Proposition 1

Let $E$ be a (unital) $A$-algebra and $f : M \to E$ an $A$-linear mapping. There exists one and only one $A$-algebra homomorphism $g : T(M) \to E$ such that $f = g \circ \phi$.

In other words, $(T(M), \phi)$ is a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1), where $\Sigma$ is the species of $A$-algebra structure, the $\alpha$-mappings being the $A$-linear mappings from the module $M$ to an $A$-algebra. Observe that here there is no question of a graduation on $T(M)$.

For every finite family $(x_i)_{1 \leq i \leq n}$ of $n$ elements of $M$, by definition of the product in $T(M)$, $x_1 \otimes x_2 \otimes \cdots \otimes x_n = \phi(x_1)\phi(x_2)\ldots\phi(x_n)$; then necessarily $g(x_1 \otimes x_2 \otimes \cdots \otimes x_n) = f(x_1)f(x_2)\ldots f(x_n)$ for $n \geq 1$ and $g(\alpha) = \alpha e$ (if $e$ is the unit element of $E$) for $\alpha \in A$, which proves the uniqueness of $g$. Conversely, note that, for all $n > 0$, the mapping

of $M^n$ into $E$ is $A$-multilinear; hence there corresponds to it an $A$-linear mapping $g_n : T^n(M) \to E$ such that Also we define the mapping $g_0 : T^0(M) \to E$ as equal to $\eta_E$ ($§ 1$, no. 3), in other words $g_0(\alpha) = \alpha e$ for $\alpha \in A$. Let $g$ be the unique $A$-linear mapping of $T(M)$ into $E$ whose restriction to $T^n(M)$ is $g_n$ ($n \geq 0$); it is immediate that $g \circ \phi = g_1 = f$ and it remains to verify that $g$ is an $A$-algebra homomorphism. By construction $g(1) = e$ and it suffices by linearity to show that $g(uv) = g(u)g(v)$ for $u \in T^p(M)$ and $v \in T^q(M)$ ($p > 0, q > 0$); now it follows from formulae (1) and (2) that this relation is true when $u = x_1 \otimes x_2 \otimes \cdots \otimes x_p$ and $v \in x_{p+1} \otimes \cdots \otimes x_{p+q}$ (where the $x_i$ belong to $E$). It is therefore true for $u \in T^p(M)$ and $v \in T^q(M)$ by linearity.