Proposition 5

Let $(E_i)_{i \in I}$ be a finite family of $A$-algebras and, for each $i \in I$, let $e_i$ be the unit element of $E_i$. For each $i \in I$, let $u_i : E_i \to E = \bigotimes_{i \in I} E_i$ be the $A$-linear mapping defined by

(i) *The $u_i$ are $\mathbf{A}$-algebra isomorphisms; further, for $i \neq j$, the elements $u_i(x_i)$ and $u_j(x_j)$ are permutable in $\mathbf{E}$ for all $x_i \in \mathbf{E}_i$ and $x_j \in \mathbf{E}_j$ and $\mathbf{E}$ is generated by the union of the subalgebras $u_i(\mathbf{E}_i)$.

(ii) Let $\mathbf{F}$ be an $\mathbf{A}$-algebra and, for all $i \in \mathbf{I}$, let $v_i : \mathbf{E}_i \to \mathbf{F}$ be an $\mathbf{A}$-algebra homomorphism, where the $v_i$ are such that, for $i \neq j$, $v_i(x_i)$ and $v_j(x_j)$ are permutable in $\mathbf{F}$ for all $x_i \in \mathbf{E}_i$ and $x_j \in \mathbf{E}_j$. Then there exists one and only one $\mathbf{A}$-algebra homomorphism $w : \mathbf{E} \to \mathbf{F}$ such that

(i) The mapping $u_i$ is an algebra homomorphism by definition of the multiplication on $\mathbf{E}$. If $i \neq j$, $x_i \in \mathbf{E}_i, x_j \in \mathbf{E}_j$, then

Clearly $x'_k x''_k = x''_k x'_k$ for all $k \in \mathbf{I}$ and hence $u_i(x_i)$ and $u_j(x_j)$ commute in $\mathbf{E}$ by formula (1) (no. 1) defining the multiplication in $\mathbf{E}$. The last assertion follows from the relation $\bigotimes_i x_i = \prod_{i \in \mathbf{I}} u_i(x_i)$.

(ii) For each $i \in \mathbf{I}$, let $x_i$ be an element of $\mathbf{E}_i$. The product $\prod_{i \in \mathbf{I}} v_i(x_i)$ is then defined in $\mathbf{F}$ independently of any ordering on $\mathbf{I}$ since the algebra $\mathbf{F}$ is associative and the elements $v_i(x_i)$ are pairwise permutable. The mapping $(x_i)_{i \in \mathbf{I}} \to \prod_{i \in \mathbf{I}} v_i(x_i)$ of $\prod_{i \in \mathbf{I}} \mathbf{E}_i$ into $\mathbf{F}$ is obviously $\mathbf{A}$-multilinear and there therefore exists one and only one $\mathbf{A}$-linear mapping $w : \mathbf{E} \to \mathbf{F}$ such that

Now, the desired $\mathbf{A}$-algebra homomorphism $w : \mathbf{E} \to \mathbf{F}$ must satisfy (5), which follows from (4) and the fact that $\bigotimes_i x_i = \prod_{i \in \mathbf{I}} u_i(x_i)$. This proves the uniqueness of $w$; it remains to show that the $\mathbf{A}$-linear mapping $w$ defined by (5) is an $\mathbf{A}$-algebra homomorphism and satisfies (4). The fact that $w$ satisfies (4) is obvious: it suffices to apply (5) to the case where $x_j = e_j$ for $j \neq i$ and we obtain $w(u_i(x_i)) = v_i(x_i)$. Finally, $w$ is an algebra homomorphism, for

since $v_i(x_i)$ commutes with $v_j(y_j)$ for $j \neq i$; hence

which, by linearity, completes the proof.

The ordered pair consisting of E and the canonical mapping $\phi : (x_i) \mapsto \bigotimes_i x_i$ of $\prod_i E_i$ into E is a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1) where $\Sigma$ is the species of A-algebra structure, the morphisms being the A-algebra homomorphisms and the $\alpha$-mappings the mappings $\prod_i u_i$ of $\prod_i E_i$ into an A-algebra such that the $u_i$ are A-algebra homomorphisms and $u_i(x_i)$ and $u_j(x_j)$ commute for $i \neq j$, for all $x_i \in E_i$ and $x_j \in E_j$.