Another (covariant) functor
January 20, 2012 1 Comment
I have one more example of a (covariant) functor that I thought is worth mentioning. It’s so important, it has it’s own name.
If , . We need to be locally small to ensure that really is a set. For a morphism in is a map from to . It sends to . That is,
We call these functors representable functors. Yes, this is complicated. The objects aren’t so hard, but the morphisms are confusing. I don’t know a good way to understand them. Beyond working through the definitions and checking that this really is a functor, I don’t know any good exercises. Try to do it yourself. If you get stuck, you can look below at what I’ve done, but as is almost always true in math, it’s best to do it yourself if you can.
We consider by . As a set map, this is simply the identity function on the set . That is, it is . Another way to write this is .
Now let and be morphisms in . This gives us . We can break this map into two pieces. First, we do the piece that composes with . Then we do the part that composes with . That is, first we have , then we apply the map to get . That is,
as desired. Another way to check this is to check that the diagram
commutes. Let’s check it again thinking about it this way. To see if a diagram commutes, we need to check that all paths give us the same result. In this case, there’s only one thing to check. Take .
One way to get from to gives us . The other way is by composing and . If we start with , and apply the map , we get . After applying the map , we get . By associativity, these are the same results, which means the diagram commutes.