Proposition 9

Tag 066S, permanent. Cite it as [Bourbaki, Tag 066S].

Let $A, B$ be two graded $K$ -algebras, $E$ an $(A, A)$ -bimodule and $F$ a graded $(B, B)$ -bimodule; let $\rho : A \to B$ be a graded algebra homomorphism and $\theta : E \to F$ a graded $A$ -homomorphism of $A$ -bimodules (relative to $\rho$ ), of degree 0. Then:

(i) For every $\varepsilon$ -derivation $d' : B \to F$ , $d' \circ \rho : A \to \rho^*(F)$ is an $\varepsilon$ -derivation of the same degree. (ii) For every $\varepsilon$ -derivation $d : A \to E$ , $\theta \circ d : A \to \rho^*(F)$ is an $\varepsilon$ -derivation of the same degree.

The two assertions follow immediately from the relations

d'(\rho(xy)) = d'(\rho(x)\rho(y)) = d'(\rho(x))\rho(y) + \varepsilon_{\delta', \xi}\rho(x)d'(\rho(y))

\theta(d(xy)) = \theta((dx)y + \varepsilon_{\delta, \xi}x(dy)) = \theta(dx)\rho(y) + \varepsilon_{\delta, \xi}\rho(x)\theta(dy)

for

x \in A

homogeneous of degree

\xi

and

y \in A

\delta

and

\delta'

denoting the respective degrees of

d

and

d'

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

Cited by

Cites

Nothing.