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Stream: theory: category theory

Topic: Double Transformations

Owen Lynch (Sep 29 2020 at 11:03):

On the ncatlab page on morphisms between double functors (https://ncatlab.org/nlab/show/vertical+transformation), they mention that the category of double categories is cartesian closed. Does anyone know of a good reference that talks about the construction of 2-cells in the exponential object $\mathbb{D}^{\mathbb{C}}$?

John Baez (Sep 29 2020 at 22:56):

This may not delight you, but you can define categories internal to C whenever C has finite limits. When C is nice enough there is a cartesian closed category Cat(C) of categories internal to C, which you cook up just by taking the construction of Cat and copying it over in Set.

Surely Cat is nice enough. So Cat(Cat) will be cartesian closed, and this will be your category of double categories.

What does all this nonsense mean "concretely"? Well, I'm saying you can take the usual construction of $D^C$ for categories and copy it, replacing sets by categories and functions by functors everywhere, and get $\mathbb{D}^{\mathbb{C}}$.

So let's see how it works for plain old categories:

A category C has a set Ob(C) and a set Mor(C) and a bunch of other stuff.

A category D has a set Ob(D) and a set Mor(D) and a bunch of other stuff.

What's the set Ob( $D^C$ )? You start with the set $\mathrm{Ob}(D)^{\mathrm{Ob}(C)} \times \mathrm{Mor}(D)^{\mathrm{Mor}(C)}$, which consists of all maps sending objects to objects and morphisms to morphisms. Then you take equalizers that say the equations in the definition of "functor" hold.

What's the set Mor( $D^C$ ) ? You start with the set $\mathrm{Mor}(D)^{\mathrm{Ob}(C)}$, which consists of all maps sending objects to morphisms. Then you take equalizers that say the equation in the definition of "natural transformation" hold.

So, you just copy all this stuff for double categories, with as little creativity as possible. I won't even change the font!

A couble category C has a category Ob(C) and a category Mor(C) and a bunch of other stuff.

A double category D has a category Ob(D) and a category Mor(D) and a bunch of other stuff.

What's the category Ob( $D^C$ )? You start with the category $\mathrm{Ob}(D)^{\mathrm{Ob}(C)} \times \mathrm{Mor}(D)^{\mathrm{Mor}(C)}$. Then you take equalizers that say the equations in the definition of "functor internal to Cat" hold. Abstractly these look just like those in the definition of "functor".

What's the category Mor( $D^C$ ) ? You start with the category $\mathrm{Mor}(D)^{\mathrm{Ob}(C)}$. Then you take equalizers that say the equation in the definition of "natural transformation internal to Cat" hold. Abstractly these look just like those in the definition of "natural transformation".

John Baez (Sep 29 2020 at 22:58):

In short: you can just turn the crank.

Reid Barton (Sep 30 2020 at 13:36):

Also, in this situation and many similar ones, the various kinds of data that make up an object of the category can be detected by mapping in corresponding "free" objects, so if you somehow know or just guess that the category is cartesian closed, you can calculate what the exponential objects have to be.
Here, there's a "walking 2-cell" double category $S$ such that, for any double category $D$, the 2-cells of $D$ are in natural correspondence with the elements of $\mathrm{Hom}(S, D)$. Namely, $S$ is just what you would draw as a 2-cell: it has a single 2-cell together with four objects, two horizontal morphisms, and two vertical morphisms that make up its boundary.
Then, by definition, whatever the exponential object $D^C$ is, it must satisfy $\mathrm{Hom}(S, D^C) = \mathrm{Hom}(S \times C, D)$, so the 2-cells of $D^C$ correspond to double functors from $S \times C$ to $D$.
In the same way, you can compute the rest of the $D^C$ using other "walking" objects in place of $S$ and the maps between them.

Owen Lynch (Sep 30 2020 at 17:04):

OK, this makes sense; I was trying to do JB's construction, but I missed the critical step of looking at $\mathsf{Mor}(D^C)$; I was only looking at $\mathsf{Ob}(D^C)$. And thank you Reid, that makes it a bit more intuitive!

Owen Lynch (Sep 30 2020 at 17:05):

And maybe a bit more delightful!