Proposition 13

— *Let $M$ and $N$ be $A$-modules, $q$ an integer $\geqslant 0$, and $f$ a mapping of $M$ into $N$. Suppose that $M$ is free, then the following conditions are equivalent: (i) There exists a $q$-linear mapping $g$ of $M^q$ into $N$ such that $f(x) = g(x, x, \ldots, x)$ for all $x \in M$. (ii) There exists a linear mapping $h$ of $TS^q(M)$ into $N$ such that $f(x) = h(\gamma_q(x))$ for all $x \in M$. (iii) There exists a basis $(e_i)_{i \in I}$ of $M$ and a family $(u_\nu)_{\nu \in N^{(I)}, |\nu| = q}$ of elements of $N$ such that

for all $(\lambda_i) \in A^{(I)}$. (iv) For each basis $(e_i)_{i \in I}$ of $M$ there exists a family $(u_\nu)_{\nu \in N^{(I)}, |\nu| = q}$ of elements of $N$ such that for all $(\lambda_i) \in A^{(I)}$. (i) $\Rightarrow$ (ii): let $g$ satisfy (i), then there exists a linear mapping $g'$ of $T^q(M)$ into $N$ such that $g(x_1, x_2, \ldots, x_q) = g'(x_1 \otimes x_2 \otimes \ldots \otimes x)$ for any $x, \ldots, x, \in M$. Then and on writing $h = g'|TS^q(M)$ we see that condition (ii) holds. (ii) $\Rightarrow$ (i) and (iv): let $h$ satisfy the conditions of (ii). By Prop. 4, (ii) (IV, p. 47) there exists a linear mapping $g'$ of $T^q(M)$ into $N$ such that $h = g'|TS^q(M)$. Let $g$ be the $q$-linear mapping of $M$ into $N$ associated with $g'$, then for any $x \in M$ we have

whence (i). On the other hand, if $(e_i)_{i \in I}$ is a basis of M, we have, by formula (6) (IV, p. 46)

on writing $e_\nu = \prod_{i \in I} \gamma_{\nu_i}(e_i)$; we thus have

(iv) $\Rightarrow$ (iii) is clear. (iii) $\Rightarrow$ (ii) : given $(e_i), (u_\nu)$ satisfying the conditions of (iii), let us write $e_\nu = \prod_{i \in I} \gamma_{\nu_i}(e_i)$ and recall that $(e_\nu)_{|\nu|=q}$ is a basis of $\mathrm{TS}^q(M)$. Let h be the homomorphism of $\mathrm{TS}^q(M)$ into N defined by $h(e_\nu) = u_\nu$; then for each $x = \sum_i \lambda_i e_i$ in M we have