Presheaves as limits of representable functors? - MathOverflow

Presheaves as limits of representable functors?

2009-10-28T23:11:34Z

If i remember correctly, i read that given a presheaf P:C^op -> Set it is possible to describe it as a limit of representable presheaves. Could someone give a description of the construction together with a proof?

Answer by Greg Stevenson for Presheaves as limits of representable functors?

2009-10-28T23:24:57Z

This follows from the Yoneda lemma - probably my favourite way of thinking about this fact is via coends in the way described here by Todd Trimble, which I think makes it quite clear what is going on.

Answer by Reid Barton for Presheaves as limits of representable functors?

2009-10-28T23:30:19Z

You mean "colimit of representable presheaves", not limit. Any limits that C has are preserved by the Yoneda embedding. So if C is, say, a complete poset like • → •, so that it is small and has all limits, you won't be able to produce any non-representable presheaves by taking limits of representable ones.

The way to write any presheaf as a colimit of representables is, like all things Yoneda-related, somewhat tautological, and should be worked out for oneself; but anyways it's explained to some extent at this nlab page. Rather than write out formulas, I usually think of the example of simplicial sets: every simplicial set X can be formed as a colimit of its simplices, i.e., a diagram of representables which is indexed on the "category of simplices of X", whose objects are pairs (n, x) where n is in the indexing category and x is an object of X_n. The same works in any presheaf category.

Answer by Urs Schreiber for Presheaves as limits of representable functors?

2009-10-29T00:08:01Z

By the way, that desciption by Todd Trimble has meanwhile been prepared at nLab:co-Yoneda lemma (this is what the question above is about) and nLab:Day convolution (for the more general statement).

Answer by Shizhuo Zhang for Presheaves as limits of representable functors?

2009-11-17T07:42:13Z

I think you mean any presheaf is colimit of representable presheaf. It follows from Yoneda Lemma. It is well known that category of presheaves is complete and cocomplete. So you can take limits and colimits.

This property has some geometric intuition. Because we can take any presheaf as as a space(this view point was first proposed by Gabriel and developed by Grothendieck and revised in full generality by Kontsevich-Rosenberg). So this statement means that any "space" can be glued (in some sense)from affine space(affine space is defined as representable presheaf).

Answer by G. Rodrigues for Presheaves as limits of representable functors?

2009-11-17T14:48:08Z

It follows by the strong form of Yoneda lemma. If the base category $V$ is symmetric monoidal closed, complete and cocomplete, then any presheaf $F:A^{\ast}\to V$ has a left Kan extension along the Yoneda embedding. The coend formula for left Kan extensions then yields

$Fa\cong \int^{b}A(b, a)\otimes Fb$

Note that the theorem says that every presheaf is a colimit of representables in a canonical way, or that the wedge $w_{b}:A(b, a)\otimes Fb\to Fa$ is universal. Or that the Yoneda embedding is dense.