Exercise 17, § 1

Let E be a monoid, $e$ its identity element, S a subset of E, and $S'$ the submonoid of E generated by S. We say that S admits a calculus of left fractions on E if the following two (tautological when E is commutative) conditions are satisfied:

(i) For every $a\in E$ and every $s\in S$, there exist $b\in E$ and $t\in S'$ such that $ta=bs$.

(ii) For every $a\in E,b\in E$, and $s\in S$ such that $as=bs$, there exists a $t\in S'$ such that $ta=tb$.

a) Suppose that there exists a homomorphism $u$ from E to a monoid $E'$ such that every element of $u(S)$ is invertible in $E'$ and that for every $a, b$ in E with $u(a) =u(b)$, there exists an $s\in S'$ such that $sa=sb$. Then S admits a calculus of left fractions on E.

b) Suppose that S admits a calculus of left fractions on E. In $E\times S'$, we denote by $(a, s)\sim (b, t)$ the relation “there exist $c$ and $d$ in E such that we have $ca=db$ and $cs=dt\in S'$.” This is an equivalence relation. We set $S^{-1}E = (E\times S')/\sim$, and we denote by $\varepsilon : E\rightarrow S^{-1}E$ the mapping that sends $a\in E$ to the class of $(a, e)$. There exists a unique monoid structure on $S^{-1}E$ such that $\varepsilon$ is a unital homomorphism, that $\varepsilon (s)$ is invertible for every $s\in S$, and that for $(a, s)\in E\times S'$, the class of $(a, s)$ in $S^{-1}E$ is equal to $\varepsilon (s)^{-1}\varepsilon (a)$. We say that $S^{-1}E$ is the monoid of left fractions of E associated with S. It coincides with that defined in I, §2, No. 4, p. 18 when the monoid E is commutative.

c) Let $a, b$ be in E. We have $\varepsilon (a) =\varepsilon (b)$ if and only if there exists an $s\in S'$ such that $sa=sb$. The mapping $\varepsilon$ is injective (resp. bijective) if and only if every element of S is right regular (resp. invertible).

d) Let $f$ be a unital homomorphism from E to a monoid F such that every element of $f(S)$ is invertible in F. There is a unique homomorphism $\overline{f}: S^{-1}E\rightarrow F$ such that $f=\overline{f}\circ s$. If $f$ is injective, then $\overline{f}$ is injective. e) We say that S admits a calculus of right fractions on E if it admits a calculus of left fractions on the opposite monoid $E^o$ of E. In this case, define the monoid of right fractions ES$^{-1}$ of E associated with S, observe that we have (ES$^{-1}$)$^o= S^{-1}(E^o)$, and rewrite b) for the monoids of right fractions. If S admits a calculus of left and right fractions on E, then the monoids $S^{-1}E$ and ES$^{-1}$ are canonically isomorphic.