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In this note by Tom Leinster the Banach space $\mathrm{L}^1[0,1]$ is recovered by "abstract nonsense" as the initial object of a certain category of (decorated) Banach spaces. So a function space, that would habitually be defined through the machinery of Lebesgue measure and integration, is uniquely described (up to isomorphism) in terms of abstract functional analysis and a bit of category theory.

I would be curious to see more results, ideally in diverse areas of mathematics, in the spirit of the above one, in which a familiar and important "concrete" mathematical object is recovered by a universal property (in the technical categorical sense) or -more generally- by a characterizing property that is abstract and general or doesn't delve into the "concrete" habitual definition of that object.

Community wiki, so put one item per answer please.

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Doesn't this amount to a collection of adjoint functor pairs? Yemon's answer has a different flavour, though. –  Johannes Ebert Dec 18 '11 at 14:15
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Not categorical, but the golden ratio is the hardest real to approximate by rationals. –  Steve Huntsman Dec 18 '11 at 15:17
    
@J.E.: the answers don't need to be "categorical", the above example by S.H. fits perfectly in the kind of answers I'm expecting to get. –  Qfwfq Dec 18 '11 at 18:30
    
@Qfwfq: so would other constants defined via extremal properties also fit? It seems like that might be too wide a scope if interpreted literally –  Yemon Choi Dec 18 '11 at 19:46
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Qfwfq: wouldn't the more interesting question be "provide examples of mathematical objects for which there's no known universal property that characterizes them"? It seems like there's almost no restriction on the objects that can appear in your list. –  Ryan Budney Dec 20 '11 at 9:58
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11 Answers 11

The Stone-Cech compactification. Neither, Stone nor Cech was thinking about category theory at the time (since it didn't exist), but of course the Stone-Cech compactification is a left adjoint to the forgetful functor from compact Hausdorff spaces to completely regular Hausdorff spaces.

If the general construction is not specific enough, then restrict my answer to $\beta \mathbb N$ which is a key object in Ramsey theory.

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Free groups. If I am not mistaken, they were first introduced by Dyck via the reduced words description. The modern universal property definition only came about later.

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The real line as "the' complete Archimedean ordered field, as opposed to a bunch of Dedekind cuts.

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This description is redundant. The real line is the only complete ordered field; the Archimedean property follows. It is also the largest Archimedean ordered field. –  Michael Hardy Dec 18 '11 at 1:28
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Perhaps Yemon meant "complete" in the sense of convergence of Cauchy sequences. In this sense there are non archimedean complete ordered fields. –  godelian Dec 18 '11 at 1:56
    
@Godelian: yes, that was what I had in mind (perhaps I am mis-using terminology?) So something like the hyper-reals would -- I think -- be complete and ordered in the sense I was thinking of –  Yemon Choi Dec 18 '11 at 2:07
    
@Godelian: Cauchy completeness makes the sequence 1/n converge to 0 and this is logically equivalent to the Archimedean property. So how can there be a non Archimedean complete ordered fields? –  Wouter Stekelenburg Sep 13 '12 at 12:25
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The natural numbers, maybe the oldest known mathematical obeject, have many universal properties in various categories. They are for example the free monoid on one generator, the initial rig, the free inductive set on one generator,...

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The category $Set$ of sets is, up to equivalence, the only locally small category $C$ whose Yoneda embedding $y: C \to Set^{C^{op}}$ admits a string of adjoint functors

$$u \dashv v \dashv w \dashv x \dashv y.$$

A precise treatment is given here.

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But this characterization uses presheaves, and thus uses the category of sets. If we're allowed to use Set in the statement, a simpler characterization is "Set is, up to equivalence, the only category equivalent to Set". –  Omar Antolín-Camarena Feb 14 '13 at 12:03
    
Why don't you write the authors to tell them that? –  Todd Trimble Feb 14 '13 at 12:24
    
Well, for small categories, you can replace "$Set^{C^{op}}$" with the "free cocompletion of $C$", which doesn't explicitly use Set anymore. –  Omar Antolín-Camarena Feb 14 '13 at 15:20
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Omar, unfortunately that's not going to work, since for large but locally small $C$, the functor category $Set^{Set^{op}}$ is not the free small cocompletion. (Also, words like "small" or "locally small" also refer to $Set$.) Anyway, I'm not very fussed about your objection. Compare for example the Gelfand-Mazur theorem, read as characterizing $\mathbb{C}$ as the unique complex Banach division algebra. We're allowed to say "complex" in the characterization. (This is the last comment I'm going to make about this. Feel free to downvote my answer, or contact the authors, or whatever you like.) –  Todd Trimble Feb 14 '13 at 17:50
    
I don't want to downvote this answer, I like this theorem (it's a lot fun to try to work out the adjunctions!), I didn't mean to make you feel defensive about it. –  Omar Antolín-Camarena Feb 14 '13 at 19:36
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My impression is that most, if not all, ''natural objects'' in linear algebra, analysis or differential geometry, ..., are usefully characterized by some \emph{symmetry} property, for eaxmple

''The exterior derivative is, up to a constant multiple, the only linear operator from $k$-forms to $k+1$-forms such that for each open embedding $f:U \to M$ and each form $\omega \in \Omega^k (M)$, the idenity $f^{\ast} d \omega = d (f^{\ast}\omega)$ holds.''

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Slightly facetious one here: the 3-sphere is, up to diffeomorphism, the unique simply connected, closed, 3-manifold.

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The space of Radon measures on the closed unit interval is the free topological vector space over the interval. It has universal property that evey continuous function on the interval has a unique extension to a continuous linear mapping. This has zillions of generalisations---Radon measures on compacta, bounded Radon measures on a completely regular space, uniform measures on a uniform space and, and ...

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The integers are the unique commutative ordered ring whose positive elements are well-ordered (thanks to Harry Altman).

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I happen to have just read Manes' theorem in the n-category café:

Theorem The algebras for the ultrafilter monad are the compact Hausdorff spaces.

The "ultrafilter monad" $X\mapsto \mathrm{U}(X)$ maps a set $X$ to the set of ultrafilters on it. The abstractness of the characterization of compact Hausdorff spaces lies in the fact that $\mathrm{U}$ is defined in purely set-theoretical (or, rather, category-theoretical) terms: it appears to be the "codensity monad" (don't ask me the meaning of this because I don't know!) of the inclusion $\mathrm{FinSet}\to\mathrm{Set}$.

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Probably, Johannes Ebert is right: (almost) all natural mathematical objects may be characterized by a universal property. The question is now what we understand exactly by the the fact that universal property is delving in the concrete habitual definition.

More concrete, let consider the usual definition of a factor structure, let say a factor group (of $G$ modulo a normal subgroup $H$). There is also a universal one: A factor group is (up to an isomorphism) an epimorphism (i.e. a surjective group homomorphism) $G\to G'$. Does the second definition delve the first? I really don't know!

Another example: Having two $R$-modules, $M$ of the right and $N$ of the left, one may define the tensor product as a factor of the free abelian group with the basis the cartezian product $M\times N$ modulo the relations which emphasize the bilinearity. Secondly, we may define the tensor $M\otimes_R-$ as the right adjoint of the functor $Hom_R(M,-)$, definition which may be extended for $M$ in a cocomplete abelian category. This time the possibility to change the settings leading to a more general definition stands as an argument that the universal definition is not delving in the usual one.

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