Presheaves as limits of representable functors?

Question

If I remember correctly, I read that given a presheaf $P:\mathcal{C}^{op} \to Set$, 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 1

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 2

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 3

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 4

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 5

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.

Answer 6

The proof is given in https://en.wikipedia.org/wiki/Density_theorem_(category_theory)#Proof as a proof of density theorem.

In very short they showed the following:

If $F:C^{op} \rightarrow Sets$ be any presheaf of sets then $F$ is the colimit of the following Diagram in the category of presheaves, $Fun(C^{op},Sets)$:

where $\pi_F:el F \rightarrow C$ is the corresponding Discrete fibration obtained by applying Grothendieck Construction on $F$. Here $el F$ is the category of elements https://en.wikipedia.org/wiki/Category_of_elements of $F$ and $Y$ is the Yoneda embedding of $C$ into $Fun(C^{op},Sets)$.

Though I am answering the question after a decade but still for the benefit of the future readers I wanted to add this reference and a small explanation.

Thanks.

Answer 7

I will give two easier-seeming facts and proofs, and then show that the "colimit of representables" idea is obtained by the essentially same arguments in a more-general setting.

If $R$ is a ring, then any $R$-module $M$ is a quotient of a free module.

Proof: Choose generators for $M$ by some intricate procedure, or by taking every element of $M$ to be a generator. For each generator $m$, build a map $R^1 \to M$ specified by sending $1 \mapsto m$. This assignment extends to at most one map of modules since $1$ generates $R^1$ as a module, and moreover, it does extend, since the formula $r \mapsto mr$ constructs a valid map of right modules.

By summing these maps over the generating set, we obtain a map from a large free module to $M$. Every generator is in the image of this map since it is hit by its corresponding basis vector. Since the image is a submodule of $M$ that contains a generating set, the map is a surjection, and so $M$ is a quotient of a free module.

Any $R$-module $M$ has a presentation by generators and relations.

Proof: Witness $M$ as a quotient of a free module, which we will call the module of generators. The kernel of the quotient map is again an $R$-module (actually, it is a submodule of the module of generators), and so is itself a quotient of a free module (the module of relations). There is a natural map from the module of relations to the module of generators given by the quotient map followed by the inclusion, and the cokernel of this map is $M$. This shows that $M$ has a presentation by generators and relations.

Ok, and here are the corresponding facts and proofs for categories. Set $\mathcal{D} = \mathcal{C}^{op}$. By an "element" of $F \colon \mathcal{D} \to \mathrm{Set}$ we mean a pair $(d \in \mathcal{D}, x \in Fd)$. Essentially, an "element of F" is an element of the disjoint union of the various $Fd$. If $f \colon d \to d'$ is a morphism of $\mathcal{D}$, then we write $(d,x) \cdot f = (d', (Ff)(x))$. This gives a right action of $\mathcal{D}$ on the elements of $F$.

If $\mathcal{D} = C^{op}$ is a category, and $F : \mathcal{D} \to \mathrm{Set}$ is a functor to sets, then $F$ is a quotient of a disjoint union of representable functors.

Proof: Choose enough elements of $F$ so that every other element may be obtained from these by use of the action of $\mathcal{D}$. For example, one may choose every element of $F$ at this stage. Call this collection of elements the "generators". Then, for each generator $(d, x)$, build a map

$$ \mathcal{D}(d,-) \to F $$

specified by $(d, 1_d) \mapsto (d, x)$ and therefore, by naturality we would have at a general element $(d', f \colon d \to d') \mapsto (d,x) \cdot f$. As this formula does give a natural transformation, the desired map exists.

By taking the coproduct of these maps over the generating set, we obtain a map from a large disjoint union of representables to $F$. Every generator $(d,x)$ is in the image of this map since it is hit by the element $(d, 1_d)$ in the corresponding summand of the disjoint union. Since the image is a subfunctor that contains a generating set, it is a surjection, and so $F$ is a quotient of a disjoint union of representables.

Any functor is a the coequalizer of a map from one disjoint union of representables to another.

Proof: By the lemma, write $F$ as a quotient of a disjoint union of representables $X$, so that $X$ the $\mathcal{D}$-set of generators. The set of pairs of elements of $X$ that map to the same element of $F$ gives a subfunctor of $X \times X$. Write this subfunctor as a quotient of a disjoint union of representables $Y$, the relations. We obtain two natural transformations $Y \to X$ by the quotient map followed by each projection. The coequalizer of these two maps identifies all pairs of elements of $X$ that map to the same thing in $F$, so the coequalizer is isomorphic to $F$.

Remarks

For a good example to work by hand, I recommend setting $R=\mathbb{Z}[x,y]$ to be a graded ring with $|x|=|y|=1$, and taking $M$ to be a monomial ideal. The objects of $\mathcal{D}$ are the grades of $R$, and there can be some nice combinatorics in the construction of the relations.

In commutative algebra, it is often useful to extend a presentation to a free resolution. The analog would be finding a simplicial resolution by coproducts of representables.