Proposition 12

Let $(\phi_i: G_i \to K)_{i \in I}$ be a family of homomorphisms of groups with operators such that, for all $i \in I$ and $j \in I$ with $i \neq j$, $x \in G_i$, $y \in G_j$, the elements $\phi_i(x)$ and $\phi_j(y)$ of $K$ commute; there exists one and only one homomorphism of groups with operators $\Phi$ of $\prod_{i \in I} G_i$ into $K$ such that $\phi_i = \Phi \circ \iota_i$ for all $i \in I$. For every element $x = (x_i)_{i \in I}$ of $\prod_{i \in I} G_i$, $\Phi(x) = \prod_{i \in I} \phi_i(x_i)$.

If $\Phi$ and $\Phi$ are solutions to the problem, they coincide on $\bigcup_{i \in I} G_i$ and hence on $\prod_{i \in I} G_i$, whence the uniqueness of $\Phi$. We now show the existence of $\Phi$: for every element $x = (x_i)_{i \in I}$ of $\prod_{i \in I} G_i$, let $\Phi(x) = \prod_{i \in I} \phi_i(x_i)$ (§ 1, no. 5). Clearly $\Phi \circ \iota_i = \phi_i$ for all $i$ and $\Phi$ commutes with homotheties; the formula $\Phi(xy) = \Phi(x)\Phi(y)$ follows from § 1, no. 5, formula (9).