Proposition 6

The notation is that of Definition 7 and $\varepsilon$ denotes the canonical homomorphism of $E$ into $E_S$. (i) Let $a$ and $b$ be in $E$; in order that $\varepsilon(a) = \varepsilon(b)$, it is necessary and sufficient that there exist $s \in S'$ with $sa = sb$. (ii) For $\varepsilon$ to be injective it is necessary and sufficient that every element of $S$ be cancellable. (iii) For $\varepsilon$ to be bijective it is necessary and sufficient that every element of $S$ be invertible.

Assertion (i) is clear and shows that $\varepsilon$ is injective if and only if every element of $S'$ is cancellable; but since the set of cancellable elements of $E$ is a submonoid of $E$ (no. 2, Proposition 2), it amounts to the same to say that every element of $S$ is cancellable.

† It is also called monoid of differences if the law on $E$ is written additively.

If $\varepsilon$ is bijective every element of S is invertible, for $\varepsilon(S)$ is composed of invertible elements of $E_S$. Conversely, suppose every element of S is invertible; then every element of $S'$ is invertible (no. 3, Corollary 2 to Proposition 4) and hence cancellable. Then $\varepsilon$ is injective by (ii) and $a/p = \varepsilon(a.p^*)$ by (1), hence $\varepsilon$ is surjective.