Localization of Ring Exists/Lemma 3

Lemma
The map $h(a/s) = g(a)g(s)^{-1}$ is a well defined ring homomorphism $A_S \to B$.

Proof
If $a/s = b/t \in A_S$, then there is $u \in S$ such that $atu = bsu$.

Therefore, by the homomorphism property, and the fact that $g(S) \subseteq B^\times$,


 * $g(a)g(t) = g(b)g(s) \Rightarrow g(a)g(s)^{-1} = g(b)g(t)^{-1}$

That $h$ is a homomorphism is trivial, because $g$ is a homomorphism, and $B$ is commutative