By Steven Dale Cutkosky

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StabΓ (a) 30 Cohomology of proﬁnite groups It follows that StabΓ (a) is open. This concludes the proof. At this point, we may deﬁne the 0th -cohomology set H 0 (Γ, A). 4. For any Γ-set A, we set H 0 (Γ, A) = AΓ . If A is a Γ-group, this is a subgroup of A. The set H 0 (Γ, A) is called the 0th cohomology set of Γ with coeﬃcients in A. 5. We will use this notation only episodically in this book, and will prefer the notation AΓ . We would like now to deﬁne the main object of this chapter, namely the ﬁrst cohomology set H 1 (Γ, A).

N ∈ Γ. Proof. For every s = (σ1 , . . ,σn . (1) ⇒ (2) Assume that α is continuous, and let s = (σ1 , . . , σn ) ∈ Γn . Then the set Us = α−1 ({αs }) is an open neighbourhood of s, since {αs } is open in A and α is continuous. By deﬁnition, α is constant on Us . (2) ⇒ (3) Assume that α is locally constant. For all s = (σ1 , . . , σn ) ∈ Γn , let Us be an open neighbourhood of s on which α is constant. By deﬁnition of the product topology, one may assume that Us = Vs(1) × · · · × Vs(n) , (i) where Vs is an open neighbourhood of σi in Γ.

It remains to prove its injectivity. Let 50 Cohomology of proﬁnite groups c, c ∈ C Γ such that δ 0 (c ) = δ 0 (c), and let α and α be the cocycles representing δ 0 (c) and δ 0 (c ) respectively. By assumption, there exists a ∈ A such that ασ = a ασ σ·a−1 for all σ ∈ Γ. If b (resp. b ) is a preimage of c (resp. c ) in B, applying f to this last equality implies that b −1 σ·b = f (a)b−1 (σ·b)(σ·f (a))−1 . It easily turns out that β = b f (a)b−1 ∈ B Γ . Hence c = g(b ) = g(b f (a)) = g(βb) = β · c.

### An Introduction to Galois Theory [Lecture notes] by Steven Dale Cutkosky

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