Theorem 5

Let $A$ be a commutative ring, $S$ a subset of $A$, $B$ a ring and $f$ a homomorphism of $A$ into $B$ such that every element of $f(S)$ is invertible. There exists one and only one $\bar{f}$ of $A[S^{-1}]$ into $B$ such that $f = \bar{f} \circ \varepsilon$.

We know (§ 2, no. 4, Theorem 1) that there exists one and only one morphism $\bar{f}$ of the multiplicative monoid $A[S^{-1}]$ into the multiplicative monoid $B$ such that $f = \bar{f} \circ \varepsilon$. Let $a, b \in A, p, q \in S'$ (stable multiplicative submonoid of $A$ generated by $S$). As the elements of $f(A)$ commute in pairs,

Hence $\bar{f}$ is a ring homomorphism.