A Short Course on Galois Cohomology - William Stein - University of ...

Definition 29. We set H 0 (G, A) = A G = {a ∈ A : s.a = a for all s ∈ G},
which is a subgroup of A. We define H 1 (G, A) as 1-cocycles modulo an
equivalence relation. Here, a 1-cocycle is a set-theoretic map G → A, s ↦→ as
such that ast = as.s(at), and we say as ∼ bs if there exists an a ∈ A with
bs = a −1 .as.s(a) for all s ∈ G.
Remark 30. H 1 (G, A) is a pointed set, with distinguished element the
image of the 1-cocycle as = 1, i.e., the map sending the all of G to 1 ∈ A.
We have described two functors: H 0 (G, −) is a functor from the category
of non-abelian G-modules to the category of groups, and H 1 (G, −) is a
functor from the same category to the category of pointed sets. For example,
this means that a G-morphism A → B induces morphisms H 0 (G, A) →
H 0 (G, B) and H 1 (G, A) → H 1 (G, B).
We now explain how the long exact sequence can be salvaged in the
context of non-abelian G-modules. Suppose we have a short exact sequence
1 → A ι −→ B ρ −→ C → 1 (†)
of non-abelian G-modules. We want to show the existence of a connecting
morphism δ : C G → H 1 (G, A), c ↦→ [as].
Let c ∈ C G and b ∈ B such that ρ(b) = c. Then s(b) ≡ b modulo ι(A)
for all s ∈ G, since
ρ(s(b).b −1 ) = s(c).c −1 = cc −1 = 1.
So let as = ι −1 (b −1 .s(b)) ∈ A. We claim that as is a well-defined cocycle.
In order to simplify notation, let us assume A ⊂ B. Then
ast = b −1 (st)(b) = b −1 s(b)s(b −1 t(b)) = as.s(at),
showing that as is a cocycle. Moreover, if ρ(b ′ ) = ρ(b) = c then b ′ = ba for
some a ∈ A by exactness and hence
a ′ s = a −1 b −1 s(b)s(a) = a −1 ass(a) ∼ as,
which shows that as is well-defined.
If we suppose that A lies in the centre of B, so that A is abelian, we
can define H 2 (G, A) as before and also obtain a morphism ∆: H 1 (G, C) →
H 2 (G, A).
Proposition 31. (i) Let 1 → A ι −→ B ρ −→→ 1 be any exact sequence of
non-abelian G-modules. Then the sequence
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