Representable

We say a Functor is representable
if it’s Isomorphic to the Hom-Functor for some ,
i.e. there is some Natural Isomorphism

We also say a Contravariant Functor is representable
if it’s Isomorphic to the the Contravariant Hom-Functor for some .

In each of the following examples,
we will find that the Natural Isomorphism at some object
looks like

where is some element of .
This is the key to Yoneda Lemma.

Example

The identity functor is representable.
In particular, take the functor .
For any set , we have Isomorphic to
The Natural Isomorphism is defined by sending
to a function
sending

(where is the only element of ).
This is clearly invertible.

Example

The Forgetful Functor is representable.
In particular consider .
For a Group we can find
defined by:

Note that we needed to “forget” that is a Homomorphism,
in order to be able to use it as a normal function between sets and .
We can then check that is a Natural Isomorphism.

We might also try .
Afterall, this will be a Natural Transformation.
Can we invert it? No.
Consider .
There is two elements of ,
but both of them give when evaluated at .
Thus we cannot differentiate them by their value at .
The special property of that allows us to define a Natural Isomorphism
is that is a Universal Element.

Example

The Contravariant Power Set Functor
is representable.
In particular, try .
For any set , we can define by:

i.e.

Example

The Covariant Power Set Functor
is not representable.
Let be any set and any Natural Transformation
Then for a set , we have that is a function:

Try , noting that .
But clearly has only one element,
so is not bijective and thus is not a Natural Isomorphism
(remember that is a Natural Isomorphism
if and only if
each of is Isomorphic)