Bourbaki

Proposition 6

Tag 067W, permanent. Cite it as [Bourbaki, Tag 067W].

Algebra, chapter III, § 11. Cogebras, products of multilinear forms, inner products and duality

Let E be a graded cogebra over A. In order that, for every commutative A-algebra B, with the trivial graduation, the graded A-algebra of type ZHomgrA(E,B)\mathbf{Z} \mathrm{Homgr}_A(E, B) (no. 2) be anticommutative (§ 4, no. 9, Definition 7), it is necessary and sufficient that, if σg\sigma_g denotes the automorphism of the A-module E⊗AEE \otimes_A E such that

σg(xp⊗xq)=(−1)pqxq⊗xp \sigma_g(x_p \otimes x_q) = (-1)^{pq} x_q \otimes x_p

for xp∈Ep,xq∈Eqx_p \in E_p, x_q \in E_q, where p and q are arbitrary elements of N\mathbf{N}, the diagram

E↘c↙cE⊗AE→σgE⊗AE \begin{array}{ccc} E & & \\ & \searrow^c & \swarrow^c \\ E \otimes_A E & \xrightarrow{\sigma_g} & E \otimes_A E \end{array}

be commutative.

The proof is analogous to that of Proposition 2 of no. 2.

When the graded cogebra E satisfies the condition of Proposition 6, it is said to be anticocommutative.

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

Cited by

Cites

Transcribed from the printed edition, 1998, Springer. Not translated.