Bourbaki

Remark 1

Tag 004A, permanent. Cite it as [Bourbaki, Tag 004A].

Algebra, VIII, p. 60, § 4. Semisimple Modules

Let V and V′V' be B-modules. The adjunction isomorphism

γ:\gamma : HomA(T(V),T(V′))⟶_A(\mathscr{T}(V),\mathscr{T}(V'))\longrightarrow HomB(V,H(T(V′)))_B(V,\mathscr{H}(\mathscr{T}(V')))

satisfies the relation γ(T(f))=βV′∘f\gamma (\mathscr{T}(f)) =\beta_{V'}\circ f for every f∈f\in HomB(V,V′)_B(V,V') because

(γ(T(f))(v))(p)=T(f)(p⊗v)=p⊗f(v)=(βV′∘f)(v)(p) (\gamma (\mathscr{T}(f))(v))(p) =\mathscr{T}(f)(p\otimes v) =p\otimes f(v) = (\beta_{V'}\circ f)(v)(p)

Let M and M′M' be A-modules; the inverse of the adjunction isomorphism

γ−1:\gamma^{-1}: HomB(H(M),H(M′))⟶_B(\mathscr{H}(M),\mathscr{H}(M'))\longrightarrow HomA(T(H(M)),M′)_A(\mathscr{T}(\mathscr{H}(M)),M')

satisfies the relation γ−1(H(u))=u∘αM\gamma^{-1}(\mathscr{H}(u)) =u\circ \alpha_M for every u∈u\in HomB(M,M′)_B(M,M'). Indeed, we have the relations

γ−1(H(u))(p⊗v)=(H(u)(v))(p)=u(v(p))=u∘αM(p⊗v) \gamma^{-1}(\mathscr{H}(u))(p\otimes v) = (\mathscr{H}(u)(v))(p) =u(v(p)) =u\circ \alpha_M(p\otimes v)

for all u∈u\in HomB(M,M′),v∈H_B(M,M'),v\in \mathscr{H} (M), and p∈Pp\in P.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.