Yoneda Lemma

Let be an Object of a Locally Small Category . Let be a Hom-Functor . and let be a Functor to Category of Sets. These are both objects in the Category of Functors .

Yoneda lemma says that the set of Natural Transformations is Isomorphic to :

Moreover, this Isomorphism is natural in both and . This natural mapping is given by:

and its inverse is given by:

where

If is Locally Small, we could say: given the functors

there is a Natural Isomorphism between them in the category . But we would be perfectly fine in just stating the naturality in elementary terms when is not Locally Small.

what is actually written here?

In a way, given from and from , there is only two ways to make a set out of them. The obvious one is . The other way is the set of Natural Transformations from to . Yoneda says that these two ways are the same “in a natural way”.

Suppose that for a Functor we could define a functor

which is also (i.e. it takes and gives us the set ) Then we could define Yoneda lemma says that there are Natural Isomorphisms between these, so we don’t need to worry about the infinite tower.

Proof

Firstly, fix and call and . We need to prove:

and

Firstly, let :

Now let be a Natural Transformation.

For any Morphism in we have:

The Naturality Square of at is:

\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
\mathcal{C}(A,A) \arrow[r,"\mathcal{C}(A{,}f)"] \arrow[d,"\alpha_{A}"]
 & \mathcal{C}(A{,}B) \arrow[d,"\alpha_{B}"] \\
FA \arrow[r,"Ff"]
 & FB
\end{tikzcd}
\end{document}

Thus

where is actually just Thus

as desired.

Now let us verify that is a Natural Transformation. Note that will automatically be a Natural Isomorphism because we have verified that is always an Isomorphism, and by uniqueness of inverses, will also be a Natural Isomorphism.

Consider first and the diagram:

\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}[column sep=huge]
{[\mathcal{C},\mathrm{Set}]}(\mathcal{C}(A,-),F) 
\arrow[r,"\bullet \cdot \mathcal{C}(f{,}-)"] 
\arrow[d,"\Phi_{(A{,}F)}"]
 & {[\mathcal{C},\mathrm{Set}]}(\mathcal{C}(A',-),F) 
 \arrow[d,"\Phi_{(A'{,}F)}"] \\
FA \arrow[r,"Ff"]
 & FA'
\end{tikzcd}
\end{document}

where

is a Natural Transformation Let be a Natural Transformation.

and

On the other hand

So is a Natural Transformation in .

Now we verify that it is a Natural Transformation in . Consider , a Natural Transformation. We have a diagram

\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
{[\mathcal{C},\mathrm{Set}]}(\mathcal{C}(A,-),F) 
\arrow[r, "\alpha \cdot\bullet"] 
\arrow[d, "\Phi_{(A{,}F)}"]
& {[\mathcal{C},\mathrm{Set}]}(\mathcal{C}(A,-),F')
\arrow[d,"\Phi_{(A{,}F')}"] \\
FA \arrow[r,"\alpha_{A}"] 
& F'A
\end{tikzcd}
\end{document}

Let be another Natural Transformation. One branch gives

The other gives

and thus is natural in .

So is natural in for any and it is natural in for any so we conclude that is natural in (we can also see this by substituting in the second square and then gluing the two squares together).