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.

  • $\begingroup$ I hope my answer didn't imply that I thought it was a silly question. I don't. I just wasn't sure that I understood it. $\endgroup$ – Tom Leinster Dec 30 '09 at 3:22
  • $\begingroup$ Here is a new version of this question: mathoverflow.net/questions/86923/… $\endgroup$ – Martin Brandenburg Feb 18 '12 at 10:07

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.

  • $\begingroup$ thank you for the answer. basically, this is not new to me, so I have specified my question, see above. besides, that elements can be treated as morphisms into A helps only in the case of universal objects defined by contravariant functors. $\endgroup$ – Martin Brandenburg Dec 30 '09 at 2:02
  • $\begingroup$ this answer has now been auto-selected although is does not answer my question at all $\endgroup$ – Martin Brandenburg Jan 6 '10 at 8:35
  • $\begingroup$ OK, sorry - nothing I can do about that! I didn't even know that could happen. $\endgroup$ – Tom Leinster Jan 6 '10 at 9:09

I'm kind of confused. It sounds like you want to do the following:

(1) start with some object X defined purely by a universal property

(2) take some canonical objects and canonical morphisms relating them to X

(3) do something with them

(4) apply the universal property to arrive at a more granular description of the elements of X

Phrasing aside, the content of this does not really sound different to me than the constructions you are trying to avoid. Your ability to actually accomplish (3) depends on knowing ahead of time an explicit construction. Is the the context where you already know the construction but you want to describe the elements differently the context you care about? Or do you want something more general?

  • $\begingroup$ you don't need to know any construction in advance. the universal property can tell you how the elements look like (and finally how a construction can be done, which does not have to be carried out). this is convenient e.g. when you want to show that words in free groups have a unique reduced form. and you arrive at these reduced words just by thinking with the universal property (compare with A+B for abelian groups A,B). this is totally different than running into a nasty construction! $\endgroup$ – Martin Brandenburg Dec 30 '09 at 3:21
  • 2
    $\begingroup$ I still don't get it. I know A+B is easy, but for the sake of discussion, how did you know that you should take those 2 "inclusion" morphisms and add them together? Can you explain roughly what your procedure is in some level of generality for taking an element-free description of a universal object and deciding what to do for steps (2) and (3) in my previous post? If you didn't already know the description or have a good guess, what would you even have as your goal? $\endgroup$ – Sean Rostami Dec 30 '09 at 7:08
  • $\begingroup$ I can elaborate this, but I don't want to go off topic and explain all what I've done so far, because actually I want to deal with localizations and I think I've made the question clear enough so that everything can start thinking about it. [yes, there IS a a general categorical approach to representabilty of subfunctors of hom-functors defined on algebraic categories; the objects you are starting with always generate the universal object and there are various motivations for this] $\endgroup$ – Martin Brandenburg Dec 30 '09 at 11:09

Here's a rough attempt of mine. I take it that you want to show that the ring $A$ and the localization $S^-1A$ are isomorphic. The way to do this is using the universal property that If $f:R\rightarrow T$ is any ring homomorphism such that $f\left(s\right)$ is a unit in $T$ for all $s\in S$ then show that there exists a unique ring homomorphism $\bar{f}:S^{-1}R\rightarrow T$ such that $\bar{f}\phi_{S}=f$ where $\phi_{S}$ is the canonical ring homomorphism $R\rightarrow S^{-1}R$ given by $r \mapsto \frac{rs}{s}$ for any $s \in S$.

I will now give my attempted solution, prefaced by the fact that I am new to category theory and this is one of the first problems that I've attempted to really prove in a category-theoretic way.

Working in the category $\mathbf{Ring}$, let $\mathcal{F}$ be the identity functor, $id_{\mathbf{Ring}}$. Then we know that $\left(S^{-1}R,\phi\right)$ is an initial morphism (i.e. a universal arrow) from $R$ to ${id}_{\mathbf{Ring}}$ because the following diagram commutes:

\xymatrix{R\ar[r]^{\phi}\ar[dr]_{f} & S^{-1}R\ar@{-->}[d]^{\bar{f}}\\ & T}

Similarly, $\left(R,{id}_{R}\right)$ is also an initial morphism because the following diagram commutes:

\xymatrix{R\ar[r]^{{id}_{R}}\ar[dr]_{f} & R\ar@{-->}[d]^{f}\\ & T}

Because $\left(S^{-1}R,\phi\right)$ and $\left(R,id_R \right)$ are both initial morphisms from $R$ to $id_{\mathbf{Ring}}$, they are unique up to a unique isomorphism. Furthermore, if we label this isomorphism $k:R\rightarrow S^{-1}R$, then plugging $\left(S^{-1}R,\phi\right)$ into the diagram for $\left(T,f\right)$, we see that $k$ satisfies $\phi=id_{\mathbf{Ring}}\left(k\right)\circ{id}_{R}$, i.e. $\phi$ is the unique isomorphism.

  • $\begingroup$ No. $A$ and $S^{-1} A$ are not isomorphic. $\endgroup$ – Martin Brandenburg Mar 8 '10 at 13:42
  • $\begingroup$ Fair enough. They are in the case where $A$ is a commutative ring with unit and $1\in A$. Should have written those pre-suppositions into my write-up above (which I have improved upon personally but not yet edited). It also is unclear in your original question exactly what isomorphism (or equality) you want to show via the universal property. $\endgroup$ – user3795 Mar 8 '10 at 21:11

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