group) and $T$ is a $G-$module, we get:

$$ 0\rightarrow H^1(G/H, A^H)\overset{inf}{\rightarrow} H^1(G, A) \overset{res}{\rightarrow} H^1(H, A)^{G/H}\overset{d}{\rightarrow}H^2(G/H, A^H)\overset{inf}{\rightarrow} H^2(G, A) $$

Just from the statement I don't understand one thing:

  1. How $G/H$ acts on $H^1(H, A)$;

I'm studying this to understand: $$ 0\rightarrow H^1(K^{ur}/K, T^{I_K})\overset{inf}{\rightarrow}H^1(K,T)\overset{res}{\rightarrow} H^1(I_K,T)^{G^{ur}_K}\rightarrow 0 $$ (if there is a simper way to see this case please let me know $\overset{o\ o}{\smile}$ )

Thank you in advance!

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    $\begingroup$ Just to clarify, your question is about the action of $G/H$ on $H^1(...)$, and not the whole spectral sequence? You can do this explicitly using cocycles. In Hochschild and Serre's original paper has detailed discussion of this. $\endgroup$
    – Donu Arapura
    Commented Jun 20, 2021 at 13:40
  • $\begingroup$ I would like to understand three things: the action of $G/H$; why the sequence is exact in $H^1(H,A)^{G/H}$; why H^2(K^{ur}/K, T^{I_K})=0. $\endgroup$
    – marco
    Commented Jun 20, 2021 at 13:45
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    $\begingroup$ There is a very clear explanation in Serre's Local fields, ch. VII, §6. $\endgroup$
    – abx
    Commented Jun 20, 2021 at 13:57
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    $\begingroup$ What you've listed is not a spectral sequence, it's a consequence of the Hochschild-Serre spectral sequence. It can be proved directly (as someone else commented), but if you really want to understand it, you can read the original paper for the proof of the actual spectral sequence, and also you can learn how to extract the exact sequence you listed from the lower-left portion of a spectral sequence diagram at the E^2_{pq} stage. $\endgroup$
    – Joe Silverman
    Commented Jun 20, 2021 at 14:54
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    $\begingroup$ It is Theorem 2.1.5 in the printed version of "Cohomology of Number Fields". In there is also one of the few places that actually shows that the differnential $d$ is the transgression map; see the theorem after. The action, inflation, restriction etc are well explained in the pages before. $\endgroup$
    – Chris Wuthrich
    Commented Jun 20, 2021 at 16:53

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