Corollary

Let M and N be two projective A-modules and $f : M \to N$ an A-linear mapping. If f is injective, then $\Lambda(f) : \Lambda(M) \to \Lambda(N)$ is injective.

We prove this first under the assumption that M is free; let $(e_\lambda)_{\lambda \in L}$ be a basis of M, so that $(e_J)_{J \in \mathfrak{B}(L)}$ (formula (18)) is a basis of $\Lambda(M)$. Suppose that the kernel of $\Lambda(f)$ contains an element $u = \sum_J \alpha_J e_J \neq 0$. Let K be an element minimal among the finite subsets J such that $\alpha_J \neq 0$ and let H be a finite subset of L disjoint from K and such that $K \cup H$ contains all the J (finite in number) such that $\alpha_J \neq 0$; for all $J \neq K$ such that $\alpha_J \neq 0$, we have therefore by definition $J \cap H \neq 0$ and consequently (no. 8, formula (20)).

belongs to the two-sided ideal of $\Lambda(M)$, the kernel of $\Lambda(f)$. We write $e_{K \cup H} = e_{\lambda_1} \wedge e_{\lambda_2} \wedge \cdots \wedge e_{\lambda_n}$; then $\alpha_K f(e_{\lambda_1}) \wedge f(e_{\lambda_2}) \wedge \cdots \wedge f(e_{\lambda_n}) = 0$; by virtue of Proposition 12, the elements $f(e_{\lambda_i})$ ($1 \leq i \leq n$) of N are linearly dependent. But this contradicts the hypothesis that $f$ is injective (II, § 1, no. 11, Corollary 3 to Proposition 17).

We now consider the general case; let $M'$ be an A-module such that $M \oplus M' = P$ is free (II, § 2, no. 2, Proposition 4). Consider the linear mapping $g : M \oplus M' \to N \oplus M \oplus M'$ such that $g(x, y) = (f(x), 0, y)$, so that there is a commutative diagram

where the vertical arrows are the canonical injections. Since $g$ is injective and $P$ is free, $\Lambda(g)$ is an injective homomorphism as has been seen above. Now, $\Lambda(j): \Lambda(M) \to \Lambda(P)$ is an injective homomorphism since $M$ is a direct factor in $P$ (no. 2). The composite homomorphism

is therefore injective and, as it is also equal to the composite homomorphism

we conclude that $\Lambda(f)$ is injective.