# Yoneda Lemma

For an object , define the covariant functor

Proposition 1.Any morphism in gives us a natural transformation

*In summary, the natural transformation is obtained by right-composing with f.*

**Proof**

Let be a morphism in . We need to show . Now run , i.e. , through both sides:

and we are done. ♦

Yoneda lemma says the converse is true.

Theorem (Yoneda Lemma).Every natural transformation is uniquely of the form for a morphism .

**Proof**

Consider ; this function takes the identity morphism to some , i.e. a morphism . Take any object ; we wish to show on , i.e.

To prove this, let . From commutativity of the following diagram:

♦

### Example

Consider the functors

where *F* is the forgetful functor which takes an *A*-algebra to the underlying set. The act of taking the trace of a matrix then gives a function . One checks that this is a natural transformation. Furthermore, both and *F* are representable functors, with respective objects:

By Yoneda lemma, this means the natural transformation tr is induced from a homomorphism of *A*-algebras. A moment of observation shows this is given by:

# Products in a Category

Henceforth *I* denotes a fixed index set. are collections of objects in a fixed category indexed by .

The following generalizes the direct product of *A*-modules.

Definition.The

productof comprises of data , where

- is an object in , and
- for each , we have a morphism ,
satisfying the following universal property. For any , where

- is an object, and
- for each , we have a morphism ,
there is a unique morphism such that for each .

The morphisms are called

projections.

Here is the diagram we used for *A*-modules which is equally relevant here.

For example, we saw earlier that for *A*-modules, the direct product and its projection maps form a product in the category of *A*-modules. Clearly, the product in the categories can be similarly constructed.

For the case of , we obtain the product topology for , where as a basis, we take products of the form where each is open and for all but finitely many . This also explains why the product topology is preferred over the box topology, where we remove the restriction that for all but finitely many .

Proposition 2.The product in is unique up to unique isomorphism, i.e. suppose

are both products in the above sense. Then there is a unique isomorphism such that for each .

**Proof**

Same as the case for *A*-modules. Left as an exercise. ♦

The product is not guaranteed to exist in ; the above only guarantees uniqueness.

### Diagonal Map

Let . Consider the case of , the product of *I* copies of *A*. If we take the collection to be all identities, then:

- we get a unique satisfying for each .

This is called the **diagonal map**. For the cases of groups, rings, *A*-modules and topological spaces, we obtain exactly what we expect, e.g. takes .

### Exercise A

1. Prove that the product exists in the category if and only if the following (contravariant) functor is representable

where , takes .

The upshot is that product in a category can be expressed in the language of representable functors.

2. Suppose the product of exists. Define, for any morphism , a *graph morphism* such that in the category of sets, .

# Functoriality of Products

Suppose , and all exist whenever they are mentioned. For the corresponding projections, we denote

Construction.Given a morphism for each , we have an induced .

For each we let be the composition

By the definition of the product , there is a unique morphism such that for each ,

Here is the construction in diagram:

Proposition 3.Suppose we have morphisms , for each . Let . By the above

- the morphisms induce ;
- the morphisms induce ;
- the morphisms induce .
Then

**Proof**

By construction is the unique morphism satisfying:

On the other hand also satisfies the same condition

Thus . ♦

Corollary 1.Suppose products always exist in . We obtain a functor of categories

which takes to .

# Coproducts in a Category

Next, we generalize the case of direct sum of *A*-modules to objects in a general category.

Definition.The

coproductof comprises of data , where

- is an object in , and
- for each , we have a morphism ,
satisfying the following universal property. For any , where

- is an object, and
- for each , we have a morphism ,
there is a unique morphisim such that .

The reader should note that it is obtained from the product by reversing all the arrows. More formally, coproduct in corresponds to product in the opposite category . Hence we can effortlessly apply all we proved about products to coproducts just by “flipping arrows around”. For example, we know immediately that the coproduct is unique up to unique isomorphism.

**Note**

In category theory, *if a certain notion is useful, its dual is likely to be useful too* (usually in rather unexpected places). For example, projective modules vs injective modules, fibration vs cofibration, etc.

## Coproduct of Groups

We end this article with a short discussion on coproduct in the category of groups.

Let be the category of abelian groups; for , their coproduct has the same underlying group as the product . To see this, note that , and we already know that the direct sum of finitely many *A*-modules is the same as their direct product.

What about coproduct in ? For two groups *G* and *H*, the answer is their *free product* . To describe this concretely, we take the set of all formal products (i.e. strings) of the form:

and product between two such elements is given by concatenation with reduction. For unique representation, we want and . Identity element is simply the empty string. E.g. we have

if (otherwise we need to simplify further). E.g. when , the result is the free group generated by 2 elements. Note that even though , their coproduct in is different from that in . Thus:

If is a subcategory (even full subcategory) of , the product / coproduct of objects in may differ from that in .

In the first proposition, is it not the case that the statement “In summary, the functor is obtained by right-composing with f” should be “In summary, the natural transformation….”?

Fixed. Thanks. 🙂

In the statement of Yondeda’s Lemma should “Every natural transformation ” be

Every natural transformation “?

Thanks – corrected it. 🙂