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)