Proposition 2

Let $A$ be a commutative ring, $M$ and $N$ two $A$-modules and $u : M \to N$ an $A$-linear mapping. There exists one and only one $A$-algebra homomorphism

such that the diagram

is commutative. Moreover, $u''$ is a graded algebra homomorphism.

The existence and uniqueness of $u'$ follow from no. 1, Proposition 1 applied to the algebra $\bigwedge(N)$ and $f = \phi_N'' \circ u : M \to \bigwedge(N)$; for $f(M) \subset N$ and hence $f$ satisfies condition (1) by definition of $\mathfrak{g}_N''$: as $f(M) \subset \bigwedge^1(N) = N$, the fact that $u''$ is a graded algebra homomorphism follows from Remark 1 of no. 1.

The homomorphism $u''$ of Proposition 2 will henceforth be denoted by $\bigwedge(u)$. If $P$ is an $A$-module and $v : N \to P$ is an $A$-linear mapping, then

for $\bigwedge(v) \circ \bigwedge(u)$ is an algebra homomorphism which renders commutative the diagram

Since $\Lambda(M)$ contains $M = \Lambda^1(M)$, $\Lambda(u)$ is sometimes called the canonical extension of $u$ to $\Lambda(M)$. The restriction $\Lambda^n(u): \Lambda^n(M) \to \Lambda^n(N)$ is such that

with the $x_i \in M$, since $\Lambda(u)$ is an algebra homomorphism and $\Lambda^1(u) = u$; the restriction $\Lambda^0(u)$ to $A$ is the identity mapping. Note that $\Lambda^n(u)$ is obtained from $T^n(u): T^n(M) \to T^n(N)$ by passing to the quotients.