Proposition 1

For every A-algebra E, the canonical mapping $\phi_E : x \mapsto x \otimes 1$ of E into $E_{(B)}$ is an A-homomorphism of algebras. Moreover, for every B-algebra F and every $A$-homomorphism $f : E \to F$, there exists one and only one $B$-homomorphism $\bar{f} : E_{(B)} \to F$ such that $\bar{f}(x \otimes 1) = f(x)$ for all $x \in E$.

The first assertion follows immediately from the definition of multiplication in $E_{(B)}$, which gives $(x \otimes 1)(x' \otimes 1) = (xx') \otimes 1$ for $x \in E$ and $x' \in E$. The existence and uniqueness of the $B$-linear mapping $\bar{f}$ of $E_{(B)}$ into $F$ satisfying the relation $\bar{f}(x \otimes 1) = f(x)$ for all $x \in E$ follow from II, § 5, no. 1, Proposition 1; here it all amounts to verifying that $\bar{f}(yy') = \bar{f}(y)\bar{f}(y')$ for $y$ and $y'$ in $E_{(B)}$; as the elements of the form $x \otimes 1$ (with $x \in E$) generate the $B$-module $E_{(B)}$, attention may be confined to the case where $y = x \otimes 1, y' = x' \otimes 1$ with $x \in E, x' \in E$; as $yy' = (xx') \otimes 1$, the relation $\bar{f}(yy') = \bar{f}(y)\bar{f}(y')$ then follows from $f(xx') = f(x)f(x')$.

It can also be said that $f \mapsto \bar{f}$ is a canonical bijection

The ordered pair consisting of $E_{(B)}$ and $\phi_E$ is therefore a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1) where $\Sigma$ is the species of $B$-algebra structure and the $\alpha$-mappings the $A$-homomorphisms from $E$ to a $B$-algebra.