Proposition 2

Let $A$ be a commutative ring, $M$ and $N$ two $A$-modules and

an A-linear mapping. There exists one and only one A-algebra homomorphism

such that the diagram

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

The existence and uniqueness of $u'$ follow from no. 1, Proposition 1, applied to the algebra $\mathbf{T}(N)$ and the linear mapping $\phi_N \circ u : M \to \mathbf{T}(N)$; as

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

The homomorphism $u'$ of Proposition 2 will henceforth be denoted by $\mathbf{T}(u)$. If P is an A-module and $v : N \to P$ an A-linear mapping, then

for $\mathbf{T}(v) \circ \mathbf{T}(u)$ is an algebra homomorphism rendering commutative the diagram

$\mathbf{T}(u)$ is sometimes called the canonical extension of $u$ to $\mathbf{T}(M)$ (which contains $M = \mathbf{T}^1(M)$). Note that the restriction $\mathbf{T}^n(u) : \mathbf{T}^n(M) \to \mathbf{T}^n(N)$ is just the linear mapping $u^{\otimes n} = u \otimes u \otimes \cdots \otimes u$ (n times), for

since $\mathbf{T}(u)$ is an algebra homomorphism and $\mathbf{T}^1(u) = u$; the restriction $\mathbf{T}^0(u)$ to A is the identity mapping. $\mathbf{T}^n(u)$ is called the $n$-th tensor power of $u$.