Proposition 9

Tag 06D6, permanent. Cite it as [Bourbaki, Tag 06D6].

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

— Let M and N be free A-modules and u an A-homomorphism of M into N ; then $\mathbf{TS}(u)$ is a bigebra homomorphism.

For we have $\Delta_N \circ u = (u,u) \circ \Delta_M$ , hence the diagram

\begin{array}{ccccccccc} \mathbf{TS}(M) & \xrightarrow{\mathbf{TS}(\Delta_M)} & \mathbf{TS}(M \oplus M) & \xrightarrow{\sigma} & \mathbf{TS}(M) \otimes \mathbf{TS}(M) \\ \downarrow \mathbf{TS}(u) & & \downarrow \mathbf{TS}(u,u) & & \downarrow \mathbf{TS}(u) \otimes \mathbf{TS}(u) \\ \mathbf{TS}(N) & \xrightarrow{\mathbf{TS}(\Delta_N)} & \mathbf{TS}(N \oplus N) & \xrightarrow{\tau} & \mathbf{TS}(N) \otimes \mathbf{TS}(N), \end{array}

where $\sigma$ and $\tau$ are canonical isomorphisms, is commutative (IV, p. 49). Thus $c_N \circ \mathbf{TS}(u) = (\mathbf{TS}(u) \otimes \mathbf{TS}(u)) \circ c_M$ .

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

Cited by

Corollary, § 4

Cites

IV, p. 49