Theorem 1

Let E be a commutative monoid, S a subset of E, $E_S$ the monoid of fractions associated with S and $\varepsilon : E \to E_S$ the canonical homomorphism. Further let f be a homomorphism of E into a monoid F (not necessarily commutative) such that every element of $f(S)$ is invertible in F. There exists one and only one homomorphism $\bar{f}$ of $E_S$ into F such that $f = \bar{f} \circ \varepsilon$.

If $\bar{f}$ is a homomorphism of $E_S$ into F such that $f = \bar{f} \circ \varepsilon$, then

for $a \in E$ and $p \in S'$, whence the uniqueness of $\bar{f}$.

Let g be the mapping of $E \times S'$ into F defined by $g(a,p) = f(a)f(p)^*$. We show that g is a homomorphism of $E \times S'$ into F. First of all,

Let $(a,p)$ and $(a',p')$ be two elements of $E \times S'$; as $a'$ and $p$ commute in E, $f(a')$ and $f(p)$ commute in F, whence $f(a')f(p)^* = f(p)^*f(a')$ by no. 3, Proposition 5. Moreover $f(pp')^* = f(p'p)^* = (f(p')f(p))^* = f(p)^*f(p')^*$ by no. 3, Proposition 4, whence We show that g is compatible with the equivalence relation R on $E \times S'$: if $(a,p)$ and $(a',p')$ are congruent mod. R, there exists $s \in S'$ with $spa' = sap'$, whence $f(s)f(p)f(a') = f(s)f(a)f(p')$. As $f(s)$ is invertible, it follows that $f(p)f(a') = f(a)f(p')$ and then by left multiplication by $f(p)^*$ and right multiplication by $f(p')^*$ Hence there exists a homomorphism $\bar{f}$ of $E_S$ into F such that $\bar{f}(a/p) = g(a,p)$, whence $\bar{f}(\varepsilon(a)) = \bar{f}(a/e) = f(a)f(e)^* = f(a)$. Hence $\bar{f} \circ \varepsilon = f$.