Proposition 1

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 $g : \bigwedge(M) \to E$ such that $f = g \circ \phi''$.

In other words, $(\bigwedge(M), \phi'')$ 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 from the $A$-module $M$ to an $A$-algebra satisfying (1).

The uniqueness of $g$ follows from the fact that $\phi''(M) = M$ generates $\bigwedge(M)$. To prove the existence of $g$, we note that by § 5, no. 1, Proposition 1 there exists an $A$-algebra homomorphism $g_1 : T(M) \to E$ such that $f = g_1 \circ \phi$; we need to prove that $g_1$ is zero on the ideal $\mathfrak{J}''$, for then if

is the canonical homomorphism, we can write $g_1 = g \circ p$, where $g : \bigwedge(M) \to E$ 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 $x \otimes x$ for $x \in M$. This completes the proof.