# Fun With Representable Functors, or Why I Like Yondea's Lemma.

Sometimes people ask, “why do you use Category Theorey? Isn't it just a set of abstract terms making things look more complicated?” Well, sometimes this is true. Often, it allows to make short and precise statements instead of listing several properties:

is a functor from the category of groups into the category of sets” (or “ is a functor”)

versus

“for every group, , is a set; if is a group homomorphism, then is a map; if , then ; if is another group homomorphism, then ”.

And it allows to apply very generic statements to a large class of specific examples. And, sometimes, it even gives new insights by abstracting results.

Today I want to discuss representable functors and Yondea's lemma which is, for example, used a lot in modern Algebraic Geometry. In the following, I will always assume that in all categories I use, the morphisms between two objects form a set. We will denote the category of sets by (and not the french ), the category of groups by , the category of abelian groups by , the category of rings by , the category of field extensions of by and, for a ring , the category of -modules by . We begin with the definition of a representable functor. For any category whose objects are sets with further structures, and whose morphisms are maps between these sets, we call concrete and we have the forgetful functor . Finally, denote by the opposite category of ; we will not use the term of contravariant functors , but treat them as functors .

Let us begin withe the definition of representable functors.

Definition.
1. We say that a functor is represented by an object if there exists a natural equivalence .

If is a functor, and is concrete, we say that is represented by if is represented by . Finally, we say that is representable if an object exists such that is represented by .

2. We say that a functor is represented by an object if there exists a natural equivalence .

If is a functor, and is concrete, we say that is represented by if is represented by . Finally, we say that is representable if an object exists such that is represented by .

This sounds rather abstract. Let us describe and in a little more in detail. To an object , they assign the set of all morphisms from to , respectively the set of all morphisms from to . And to a morphism they assign the map

respectively

We want to make both concepts more clear with two important examples.

Example.

Let us consider the category of -algebras, where is a fixed ring (or field, if you feel more comfortable). Let be the ring itself, and let and , where are polynomials. Now every -homomorphism is a substitution homomorphism, i.e. there exists a uniquely determined with . Conversely, for any tuple , there exists a homomorphism , . Hence, we can identify with .

Now the homomorphisms correspond to the homomorphisms whose kernel contain . Let , be a homomorphism. Then

where

is the variety defined by the polynonials . Hence, we can identify with . Now consider the projection , . If corresponds to the point , then also corresponds to , but this time ! Hence, is the inclusion map .

Now, if one replaces by , where is a -algebra, we can identify with and with

and again is the inclusion map . Note that this is not just a toy example, but a very fundamental concept used with affine schemes. In fact, there, one fixes a -algebra , say our , and looks at the functor ; this functor is called the functor of points as it assigns to every -algebra (indirectly) the solutions of the polynomial equations with . In fact, if we consider the functor

with , for , then we obtain a natural transformation

which assigns to the map

Since by the above, this is a bijection, it turns out that is in fact a natural equivalence; hence, represents the functor .

Example.

Another example also comes from algebraic geometry, namely elliptic curves. If is an elliptic curve defined over a field , i.e. , then for every field extension we can define

For every -homomorphism , we obtain a map , , . Then is a functor into the category of abelian groups!

Now one can ask whether the functor is representable, i.e. whether there exists a field extension of such that can be identified in a natural way with for every field extension – that is exactly what it means for being represented by . Unfortunately, it turns out not to be possible.

For two categories , , we can consider the category , whose objects are functors and whose morphisms are natural transformations of such functors.

If we fix a category and an object , we obtain the functor , which is an element of . If we have another object together with a morphism , we obtain a natural transformation by assigning the morphism , . Hence, can be seen as a functor from to . This functor is also called the (contravariant) Yoneda embedding.

Theorem (Yoneda's Lemma).

Let be a category and , . Then there is a natural bijection

with the following property:

If , then is mapped onto the natural transformation , which satisfies that is mapped onto .

In other terms:

is a natural equivalence.

This looks rather complicated. A simpler corollary is:

Corollary.

The Yoneda embedding is fully faithful, i.e. for every , the map is bijective. In fact, is a full embedding.

Before we continue with the proof of Yoneda's lemma and the corollary, let us first take apart the notation from Yoneda's lemma. If , it means that is a functor . The class (which, by Yoneda's lemma, is a set) is the class of natural transformations . Now, Yoneda's lemma basically says that there is a natural bijection between the set and the set of natural transformations .

The property says that the bijection has to look as follows: to an element , we want to assign a natural transformation which is defined as follows: for an object , we want that is defined by , i.e. we want to map to evaluated at , which was an element of .

Still sounds really complicated, doesn't it? Well, lets start with the proof, which includes some fancy diagrams.

Proof (of Yoneda's Lemma).

First, let . For , define by . Now let be a morphism; then we have the following diagram:

But since is a contravariant functor, ; therefore, . This shows that is a natural transformation, i.e. .

Next, we have to show that

is a natural equivalence between functors

and

For that, let , and a morphism and a natural transformation, and consider the following two diagrams:

Now and are both natural transformations of functors . Hence, let and . Then

as is contravariant, and

Now consider the following diagram:

Now and are both natural transformations of functors . Hence, let and . Then

and

Hence, the condition that these two elements are the same is equivalent to the fact that the diagram

commutes; but that follows from the fact that is a natural transformation . Hence, we have shown that is a natural transformation.

It is left to show that is in fact an equivalence, i.e. that for , , the map

is bijective. First, we show that it is injective; for that, note that for we have , whence uniquely determines . To show that is surjective, let . Set ; if is an object and , consider the following commutative diagram:

Hence, . Since and were arbitrary, we get .

Wow, looks like a huge collection of abstract nonsense, eh? Well, the good news is that the worst part is done. Now, let us prove the corollary.

Proof (of the Corollary).

Let , and let ; then , and by Yoneda's Lemma, the map

is bijective. We have to check that equals the map ; for that, let be a morphism, and . Then

Finally, we have to show that is injective on objects. For , we have . Since morphism sets for distinct objects are assumed to be disjunct, it follows that we can get back from .

Well. After all this abstract nonsense, let us do some more concrete abstract nonsense. First, a lengthy definition which will turn out to be quite cool.

Definition.

Let be a category with a final object in which finite products  and exist for all objects .

1. An object together with morphisms , and is called a group object if the following diagrams commute:

2. We say that a group object is commutative if , where switches its operands.
3. Let and be group objects. A morphism is called a homomorphism of group objects if , and .

Let us first give some examples:

Example.

Let ; then is a final object in . The group objects in are exactly the groups. A group object is commutative if the corresponding group is commutative, and homomorphisms of group objects are nothing else than group homomorphisms.

Example.

If is the category of topological spaces, with continuous maps as morphisms, then the group objects in are exactly the topological groups. Homomorphisms of group objects are again nothing else than continuous group homomorphisms.

Example.

If is the category of all groups, the group objects in are a bit more interesting. A final object is given by , a group of one element (in fact, is an initial object as well).

First, let be an abelian group. Then , is a group homomorphism. So is , , and the map , is a group homomorphism as well. Hence, every abelian group gives rise to a commutative group object .

Now, let be a group object. As is a group homomorphism, we have . Next, since is a group homomorphism, we get

i.e. . Now, must be a group homomorphism as well, and the group object axioms force for all ; hence, for all ,

Therefore, forces to be abelian, and the argument shows that , and are uniquely determined by .

Therefore, the group objects in are exactly the abelian groups, and all of them are commutative. The group object homomorphisms between two abelian groups (with their only possible group object structure) are exactly the morphisms in between the two objects.

Now, one can do other stuff with group objects . Namely, given any other object , we can turn into a group:

Proposition.

Let be a group object in . For , set and define

and set , where is the unique morphism to the final object . Then is a group whose inverses are given by and whose neutral element is . If is commutative, then is abelian.

Proof.

First, we show that is associative. For that, let . Then

We can rewrite this to

but from the definition of a group object, we know . Therefore, both expressions are the same.

Next, let us show that is a neutral element. For that, let . Then

and, analogous, . But from the commutative diagrams of the definition of a group object, it turns out that both are the same .

Now let us show that is the inverse map. For that, let . Then

and . Now the definition of a group object gives , whence

Finally, assume that is commutative. Let ; then

hence, is abelian.

For and being a group (identified with the associated group object in ), we get that the group defined in the proposition is exactly the set , where for elements we define ; then and .

In , this seems to be not too interesting, but now assume that is the category of topological spaces (with continuous maps). If , then gives exactly the group structure of , throwing away the additional information (i.e. that the group operations are continuous with respect to the topology on ). Now, if with the usual topology, the elements of are continuous paths in . Hence, in , we can add continuous (parameterized) paths in (using pointwise addition)!

Now the interesting result is the following application of Yondea's lemma:

Theorem.

Let be a category with finite products and a final object .

1. If is a group object, then the assignment

is a functor . Moreover, is commutative if, and only if, the image of this functor lies in the subcategory of .

2. Let and be two group objects in , and let be the corresponding functors. The group object homomorphisms correspond one-to-one to natural transformations .

In particular, two group object structures on an object give two naturally equivalent functors if, and only if, the group object structures are isomorphic.

3. Let . There is a one-to-one correspondence between the isomorphism classes of group object structures on and natural equivalence classes of functors which are represented by :

If is a group object structure, then

is the corresponding functor.

Before showing the theorem, note that the functor preserves products, i.e. if and exists, then exists in and in a natural way.

Proof.
1. First, we have to show that for morphisms , the induced map is a group homomorphism. For that, let ; then

Finally, we are left to show that the fact that the image of the functor lies in implies that is commutative, i.e. that . For that, consider the natural transformations with and with . (The fact that this is natural follows from the fact that we interpreted as a functor .) Then we have the commutative diagram

since for every , is abelian. But now , whence this diagram is the image of the diagram

under : obviously, and . But since is faithful, it follows from that we also have , i.e. that is commutative.

2. By Yoneda's lemma, we get a bijection between the morphisms and the natural transformations . Hence, we have to sow that a morphism is a homomorphism of group objects if, and only if, the corresponding natural transformation is actually a natural transformation .

First, note that is a homomorphisms of group objects and if, and only if, : in case this condition holds, one can obtain and since are uniquely determined by and are uniquely determined by . (Just consider the same statement for groups: given the group operation, there is exactly one neutral element with respect to that operation and the inverses are unique as well.)

Now, since is faithful, the diagram

commutes if, and only if, the diagram

commutes. That the first diagram commutes is equivalent to the fact that is a homomorphism of group objects. That the second diagram commuts is equivalent to the fact that is a group homomorphism for every , i.e. that is a natural transformation for functors .

3. Part 1. shows that every group object structure on induces a functor . Part 2. shows the statement that the group structures are isomorphic if, and only if, the functors are naturally equivalent.

Let be a functor which is represented by , i.e. there exists a natural equivalence . For , define

As is a natural transformation, it turns out that gives a natural transformation

Since in a natural way, we have that gives a natural transformation . By the corollary on Yoneda's lemma, this natural transformation corresponds to a morphism .

Let us show that satisfies the associativity diagram, i.e. we have that

commutes.

Consider the natural transformations

From the definition of it follows that the following diagram commutes:

(Simply plug in an object , and then use the definition of and the fact that is a group.) Now, since and , one can see that this diagram is the image of the previous diagram under . But since is faithful, the previous diagram also has to commute.

Now, one can obtain and in the same manner and show that they satisfy the conditions they have to such that is a group object.

In a nutshell, this result says that representable functors are the same as group objects in , and that representable functors are the same as commutative group objects in . This is somewhat surprising, as one expects that constructing a group object structure on an object in is hard, while coming up with a (representable) functor (or ) sounds easier. In fact, constructing a functor is usually easier, but showing that the constructed functor is representable is hard. (Just consider the construction of Picard schemes or Hilbert schemes.)

If you are interested in literature, see, for example, the book “Néron Models” by S. Bosch, W. Lütkebohmert and M. Raynaud (Ergebnisse der Mathematik und ihrer Grenzgebiete no. 21, Springer, 1990), and the book “Commutative Group Schemes” by F. Oort (Lecture Notes in Mathematics no. 15, Springer, 1966).