Yoneda Lemma

Let us take a look at Yoneda Lemma, which might be the most trivial yet the hardest part of Category theory (and algebraic geometry). I would not be drawing any commutative diagrams.


Take a (small locally presumably poset) category $\mathcal{C}$ and hom-functors $h$ on it to ${\bf Set}^C$. So if we have a set of morphism $mor(A,B)$ ($\pi \colon A \rightarrow B$) for $A,B \in \mathcal{C}$, I can construct a functor to ${\bf Set}^C$ out of set of morphism which I write as $H(A, B)$. One does this for every object inside $\mathcal{C}$; in this way, we get many sets of morphisms to form $H(A, X)$. We now find the normal (representation) isomorphism of this functor
$$\xi \colon F \rightarrow Hom(A,X)$$
and this means that an object $A$ is determined up to isomorphism by the pair $(\xi, F)$. We can also say $F$ is the $Hom(Hom(A,X))$. 

Yoneda Lemma states that any information about the local category is encoded in ${\bf Set}^C$. The set of the morphism becomes the objects for ${\bf Set}^C$, and morphism is given by the natural representation of the functor. So any functor in $\mathcal{C}$ can be sent to its functor category ${\bf Set}^C$, which sends $A$ to $h$. Note that we did not say if $h$ is a covariant or contravariant functor, the result is the same for either.

The philosophy of Yoneda Lemma is also encaptured in this video, essentially meaning why only one view is wrong. Another good exercise is to realize how this is a universal property and why taking maps to and from $A$ is important to understand a category.

This entry was posted in . Bookmark the post. Print it.

2 Responses to Yoneda Lemma

  1. Harsh says:

    When category theory got you? Remember me?

  2. Harsh,
    Yes, I do remember.

Leave a Reply