Proposition 3

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 $u' : S(M) \to S(N)$ such that the diagram

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

The existence and uniqueness of $u'$ follow from no. 1, Proposition 2 applied to the commutative algebra $S(N)$ and $f = \phi'_N \circ u : M \to S(N)$; as

the fact that $u'$ is a graded algebra homomorphism follows from no. 1, Remark 1.

The homomorphism $u'$ of Proposition 3 will henceforth be denoted by $S(u)$. If $P$ is an $A$-module and $v : N \to P$ an $A$-linear mapping, then

for $S(v) \circ S(u)$ is an algebra homomorphism rendering commutative the diagram

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

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