Bourbaki

Remark

Tag 06NL, permanent. Cite it as [Bourbaki, Tag 06NL].

Algebra, V, p. 76, § 10. Galois extensions

a) It is easy to show that for every KK-algebra homomorphism φ\varphi of Θ(X)\Theta(X) into KK, there exists a unique element xx of XX such that φ(f)=f(x)\varphi(f) = f(x) for all f∈Θ(X)f \in \Theta(X).

2) Let XX and YY be two finite Γ\Gamma-sets. Let TΓ(X,Y)\mathfrak{T}_\Gamma(X, Y) be the set of mappings uu of XX into YY such that u(σx)=σu(x)u(\sigma x) = \sigma u(x) for all σ∈Γ\sigma \in \Gamma and all x∈Xx \in X. For u∈TΓ(X,Y)u \in \mathfrak{T}_\Gamma(X, Y) we define a KK-algebra homomorphism u∗:Θ(Y)→Θ(X)u^*: \Theta(Y) \to \Theta(X) by u∗(f)=f∘uu^*(f) = f \circ u. For every homomorphism Ψ\Psi of Θ(Y)\Theta(Y) into Θ(X)\Theta(X) there exists a unique element uu of TΓ(X,Y)\mathfrak{T}_\Gamma(X, Y) such that Ψ=u∗\Psi = u^*.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2003, Springer. Not translated.