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32. The sets H n (Γ/U, AU ) together with the maps inf U,U form a directed system of pointed sets (resp. of groups if A is abelian). Moreover, we have fU = fU ◦ inf U,U . We now come to the main result of this section. 33. Let Γ be a proﬁnite group, and let A be a Γ-group. Then we have an isomorphism of pointed sets (resp. an isomorphism of groups if A is abelian) n U lim −→ H (Γ/U, A ) H n (Γ, A). U ∈N If [ξU ] ∈ H n (Γ/U, AU ), this isomorphism maps [ξU ]/∼ onto fU ([ξU ]). Proof. We ﬁrst prove that there exists a well-deﬁned map n U n f: − lim → H (Γ/U, A ) −→ H (Γ, A), U ∈N which sends the equivalence class of [ξU ] ∈ H n (Γ/U, AU ) onto fU ([ξU ]).

We would like to observe now that the map ϕ∗ depends on ϕ only up to conjugation. For, let ρ ∈ G and set ψ = Int(ρ) ◦ ϕ. Then ψ ∗ ([α]) is represented by the cocycle γ deﬁned by γ: Γn −→ A (σ1 , . . ,ρϕ(σn )ρ−1 . 21. Assume now that Γ also acts trivially on G, so that H 1 (Γ, G) is nothing but the set of conjugacy classes of continuous morphisms ϕ : Γ −→ G. The previous observations then imply that for each class [α] ∈ H n (G, A), we have a well-deﬁned map H 1 (Γ, G) −→ H n (Γ, A) [ϕ] −→ ϕ∗ ([α]).

5. There is a natural bijection of pointed sets between the orbit set C Γ /B Γ and ker(H 1 (Γ, A) −→ H 1 (Γ, B)). More precisely, the bijection sends the orbit of c ∈ C Γ onto δ 0 (c). Proof. 4, we have ker(H 1 (Γ, A) −→ H 1 (Γ, B)) = im(δ 0 ). Hence we have to construct a bijection between C Γ /B Γ and im(δ 0 ). Let c, c ∈ C Γ lying in the same orbit, that is c = β·c for some β ∈ B Γ . Then c = g(βb), for some preimage b ∈ B of c, and βb is a preimage of c . Since we have (βb)−1 σ·(βb) = b−1 β −1 (σ·β)(σ·b) = b−1 σ·b, it turns out that δ 0 (c ) = δ 0 (c).