Proof that the localization $R_S$ is naturally isomorphic to the localization at the saturation $R_{\overline{S}}$? Localizations have the universal property that if $S$ is a multiplicative subset of a commutative ring $R$, and $i\colon S\to R$ is the canonical embedding, then if $g\colon R\to T$ is any map such that $g(s)$ is a unit in $T$ for all $s$, then there exists a unique homomorphism $h\colon R_S\to T$ such that $g=hf$. 
I was reading a handout of Brian Conrad's that said that the localizing at $S$ is the same as localizing at the saturation $\overline{S}$ of $S$, as can be seen "instantly" by the universal property, since the universal property of $R_S$ implies that this ring also has the universal property to be the localization with respect to $\overline{S}$.
How can you use the universal property to see this instantly? Taking $f: R\to R_S$ and $\bar{f}:R\to R_\overline{S}$ to be the standard embeddings, and $g:R\to R_\overline{S}:r\mapsto rs/s$ and $\bar{g}:R\to R_S:r\mapsto r\bar{s}/\bar{s}$ to essentially also be standard embeddings, so that they send elements of $S$ and $\overline{S}$ to units, I found unique $h$ and $\bar{h}$ such that $g=hf$ and $\bar{g}=\bar{h}\bar{f}$ in hopes that $h$ and $\bar{h}$ would be isomorphisms between $R_S$ and $R_\overline{S}$, but it doens't seem to be working.
 A: Since $\bar f$ maps the elements of $S$ to invertible elements, it factorizes uniquely through $f$ by the universal property of the latter. On the other hand, since every element of $\bar S$ is mapped by $f$ to an invertible element of $R_S$, $f$ factors uniquely through $\bar f$. 
You can check that the two compositions of the maps which establish these factorings are identities, using again the universal properties.
A: As Brian writes in his handout, use the universal property: on the one hand, $\overline{S}$ has invertible image in $R_S$, by definition.
Now given any morphism $R \to R'$ for which the elements of $\overline{S}$ become
 invertible in $R'$,
we see in particular that the elements of $S$ become invertible in $R'$.  Thus (by the universal property) we get a uniquely determined morphism $R_S \to R'$. 
Thus $R_S$ satisfies the same universal property as $R_{\overline{S}}$, and so
they are canonically isomorphic.  
(Note that Mariano's proof is just a proof of this canonical isomorphism adapted to the case at hand.)
