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Wikipedia gives a formal definition of universal property: a universal property is associated with an initial (resp. terminal) object in the comma category $X \downarrow U$ (resp. $U \downarrow X$) for object X and functor U.

Is the above definition of universal properties overly narrow? (Similar questions have come up here a few times, but without real resolution -- see links below.)

For instance, how can you see the universal property of ring localization or the universal property of quotient groups as a case of the the initial/terminal morphisms in the Wikipedia definition?

One resolution that works for the above two examples is to modify the definition of universal property to consider the initial object in a subcategory of $X \downarrow U$. But then we're not really working with a comma category anymore, so we don't get an adjunction for free, and we lose the global perspective that an adjunction would give us.

A better solution for the above examples is to keep the Wikipedia definition and do a little more work. For instance, for quotient groups, define the category $C$ whose objects are pairs $(N,G)$ for group $G$ and $N \triangleleft G$, and where morphisms between $(N,G)$ and $(K,H)$ are maps $f : G \to H$ s.t. $f(N)=1$. Then for functor $U : \mathbf{Group} \to C, G \mapsto (\emptyset,G)$, we have that $G \to G/N$ is initial in $G \downarrow U$. Equivalently, we have an adjunction $F \dashv U$ for $F : C \to \mathbf{Group}$, $(N,G) \mapsto G/N$. A similar construction formalizes ring localization as a "true" universal propertry and also an adjunction.

Another questionable example is the universal property of tensor products, as discussed e.g. here and here. The usual universal property refers to bilinear maps, but that statement doesn't directly fit the Wikipedia definition. Alternatively, if you consider the tensor-hom adjunction and work out the associated universal property you get from an adjunction, you do get a "true" universal property, but you lose the reference to bilinear maps. (Though more optimistically, you end up defining bilinear maps from $X \times Y \to Z$ as maps $X \to \textrm{hom}(Y,Z)$.)

So far, we've been able to make all of our examples fit the Wikipedia definition, but it's taken some work. Is there an alternative definition that doesn't require us to work as hard? (Admittedly, the "work" is illuminating in a way.) Or are there any examples that one wants to call universal properties that really don't fit the definition?

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    $\begingroup$ An object satisfies a universal property if it represents some functor. In this sense every object tautologically has a universal property, but the point in practice is that we can sometimes name representable functors before we know how to name the objects representing them. $\endgroup$ – Qiaochu Yuan Sep 28 '15 at 5:55
  • $\begingroup$ Is that your definition -- having a universal property means representing a functor? Whereas the Wiki definition would say the object must represent a functor specifically of the form Hom(X,U-), right? $\endgroup$ – Daniel Ranard Sep 28 '15 at 6:31
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    $\begingroup$ All set-valued functors are of such form: U ≅ Hom(1, U-). $\endgroup$ – user54748 Sep 28 '15 at 7:02
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    $\begingroup$ Also, representability of a functor $F : \mathcal{C} \to \mathbf{Set}$ is the same as the existence of an initial object in the comma category $(1 \downarrow F)$. $\endgroup$ – Zhen Lin Sep 28 '15 at 7:17
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    $\begingroup$ The answer is pretty much unequivocally yes. The problem is that you haven't defined what you consider to be a universal property... Write down any definition you can think of, and you'll see it will be equivalent to the one given by Wikipedia. As for bilinear maps: what about the functor $F : \mathsf{Vect} \to \mathsf{Set}$ given by $F(Z) = \{ \text{bilinear maps } X \times Y \to Z \}$? A representative for this functor (AKA initial object of $(1 \downarrow F)$) is the tensor product $X \otimes Y$. $\endgroup$ – Najib Idrissi Sep 28 '15 at 7:58
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To summarize the discussion from the comments: yes, every example of a thing you want to call a universal property (for an object in a category) has this form, even if it sometimes takes some thinking to cook up the relevant functor. (But I promise that this is not work relative to what work looks like in the rest of mathematics.)

I prefer to think of the relationship between the functor and the universal object as being that the universal object represents the functor; this is equivalent to the comma category definition. For example,

  • the quotient group $G/N$ represents the functor of homomorphisms out of $G$ which are trivial on $N$,
  • the localization $R[S^{-1}]$ represents the functor of homomorphisms out of $R$ which invert every element of $S$, and
  • the tensor product $X \otimes Y$ represents the functor of bilinear maps out of $X \times Y$.
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  • $\begingroup$ Thanks! This helps. What's different about these formulations is that the associated category $X \downarrow U$ always has $X = \{*\} \in \mathbf{Set}$, rather than letting the choice of $X$ encode the"input data" (e.g. the group and normal subgroup, or ring and multiplicative subset) that goes into the universal construction. (Compare to the quotient group example I gave in the question.) Relatedly, you also miss the possible adjunction. $\endgroup$ – Daniel Ranard Sep 28 '15 at 19:09
  • $\begingroup$ @Daniel: all of these formulations are the same. A functor $C \to \text{Set}$ is representable iff it has a left adjoint. $\endgroup$ – Qiaochu Yuan Sep 28 '15 at 21:09
  • $\begingroup$ I just mean to point out that the adjunction you just mentioned (say for quotient group) is different from the one I described in the question -- if I understand right. Here, you have one adjunction for every quotient G/N; in my example, there's a single adjunction encoding all of this. $\endgroup$ – Daniel Ranard Sep 28 '15 at 22:01
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    $\begingroup$ @Daniel: sure, you can do that too, especially if you want to investigate the naturality of these constructions in their input data. But the simplest thing you can say doesn't require this, and it's best adapted to the case when the universal construction doesn't always exist. $\endgroup$ – Qiaochu Yuan Sep 28 '15 at 22:11
  • $\begingroup$ (Great, that's what I was getting at, and I suppose the simpler construction is more general.) $\endgroup$ – Daniel Ranard Sep 28 '15 at 22:25

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