Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there any interesting examples where one proves something about a representable functor $\mathrm{Hom}(-,X)$ by using the functor $\mathrm{Hom}(X,-)$?

By Yoneda's lemma, these two functors contain the same information as $X$ itself, so anything about one can be expressed in some uninteresting way as a property of the other. (For example, a nonempty topological space $X$ is connected if and only if every map from $X$ to the two-point discrete set factors through one of the points. This is best expressed in terms of the functor $\mathrm{Hom}(X,-)$, but using Yoneda's lemma, you could also do it in a silly way in terms of the functor $\mathrm{Hom}(-,X)$.) I'm not interested in these examples, but to rule them out, I'd have to know a way of formalizing the vague concept of Yoneda property, which I don't. I want genuine examples where one proves something most naturally expressed in terms of maps into $X$ by using things which are most naturally expressed in terms of maps out of $X$.

This question was motivated by discussions in the comments here and here.

share|improve this question
    
If you have such a theorem, then by Yoneda, you can anyway express it in terms of properties of $Hom(_,X)$. –  Regenbogen Mar 14 '10 at 11:29
2  
Jim, how about this: we often need to study the "points" of a coset space $G/H$ (valued in various rings, like interesting fields), and sometimes do so by working etale-locally or fppf-locally and lifting to points of $G$. The lifting step involves the viewpoint of the quotient sheafification by which $G/H$ is characterized by maps out of it (whereas the "points" considered above involve maps into $G/H$). And likewise in many other quotient situations (for algebraic spaces, sheaves of various sorts, etc.) –  BCnrd Apr 25 '10 at 0:48
1  
Yeah, I think that's an excellent general remark. For quotients in sheaf theory, maps out and maps in are not so far apart. I wonder if this is in some sense part of the reason why sheaf theory is such a useful formalism. –  JBorger Apr 25 '10 at 0:58

3 Answers 3

An absolute Galois group is an inverse limit of finite Galois groups over a system of finite Galois extensions of fields, so it represents a functor on groups defined by a compatible system of homomorphisms. As you no doubt know, many mathematicians like to describe Galois representations, i.e., maps from such a group to groups of linear transformations, and such information arises as part of the functor the group corepresents. I think this provides a good collection of examples, since careful study can produce interesting information concerning the Galois groups over our base field (together with large chunks of number theory and arithmetic geometry).

share|improve this answer
    
That's pretty good. It would be nice to carry it a bit further and find a nice example where some property of the functor $\mathrm{Hom}(-,G)$ is established by studying its representations. Here $G$ is the absolute Galois group. –  JBorger Apr 24 '10 at 21:50
    
Actually, here's something for abstract groups. Being abelian is a good Yoneda property ($G$ is abelian if and only if every map from the free group $F_2$ factors through the quotient $\mathbf{Z}^2$). Then a finite group is abelian if and only if every irreducible complex representation is 1-dimensional. (BTW, how necessary is the finiteness assumption here?) –  JBorger Apr 24 '10 at 21:53

This is rather a (long) comment.

I don't think that something like this exists or is at least useful. The only chance could be if the category has an anti-autoequivalence (e.g. finite abelian groups, $A \mapsto Hom(A,S^1)$). I want to comment on

By Yoneda's lemma, these two functors contain the same information as X itself

This is not true. For a covariant functor $F$, morphisms $Hom(X,-) \to F$ correspond to elements of $F(X)$, and for a contravariant functor $G$, morphisms $Hom(-,X) \to G$ correspond to elements of $G(X)$. But what about morphisms in the other direction? I think that these hom-functors, regarded as objects in the functor category, contain much more information than $X$, and they are not related at all. Of course, you could restrict yourself to the category of representable functors, but then somehow it is artificial to talk about these functors, right?

I think it would be the best if you give us at least one example?

share|improve this answer
    
I just meant that the Yoneda functor is fully faithful, so passing from a category to its image under the Yoneda functor involves no loss of information. In other words, you can recover an object, up to unique isomorphism, from the functor it represents. I don't know of any nice examples. That's why I'm asking the question! There must be lots, but I don't usually think of things in these terms. –  JBorger Apr 20 '10 at 3:30
    
OK, here's an example: a group is initial if and only if it is terminal. It would be nice to have a meatier example. For instance, it might be reasonable to say that Lefschetz fixed-point theorem gives an example. If you think of the cohomology of $X$ as being $\mathrm{Hom}(X,E)$, where $E$ is some generalized space, then for $X$ over a finite field, you can prove there exist rational points, i.e. certain maps into $X$, by analyzing the cohomology of $X$. It would be nice to have lots of other examples so that maybe we could draw some general lessons. –  JBorger Apr 20 '10 at 3:38

An endomorphism of a finite set (or finite-dimensional vector space) is an injection if and only if it is a surjection.

For sets and vector spaces, injections are the same as monomorphisms, and surjections are the same as epimorphisms. So these are very naturally representable/co-representable properties. (A map $X\to Y$ is said to be a monomorphism if the maps $\mathrm{Hom}(Z,X)\to\mathrm{Hom}(Z,Y)$ are injective for all $Z$, and it's an epimorphism if the maps $\mathrm{Hom}(X,Z)\to\mathrm{Hom}(Y,Z)$ are all injections.)

NB This isn't technically an answer to the question, since these are properties of morphisms and not of objects, but it has the right spirit.

share|improve this answer

Your Answer

 
discard

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.