Let $A$ be a ring, with $S\subset A$ multiplicatively closed subset, and let $f:A\to A_S$ be the canonical map. Suppose that there is a ring $B$, together with two maps $g:A\to B$, $h:B\to A_S$, such that $f=hg$ and, for all $b\in B$, exists $s\in S$ with $b g(s)\in g(A)$. Finally, define $T$ as $h^{-1}(A_S^\times)$. I shall prove that $T$ is the saturation of $ g(S)$; i.e. that, for any $t\in B$, $h(t)$ is a unit iff exists $b\in B$ with $b t\in g(S)$.
One direction is trivial: if $bt\in g(S)$, then $h(b)h(t)\in f(S)$, so $h(t)$ is a unit. I really don't see how to prove that, assuming $h(t)\in A_S^\times$, exists $b\in B$ with $bt\in g(S)$; can you give me a hint?
EDIT 1. I looked back at this problem and I think that I have a solution. For $b\in B$, by hypothesis there are $a\in A,s\in S$ s.t. $g(a)=bg (s)$, so that $h(b)=\frac as$. If $\frac as\in A_S^\times$, exists $a'\in A$ s.t. $aa'\in S$, so $g(aa')=bg(sa')$, meaning that $b$ divides $g(S)$. Conversely, $h(g(S))\subseteq A_S^\times$, and so $h$ sends any divisor of $g(S)$ to a unit.
EDIT 2. I still have a question related. Matsumura proves that $h$ is isomorphic to the localization map $B\to B_T$, saying that the fact that $h$ is isomorphic to $B\to B_{g(S)}$ can be proved separately (similarly), or deduced proving that $T$ is the saturation of $g(S)$. If instead one proves that $h$ is isomorphic to $B\to B_{g(S)}$, does it follow already (without the saturation check) that $h$ is isomorphic to $B\to B_T$? I would say yes, because I think that in general if $R\to R_S$ is a localization map, the saturation of $S$ is always $R_S^\times\cap R$. Do you agree?