3
$\begingroup$

I just started reading Awodey's category theory text book and I came across functors and the category of categories. It seems to me that one can define a functor-like object that takes categories of categories to categories of categories:

A bigfunctor $$ \mathcal{F} : \textbf{C} \rightarrow \textbf{D} $$ between categories of categories is a mapping of categories to categories and functors to functors such that

  • $\mathcal{F}(F: \textbf{C} \rightarrow \textbf{D}) = \mathcal{F}(F): \mathcal{F}(\textbf{C}) \rightarrow \mathcal{F}(\textbf{D}) $
  • $\mathcal{F}(G \circ F) = \mathcal{F}(G) \circ \mathcal{F}(F) $
  • $\mathcal{F}(1_{\textbf{A}} ) = 1_{\mathcal{F}(\textbf{A})} $

where the boldletters are categories from the categories of categories and $F, G$ are functors between categories.

Could we then speak of categories of categories of categories? Is this of any importance or does it somehow "collapse" to the category of categories? Seems like you can go on to make categories of categories of categories of.... of categories and so on. Are these interesting from any point of view?

$\endgroup$
2
  • $\begingroup$ We do not have full LaTeX here, only the maths part. $\endgroup$
    – user856
    Oct 17, 2019 at 18:38
  • $\begingroup$ Aren't you simply talking about functors ${\bf Cat}\to{\bf Cat}$? (Or, perhaps ${\bf C}\to {\bf Cat}$ where $\bf C$ is a subcategory of ${\bf Cat}$?) $\endgroup$
    – Berci
    Oct 17, 2019 at 22:27

1 Answer 1

1
$\begingroup$

A "bigfunctor" $\mathscr F:\textbf{C} \rightarrow \textbf{D}$ is just a functor of categories since $\textbf{C}$ and $\textbf{D}$ both are categories in the usual sense.

Something you may be interested in though are $n-$categories. They consist of a collection of objects $ob(\textbf X)$ along with $1-$morphisms between two objects. And it has additional structure such that for two objects a,b in $\textbf{X}$, $\text{Hom}(a,b)$ has the structure of a category whose objects are $1-$morphisms and arrows are called $2-$morphisms. $2-$morphisms are "arrows between arrows" and if $f,g$ are two $1-$morphisms in $\textbf X$, $\text{Hom}(f,g)$ also has the structure of a category whose objects are $2-$morphisms and arrows are called $3-$morphisms or "arrows between arrows between arrows". Then you also have to have the category whose objects are $3-$morphisms and morphisms called $4-$morphisms.

Anyway, if your category meets all of these requirements up to $n-$morphisms it's an $n-$category.

A category of categories is a $2-$category because for two categories you can form the functor category $\text{Funct}(C,D)$ whose $2-$morphisms are the natural transformations. This makes the set $\text{Hom}(C,D)$ the objects of $\text{Funct}(C,D)$ so this indeed is a $2-$category.

If you know about enriched categories you can define $n-$categories inductively as categories enriched over the category of $(n-1)-$categories and defining $1-$categories to be usual categories.

$\endgroup$
1
  • 1
    $\begingroup$ or start by defining $0$-categories: the sets. $\endgroup$
    – Berci
    Oct 17, 2019 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.