I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres <i>Trees</i> (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.
 
so I <i>define</i> $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).
 
the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, <i>only using the universal property?</i> I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.
 
I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits, the prime ideal structure of $S^{-1} A$), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements.