MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C$ be a category. I'd like to say that a property $P$ of objects of $C$ (or rather isomorphism classes of objects) is a "Yoneda property" or a "maps-in property" if there is a property $P'$ of contravariant functors $h:C\to\mathrm{Set}$ such that the functor $\mathrm{Hom}(-,X)$ has $P'$ if and only if $X$ has $P$. We might also say a property is "co-Yoneda" or "maps-out" if the induced property on $C^{\mathrm{op}}$ is Yoneda.

But this definition is useless because every property is a Yoneda property (and, hence, also a co-Yoneda property) -- just take $P'$ to be the property "$h$ is isomorphic to the functor $\mathrm{Hom}(-,X)$, for some object $X$ with property $P$". So my question is: Is there a good definition of "Yoneda property"?

Here are some examples of what I have in mind. In the category of modules over a given ring, injectivity should be a Yoneda property and projectivity should be a co-Yoneda property. In any category, being a terminal object should be a Yoneda property and being an initial object should be a co-Yoneda property. We could do the same thing with maps instead of objects, and then in the category of schemes being proper should be a Yoneda property (by the valuative criterion), as would being separated, formally smooth, formally unramified, locally of finite presentation, and so on.

A few more remarks:

  1. It seems that we'd want to keep the definition I gave above but make some restriction on properties of functors we allow. Quantifying existentially over all objects of the category (which is what breaks the definition above) probably should not be allowed. But what exactly should be allowed?

  2. There appear to be different kinds of Yoneda properties. For example, in the category of schemes, the definition of formally smooth is of the form "for all diagrams of type $Y$, there exists a map $f$ such that $Z$ holds", and the definition of formally unramified is of the form "for all diagrams of type $Y$ and all maps $f, f'$ such that $Z$ holds, we have $f=f'$". Maybe it would be better to distinguish these different kinds of properties. So it might be more natural to define separately "Yoneda properties of existence type" (e.g. formal smoothness), "Yoneda properties of uniqueness type" (formal unramifiedness), and maybe others.

share|cite|improve this question
up vote 3 down vote accepted

Hi Jim, not sure if this is the sort of thing you are after, but here is one possibility.

Let $K$ be a category and $F=(f_i:A_i\to B_i)_{i\in I}$ a family of morphisms in $K$. Say that an object $X$ is injective to $F$ if for each $i\in I$ and each $a:A_i\to X$ there exists a morphism $b:B_i\to X$ whose restriction along $f_i$ is $a$. The collection of all such objects $X$ (for given $F$) is called an injectivity class in $K$.

Any injectivity class in $[C^{op},Set]$ defines a property of objects of $C$: those objects $c$ for which the representable functor $C(-,c)$ lies in the injectivity class. Your various examples arise in this way:

  • For injectivity, just take all maps of the form $C(-,i):C(-,a)\to C(-,b)$ with $i:a\to b$ mono.
  • For terminal object, take all the maps $0\to C(-,a)$ and $\nabla:C(-,a+a)\to C(-,a)$ (where $\nabla$ is the codiagonal)
  • and similarly being formally smooth or formally unramified.

Your properties "of uniqueness type'' can be seen as a special case of those "of existence type'', by using codiagonal maps, as in the case of terminal object.

Of course in the first example, of injectivity, you could just as well work in the original category $C$ itself, rather than $[C^{op},Set]$. More generally, you would always be able to do this if $C$ had colimits.

You will get better properties if the family of morphisms defining the injectivity class is, or can be taken to be, small. Not sure if this is important for your purposes.

share|cite|improve this answer
Very nice. Thanks! I especially like how you can get things of existence type as a special case of those of uniqueness type. But before I can be convinced that it really captures what I have in mind, I have a few questions. Do the finitely presentable objects (ie the $c$ for which $C(c,-)$ commutes with filtered colimits) form an injectivity class? Also, are there any properties which you can prove cannot be expressed as injectivity classes? – JBorger Jan 4 '11 at 13:20
For instance, is it true that the class of initial objects is not (or may not be) an injectivity class. What about the class of projective modules? – JBorger Jan 4 '11 at 13:21
In general, injectivity classes are closed under products and under retracts. As your comment suggests, the initial objects can be an injectivity class: for example if the category is pointed, so that initial an terminal objects coincide. On the other hand, if C has a terminal object 1, then C(-,1) is terminal in $[C^{op},Set]$, so is contained in any injectivity class. Thus being an initial object could only be a Yoneda property in this sense if initial and terminal objects coincide. – Steve Lack Jan 5 '11 at 23:00
As for projective modules, this should depend on the ring. In a boring case where all modules are projective, projectivity is both Yoneda and co-Yoneda. Over the integers, where projective=free, the projectives are not closed under infinite products, so cannot be Yoneda. – Steve Lack Jan 5 '11 at 23:07
In general, the functors $F:C\to Set$ which preserve certain limits form an orthogonality class in $[C,Set]$: this is like injectivity except that you have existence and uniqueness. Since, as we saw, uniqueness properties can be re-expressed as existence properties, orthogonality is a special case of injectivity. Thus functors preserving certain limits are also an injectivity class. Thus being finitely presentable should be a coYoneda property. I would guess that once again it is usually not Yoneda. (I'm using these names, but I'm not sure if they really feel right.) By the way, for more – Steve Lack Jan 5 '11 at 23:12

Well, I'll take a stab at it. I think you're going to want properties $P$ of functors $F$ of the form

$F$ has property $P$ if and only if $\forall X$, $F(X)$ has property $P'$

as you'd suggested above. I think we can avoid $P'$ being a natural isomorphism with $h_X$ by making sure it's a property of all of these sets, rather than something we say about it as a functor (on the other hand, if we want to define Yoneda and CoYoneda properties of morphisms, this becomes a problem, and so this is really bothering me philosophically...)

But this definition handles "is a group", "is initial", "is terminal" by "every $F(X)$ is a group", and statements that for a contravariant or covariant (as appropriate) we have a single element. We're going to want to allow products as well to be Yoneda, presumably. Now, $X\times Y$ is a product if for any $Z\to X, Z\to Y$ we get $Z\to X\times Y$ commuting with projection. So I guess here we might want to have the property "There exist functors $G,H$ such that for all $X$, we have $F(X)=G(X)\times H(X)$" perhaps? It avoids quantifying over the category in question, but instead does so over the category of functors, which I haven't thought through terribly well.

Mostly I'm just thinking aloud here, but no one else had much, so I thought I'd throw my two cents in.

share|cite|improve this answer
@fpqc: care to elaborate? Your comment is a bit cryptic... – Alberto García-Raboso Dec 6 '09 at 18:08
I said something stupid and deleted the comments to save face. – Harry Gindi Dec 6 '09 at 19:21
Thanks for giving it a go. I would hope that your condition would be a sufficient but not necessary condition for something to be a Yoneda property. The reason that it's probably too restrictive is that F(X) is a set, so any property of F(X) will be a property of sets. But there aren't that many properties of sets: a property of (isom classes of) sets is (tautologically) a statement about cardinality. So your proposed definition doesn't allow the examples I gave from module theory and scheme theory. – JBorger Dec 6 '09 at 23:52
Also, your property "X is a group", for sets X, is probably better regarded as a structure. It must be true that every nonempty set admits a groups structure, but this is probably not what you meant. I haven't considered whether there is a good definition of a Yoneda structure, rather than a Yoneda property. If you find one, let me know! – JBorger Dec 6 '09 at 23:58
Ahh, good points. My answer doesn't seem to get very far. Well, was worth taking a crack at it. Was curious what I could manage, because I'm not entirely sure that you CAN define a sensible notion of what you want here. – Charles Siegel Dec 7 '09 at 2:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.