Lemma 1

There exists an A-bilinear map from Alt$^s(M'')\times$ Alt$^r(M')$ to Alt$^{s+r}(M)$, denoted by $(u, v) \rightarrow u\cap v$, and characterized by either of the following two properties:

a) Denote by $u_1\in$ Alt$^s(M)$ the form $(x_1, . . . , x_s) \rightarrow u(p(x_1), . . . , p(x_s))$, and let $v_1\in$ Alt$^r(M)$ be a form such that $v_1(i(x'_1), . . . , i(x'_r)) =v(x'_1, . . . , x'_r)$ for $x'_1, . . . , x'_r$ in $M'$; then $u\cap v=u_1\wedge v_1$.

b) For all $x_1, . . . , x_s$ in M and $x'_1, . . . , x'_r$ in $M'$,

The map $\varphi :$ Alt$^s(M'')\otimes_A$Alt$^r(M')\rightarrow$ Alt$^{s+r}(M)$ such that $\varphi (u\otimes v) =u\cap v$ is an isomorphism of free A-modules of rank one.

The existence of a form $v_1$ satisfying condition a) follows from the fact that $\wedge \wedge^r(i)$ induces an isomorphism from $\wedge \wedge^r(M')$ to a direct factor submodule of $\wedge \wedge^r(M) ($Algebra, Chap. III, §7, no. 2). Let $v_1$ be such a form; put $u\cap v=u_1\wedge v_1$. Formula (1) is then satisfied, since if we put $i(x'_k) =x_{s+k}$ for $1\leq k\leq r$, the only element $\sigma$ of $\mathfrak{S}_{s,r}$ such that $p(x_{\sigma(i)})\not= 0$ for $1\leq i\leq s$ is the identity permutation. On the other hand, formula (1) determines $u\cap v$ uniquely: indeed, let $(e'_1, . . . , e'_r)$ be a basis of $M', (f''_1, . . . , f''_s)$ a basis of $M''$, and $f_1, . . . , f_s$ elements of M such that $p(f_i) =f''_i$ for $1\leq i\leq s$. Then $(f_1, . . . , f_s, i(e'_1), . . . , i(e'_r))$ is a basis of M (Algebra, Chap. II, §1, no. 11, Prop. 21), and formula (1) can be written

but an element of Alt$^{s+r}(M)$ is determined by its value on a basis.

It follows from the preceding that each of the conditions a) and b) determines the product $u\cap v$ uniquely; it is clear that this product is bilinear. Finally, the last assertion of the lemma follows from formula (2).