Lemma 2

Let

be a commutative diagram of sets and mappings such that $v \circ u$ and $v' \circ u'$ are the identity mappings of $X$ and $X'$ respectively. Then, if $g$ is injective (resp. surjective, resp. bijective), so is $f$.

$u$ is injective since $v \circ u$ is, hence, if $g$ is injective, $u' \circ f = g \circ u$ is injective and therefore $f$ is injective. Similarly $v'$ is surjective since $v' \circ u'$ is; hence, if $g$ is surjective, $f \circ v = v' \circ g$ is surjective and therefore $f$ is surjective.

The last assertion of Proposition 7 follows from the fact that $\Lambda(M)$ is then a finitely generated A-module (§ 7, no. 3, Proposition 6 and II, § 11, no. 6, Remark).

We now examine what can be said concerning the homomorphism $\theta_S$ when $M$ is projective and finitely generated. Suppose first that $M$ admits a finite basis $(e_i)_{1 \leq i \leq m}$. In the notation at the beginning of the chapter the A-module $S^n(M)$ admits as basis the family of elements $e^\alpha$ such that $|\alpha| = n$. Let $u_\alpha$ (for $|\alpha| = n$) denote the element of index $\alpha$ in the basis of $(S^n(M))^*$ dual to $(e^\alpha)$. The elements $u_\alpha$, for $\alpha \in \mathbf{N}^m$, therefore form a basis of the algebra $S(M)^{*gr}$ and we shall obtain the multiplication table of this basis explicitly. We write

Then by definition

where $m : A \otimes A \to A$ defines the multiplication on $A$ and $c$ is the coproduct of $S(M)$. In other words, $a_{\alpha \beta \gamma}$ is just the coefficient of $e^\alpha \otimes e^\beta$ when $c(e^\gamma)$ is written in terms of the basis of $S(M) \otimes S(M)$ consisting of the $e^\xi \otimes e^\eta$, where $\xi$ and $\eta$ run through $\mathbf{N}^m$. But since $c$ is an algebra homomorphism,

by formula (4) of no. 1; this gives

(31)

where we write

(32)

Hence we obtain the multiplication table

(33)

On the other hand, if $(e_i^*)_{1 \leq i \leq m}$ is the basis of $M^*$, dual to $(e_i)$, it follows from formula (29 bis) that, for all $\alpha \in \mathbf{N}^m$,

(34)

in the notation of § 6, no. 6. Hence the homomorphism $\theta_S$ is bijective if and only if the $\alpha! u_\alpha$ form a basis of $S(M)^{*gr}$, or also if the elements $\alpha! 1$ are invertible.