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](https://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach).



equality of elements in localization via universal property [unsolved!]
I'm not sure I understand the question. More exactly, I think I disagree with what seems to be an assumption built into it: that there are sharp lines to be drawn between 'only using the universal property' and not, or between working 'without elements' and not.


Generally, a universal property describes how something interacts with the world around it. For example, if you say that an object $I$ of a category $\mathcal{C}$ is initial, that describes how $I$ interacts with other objects of $\mathcal{C}$. If you don't know much about the objects of $\mathcal{C}$, it doesn't tell you much about $I$. Similarly, you're not going to be able to deduce anything about the ring $S^{-1}A$ without using facts about rings. I don't know which of those facts are ones you'd be happy to use, and which aren't.


There's nothing uncategorical about elements. For example, you're dealing with rings, and an element of a ring $A$ is simply a homomorphism $\mathbb{Z}[x] \to A$. (And $\mathbb{Z}[x]$ can be characterized as the free ring on one generator, that is, the result of taking the left adjoint to the forgetful functor $\mathbf{Ring} \to \mathbf{Set}$ and applying it to the terminal set 1.) 


So I'm unsure what exactly your task is. But I'd like to suggest a different universal property of localization, which might perhaps make your task easier. Here it is.


Let $\mathbf{Set}/\mathbf{Ring}$ be the category of rings equipped with a set-indexed family of elements. Formally, it's a 'comma category'. An object is a triple $(S, i, A)$ where $S$ is a set, $A$ is a ring, and $i$ is a function from $S$ to the underlying set of $A$. (You might think of $i$ as an including $S$ as a subset of $A$, but $i$ doesn't have to be injective.) A map $(S, i, A) \to (S', i', A')$ is a pair $(p, \phi)$ consisting of a function $p: S \to S'$ and a homomorphism $\phi: A \to A'$ making the evident square commute.


There is a functor $R: \mathbf{Ring} \to \mathbf{Set}/\mathbf{Ring}$ given by
$$
R(A) = (A^\times \to A)
$$
where $A^\times$ is the set of units of $A$ and the arrow is the inclusion. Then $R$ has a left adjoint $L$, given by $L(S, i, A) = (iS)^{-1}A$. In other words, the left adjoint to $R$ is localization.


