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.
-
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
.
-
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.
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
.
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.
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:
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.
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.
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.
Let be a category with a final object
in which finite products
and
exist for all objects
.
-
An object
together with morphisms
,
and
is called a group object if the following diagrams commute:
- We say that a group object
is commutative if
, where
switches its operands.
- Let
and
be group objects. A morphism
is called a homomorphism of group objects
if
,
and
.
Let us first give some examples:
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.
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.
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:
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.
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:
Let be a category with finite products and a final object
.
-
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
.
-
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.
-
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.
-
First, we have to show that for morphisms
, the induced map
is a group homomorphism. For that, let
; then
what we had to show.
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.
-
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
.
-
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).
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