Proposition 2

Let E be an A-algebra and $f : M \to E$ an A-linear mapping such that

There exists one and only one A-algebra homomorphism $g : S(M) \to E$ such that $f = g \circ \phi'$.

In other words, $(S(M), \phi')$ 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 linear mappings of the A-module M to an A-algebra satisfying (1).

The uniqueness of g follows from the fact that $\phi'(M) = M$ generates S(M). To prove the existence of g, note that by virtue of § 5, No. 1, Proposition 1 there exists an A-algebra homomorphism $g_1 : T(M) \to E$ such that $f = g_1 \circ \phi;$

all that needs to be proved is that $g_1$ is zero on the ideal $\mathfrak{J}'$, for then, if $p : T(M) \to S(M) = T(M)/\mathfrak{J}'$ is the canonical homomorphism, we can write $g_1 = g \circ p$, where $g : S(M) \to E$ is an algebra homomorphism, and the conclusion will follow from the fact that $p \circ \phi = \phi'$. Now the kernel of $g_1$ is a two-sided ideal which, by virtue of (1) and the relation $g_1 \circ \phi = f$, contains the elements $x \otimes y - y \otimes x$ for $x, y$ in $M$. This completes the proof.