Proposition 11

The canonical homomorphism $\theta_T : T(M^*) \to T(M)^{*gr}$ (resp. $\theta_S : S(M) \to S(M)^{*gr}$, resp. $\theta_\Lambda : \Lambda(M^*) \to \Lambda(M)^{*gr}$) is a right $T(M)$-*module* (resp. right $S(M)$-*module*, resp. left $\Lambda(M)$-*module*) homomorphism.

We show first that, for $z^* \in T(M^*)$ and $t \in T(M)$,

Since $M$ is a generating system of the algebra $T(M)$, we need only prove (69) when $t = x \in M$; moreover we can restrict our attention to the case where $z^* = x_1^* \otimes x_2^* \otimes \cdots \otimes x_p^*$, where the $x_j^* \in M^*$, and then, by (66) with the roles of $M$ and $M^*$ interchanged, $z^* \wedge x = \langle x, x_1^* \rangle x_2^* \otimes \cdots \otimes x_p^*$. Therefore, for all $y_2, \ldots, y_p$ in $M$,

whence (69).

We prove secondly that for $z^* \in S(M^*)$ and $t \in S(M)$,

As above, we can limit ourselves to the case $t = x \in M$. But further, here $i(x)$ is a derivation of $S(M^*)$ and a derivation of $S(M)^{*gr}$. Therefore (§ 10, no. 7, Corollary to Proposition 9) it suffices to verify (70) for $z^* = x^* \in M^*$, since $M^*$ is a generating system of $S(M^*)$; but this is trivial, the two sides then being equal to $\langle x^*, x \rangle$. A similar argument proves the relation

for $z^* \in \Lambda(M^*)$ and $t \in \Lambda(M)$: observe then that, for $x \in M$, $i(x)$ is an antiderivation in $\Lambda(M^*)$ as well as in $\Lambda(M)^{*gr}$ and use § 10, no. 7, Corollary to Proposition 9. There is an analogous result for left inner products.