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I am learning the basics of category theory, so this question is probably obvious to anyone who knows the subject.

The resources I've seen all take the following approach:

0) A category is a collection of objects and morphisms between those objects that satisfy some rules.

1) A functor is a morphism in the category of categories.

2) A natural transformation is a morphism in the category of functors.

But they all stop right there. What about:

3) the morphisms in the category of natural transformations?

4) Or the "morphisms in the category of the morphisms in the category of natural transformations"

5) ...

Are these uninteresting? Why does the "meta-ness" stop at 2 levels deep?

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2 Answers 2

up vote 6 down vote accepted

They are not uninteresting. Just look up the term n-category (e.g. Baez's introduction).

Nevertheless, it's true that functors and natural transformations already suffice for most common ideas and constructions used across mathematics, so this is why one usually stops there.

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I want to point out something potentially misleading about Marek's answer. The n-categories he mentions are not categories, but generalizations of them, so the question still remains, why do categories only form a 2-category, that is, why do people stop after categories, functors, and natural transformations? Why don't people define modifications of natural transformations?

I think it is good to realize that categories really are in an essential way only 2-categorical, if you want interesting higher morphisms you do need to define something like a higher category. One way to think about it is this: natural tranformations are basically homotopies. To make this precise, take I to be the category with two objects, 0 and 1, one morphism from 0 to 1 and the identity morphisms. Then, it is easy to check that to specify a natural tranformation between two functors F and G (both functors C → D) is the same as specifying a functor H : C × I → D which agrees with F on C × {0} and with G on C × {1}.

So then we could get higher morphisms by saying they are homotopies of homotopies, i.e., functors C × I × I with appropriate restrictions. This works, and we indeed get some definition of modification, but it is not interesting as it reduces to just a commuting square of natural transformations, i.e., it can be described simply in terms of the structure we already had.

This is similar to what happens for, say, groups: you can think of a group as a category with a single object where all the morphisms are invertible (the morphisms are the group elements and the composition law is the group product). Then group homomorphisms are simply functors. This makes it sound as if groups now magically have a higher sort of morphisms: natural tranformations between functors! And indeed they do, they are even useful in certain contexts, but they're not terribly interesting: a natural transformation between to group homomorphisms f and g is simply a group element y such that f(x) = y g(x) y -1 . Again, this is described in terms of things we already new about (the group element y and conjugation), and is not really a brand new concept.

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Indeed, good points. – Marek Jun 14 '11 at 15:36
In the last two lines you're claiming $\mathrm{nat}(g,f)\cong \mathrm{hom}(G,G)$. Is this stronger than the Yoneda lemma $\mathrm{nat}(\mathrm{Hom}(G,-),f)\cong fG$? Maybe you use that there is only one object in the domain category and that all morphisms are invertible. Anyhow you seem to imply all functors are representable or something. – NikolajK Oct 28 '14 at 12:59
I don't understand your comment, @NikolajK, in particular I don't know where you got the $\hom(G,G)$ from. In the last two lines I say that for $f,g : G \to H$ we have $\mathrm{Nat}(f,g) = \{ y \in H : \forall x\in G, f(x) = y g(x) y^{-1}\}$ which is very easy to see: each natural transformation has a single component $y$ (because $G$ has a single object) and the stated condition is just naturality. – Omar Antolín-Camarena Oct 28 '14 at 13:19
I wrote $\mathrm{hom}(G,G)$ because it's the set of arrows on the one object $G$, which are exactly the group elements of the group $G$ - you take $y$ only out of $H$ (it wasn't in the answer, but indeed it makes sense, otherwise the multiplication $y·g(x)·y^{-1}$ can't be made sense of). The thing still looks so close to the Yoneda lemma expression - can any functor $g:G\to H$ be represented somehow? – NikolajK Oct 28 '14 at 14:05
It only makes sense to talk about representing functors $G \to \mathrm{Set}$, the codomain of a representable functor is $\mathrm{Set}$, @NikolajK. – Omar Antolín-Camarena Oct 28 '14 at 14:06

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