# Yoneda’s Lemma (part 2)

February 20, 2012 Leave a comment

#### Theorem (Yoneda):

Let be a locally small category. Let be an object in , and let . Then,

,

and this isomorphism is natural in and in . (we won’t prove this part here)

#### Proof:

Given a natural transformation from to , we need to produce an element in . There aren’t many things we could try. We have a map . We can use that to produce an element in . Of course, we’d have to plug something in to , and there could be many choices there. There is an obvious one though. In fact, it’s the only element of we know exists: the identity map. So for a natural transformation , the associated element of we’ll produce is going to be . I’ll write this as function as . so . We need to check that is an isomorphism.

First, let’s check that has an inverse. One thing we could do is check that if we knew , we could recover . So let’s say we have some , and we want to find a natural transformation such that . What do we need to specify ?

A natural transformation is defined by all of the maps (we have one for each in ). So if I can tell you what each is going to be, without making any choices, then I’ve uniquely defined a natural transformation from . Great, so how do I specify ? Well that’s just a map , so I need to tell you what it does to each element in the domain (that is, each ).

So I’m going to define the transformation, and then explain why it works. Let , and let . Then should be an element of . I define it to be:

How do we check if this defines a natural transformation? We need to check the commuting stuff. So take a look at this diagram (for an arbitrary ):

Starting with an element , and going down and then across is

Going across and then down is

Great, so the diagram commutes. So really is a natural transformation. We just need to check that it’s the inverse of the map. That is, we need to check that . We also need to check the other direction, that if we start with some , compute , and then do this construction, we get back the original . I’m lazy, so I’ll only do the first one, though neither are difficult. You simply need to wade through the definitions.

Suppose we construct as above from a given . We hope that . Indeed, . From the definition of ,

.