Bourbaki

Proposition 8

Tag 06D5, permanent. Cite it as [Bourbaki, Tag 06D5].

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— Let M be a free A-module, then with its algebra and cogebra structures, TS (M) is a graded commutative and cocommutative bigebra. The counit is the A-linear mapping ε:TS(M)→TS0(M)=A\varepsilon : TS(M) \to TS^0(M) = A which is zero on TSp(M)TS^p(M) for p>0p > 0 and such that ε(1)=1\varepsilon(1) = 1.

We know that the A-algebra TS(M)TS(M) is associative, commutative and unital. On the other hand, the coproduct is by construction a homomorphism of graded algebras; now the fact that the cogebra TS(M)TS(M) is coassociative and cocommutative follows by an easy calculation from the Formula (10). The mapping ε\varepsilon of TS(M)TS(M) into A is a homomorphism of graded algebras such that ε(l)=l\varepsilon(l) = l. Finally for every x∈Mx \in M we have ε(γp(x))=0\varepsilon(\gamma_p(x)) = 0 if p>0p > 0, ε(γ0(x))=l\varepsilon(\gamma_0(x)) = l; if we bear in mind (lo), this shows that (F⊗1)∘c=(l⊗E)∘c=IdTS(M)(F \otimes 1) \circ c = (l \otimes E) \circ c = \mathrm{Id}_{TS(M)}; thus EE is the counit of TS(M)TS(M).

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.