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 *Trees* (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 *define* $S^{-1} A$ 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, *only using the universal property?* I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), 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. a major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.