By Samuel Moy

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A1 a2 ) = (σ·a1 )(σ·a2 ) for σ ∈ Γ, a1 , a2 ∈ A. A Γ-group which is commutative is called a Γ-module. A morphism of Γ-sets (resp. Γ-groups, Γ-modules) is a map (resp. a group morphism) f : A −→ A satisfying the following property: f (σ·a) = σ·f (a) for all σ ∈ Γ and all a ∈ A. 2. (1) Assume that Γ is a finite group. Then any discrete topological set A on which Γ acts on the left is a Γ-set. Indeed such an action is continuous since any finite set is open for the discrete topology. (2) Any discrete topological set A on which Γ acts trivially is a Γ-set.

Since f2 ◦ f1 = f4 ◦ f3 and ϕ1 ◦ ϕ2 = ϕ3 ◦ ϕ4 by assumption, we get the desired result. 20, unless specified otherwise. 3 Cohomology sets as a direct limit In this paragraph, we would like to relate the cohomology of profinite groups to the cohomology of its finite quotients. 6, which says more or less that an n-cocycle α : Γn −→ A is locally defined by a family of n-cocycles α(U ) : (Γ/U )n −→ AU , where U runs through the set of open normal subgroups of Γ. ,σn for all σ1 , . . , σn ∈ Γ, it easily implies that for all U, U ∈ N , U ⊃ U , we have inf U,U ([α(U ) ]) = [α(U ) ].

G(b) = c. By assumption, we have c = σ·c for all σ ∈ Γ. Therefore, we have g(σ·b) = σ·g(b) = σ·c = c = g(b). By assumption on g, there exists a unique element ασ ∈ A such that f (ασ ) = b−1 σ·b. 2. The map α : Γ −→ A is a 1-cocycle, and its class in H 1 (Γ, A) does not depend on the choice of b ∈ B. Proof. Let us prove that α is a cocycle. By definition of α, for all σ, τ ∈ Γ, we have f (αστ ) = b−1 στ · b = b−1 σ·(bb−1 τ ·b) = (b−1 σ·b) σ·(b−1 τ · b). Hence we have f (αστ ) = f (ασ )σ·f (ατ ) = f (ασ )f (σ·ατ ) = f (ασ σ·ατ ).

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