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Stream: learning: questions

Topic: nCat as an (n+1)-category

John Baez (Nov 21 2020 at 17:56):

I got an email that required a long answer. I thought some other people might like to see the answer.

Dear John
I have problem with understanding the 3cells in 3Cat:
can we evaluate a 3-cell in 3Cat at all of these: an object, an arrow (1-cell) and 2-cell (a natural transformation) ?
And yet: how is the 3-cell determined by what data and between what type (for which k in k-cell) of objects it leads ?
Best, Jan Pax

My reply:

John Baez (Nov 21 2020 at 17:58):

It's easiest to think about it in general:

A 0-cell in nCat is an n-category.

A 1-cell in nCat is a map between n-categories, say f: A $\to$ B. This sends any k-cell in A to a k-cell in B.

A 2-cell in nCat goes between a map f: A $\to$ B and a map g: A $\to$ B. It sends any k-cell in A to a (k+1)-cell in B...

... but what happens when k+1 > n? When k+1 = n+1, our 2-cell sends any k-cell in A to an equation that must hold in B.

For example, take n=1. Then a 2-cell in nCat is a natural transformation between functors between categories. It sends any object in A to a morphism in B, but it sends any morphism in A to an equation saying in B saying that a square commutes: the "naturality square".

Secretly the reason is that an n-category has (n+1)-cells that are equations between n-cells. It also has (n+2)-cells and so on, but these are "equations between equations", so they convey no information at all.

So, the pattern works like this:

A 2-cell in nCat goes between a map f: A $\to$ B and g: A $\to$ B. It sends any k-cell in A to a (k+1)-cell in B if k+1 ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+1 = n+1. And it does nothing at all to k-cells with k+1 > n+1.

Next:

A 3-cell in nCat goes between natural transformations between a map f: A $\to$ B and a map g: A $\to$ B. It sends any k-cell in A to a (k+2)-cell in B if k+2 ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+2 = n+1. And it does nothing at all to k-cells with k+2 > n+1.

Now we see the general pattern. A j-cell in nCat sends any k-cell in some n-category A to some (k+j)-cell in some n-category B if k+j ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+j = n+1. And it does nothing at all to k-cells with k+j > n+1.

Dan Doel (Nov 21 2020 at 18:32):

A j-cell sends a k-cell to a (k+j-1)-cell, right?

Dan Doel (Nov 21 2020 at 18:33):

Or perhaps better to start with a (j+1)-cell, since the 0-cells don't send things.

John Baez (Nov 21 2020 at 19:11):

Oh damn, did I screw up the counting?

John Baez (Nov 21 2020 at 19:12):

Okay, I just need to fix this last part:

Now we see the general pattern. A (j+1)-cell in nCat sends any k-cell in some n-category A to some (k+j)-cell in some n-category B if k+j ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+j = n+1. And it does nothing at all to k-cells with k+j > n+1.

Dan Doel (Nov 21 2020 at 19:13):

Yeah, I think that's the only change necessary.

Jan Pax (Nov 22 2020 at 17:55):

Dear John,
could you kindly add to your answer how does the equation look like for a general n,k and j ?
For which n,k and j it is the classical naturality square ?

John Baez (Nov 22 2020 at 19:12):

For this we need to start by deciding whether we're talking about strict or weak n-categories.

Jan Pax (Nov 22 2020 at 19:15):

I would like to see both cases, indeed. We can start with strict they should be easier to cope with.

John Baez (Nov 22 2020 at 19:16):

Okay, so think about strict 2-categories. There's a strict 3-category 2Cat consisting of

[strict 2-categories, strict functors, strict natural transformations, strict modifications]

John Baez (Nov 22 2020 at 19:17):

(People don't always say "strict" all the time.)

John Baez (Nov 22 2020 at 19:17):

Do you know what a modification between natural transformations looks like, in this strict context?

Jan Pax (Nov 22 2020 at 19:19):

not sure at all ...

John Baez (Nov 22 2020 at 19:20):

Well then you should start by figuring that out, or talking about it. There's no way to study n-categories without first doing 2-categories.

John Baez (Nov 22 2020 at 19:21):

Can you draw what happens when you have two 2-categories $A,B$, two functors $f,g : A \to B$, and two natural transformations $\alpha, \beta : f \Rightarrow g$?

John Baez (Nov 22 2020 at 19:22):

Can you see how to create something "going from $\alpha$ to $\beta$"? It helps to draw diagrams of naturality squares...

Jan Pax (Nov 22 2020 at 19:22):

what do you mean by "draw" ? I'm not a native English speaker.

John Baez (Nov 22 2020 at 19:22):

Draw = create a picture.

Jan Pax (Nov 22 2020 at 19:22):

Oh that would be nice if I could draw here diagrams, I cannot.

John Baez (Nov 22 2020 at 19:23):

Right, but you can draw them on paper.

John Baez (Nov 22 2020 at 19:23):

I'm assuming you know how the natural transformation $\alpha$ gives a bunch of squares in $B$, one for each object in $A$.

John Baez (Nov 22 2020 at 19:24):

So you can draw one.

Jan Pax (Nov 22 2020 at 19:24):

they should be n.t. at all objects $A$ yes

John Baez (Nov 22 2020 at 19:24):

You need to use two dollar signs here, not one.

John Baez (Nov 22 2020 at 19:24):

Then you can draw what $\beta$ does, in the same picture.

John Baez (Nov 22 2020 at 19:25):

It's always good to draw lots of things in the same picture so you can see how they interact.

Jan Pax (Nov 22 2020 at 19:25):

yes, and I can compose $\alpha,\beta$ at $A$ ...

John Baez (Nov 22 2020 at 19:26):

Well, $\alpha : f \Rightarrow g$ and also $\beta : f \Rightarrow g$, so it's not a good idea to try to compose $\alpha$ and $\beta$.

John Baez (Nov 22 2020 at 19:27):

What you're trying to do is figure out what a "modification" is: something that goes from $\alpha$ to $\beta$.

John Baez (Nov 22 2020 at 19:27):

Modifications will be the 3-cells in 2Cat.

John Baez (Nov 22 2020 at 19:27):

You were asking me if they look like squares, so I'm trying to get you to draw one.

Jan Pax (Nov 22 2020 at 19:28):

I'm listening

John Baez (Nov 22 2020 at 19:28):

I want you to be drawing, not listening. :upside_down: I think it'll work better if you do a lot of the work here. I can talk about this for about fifteen minutes a day, and other people can join in too.

Jan Pax (Nov 22 2020 at 19:29):

well, I cannot draw a good diagram now. I have a one which says what are n.t. for a morphism $f$ and object $A$.

Jan Pax (Nov 22 2020 at 19:30):

$\alpha: f\implies g$ at $A$

John Baez (Nov 22 2020 at 19:30):

Good. Draw what both $\alpha$ and $\beta$ do to a morphism in $A$, in the same picture, and see where there's room to have something going from $\alpha$ to $\beta$: a modification.

John Baez (Nov 22 2020 at 19:31):

As a clue, you already know that a modification will give some sort of 2-cell in $B$ for each object of $A$, and some sort of equation for each 1-morphism in $A$.

John Baez (Nov 22 2020 at 19:32):

So you need to figure out what this 2-cell looks like, and what this equation looks like. The picture will tell you.

Jan Pax (Nov 22 2020 at 19:33):

$\alpha_A \circ f_A$

Jan Pax (Nov 22 2020 at 19:36):

$\alpha_A\circ f=g\circ \beta_A$

Jan Pax (Nov 22 2020 at 19:41):

AYT, John ?

John Baez (Nov 23 2020 at 00:47):

Jan Pax said:

$\alpha_A\circ f=g\circ \beta_A$

I'm not sure what this means. We said $\alpha$ and $\beta$ are natural transformations from $f: A \to B$ to $g: A \to B$. This means that for every morphism $h : a \to b$ in $A$ these equations hold:

$\alpha_b \circ f(h) = g(h) \circ \alpha_a$

and

$\beta_b \circ f(h) = g(h) \circ \beta_a$

But the question is, what should a modification from $\alpha$ to $\beta$ do? I suggested taking the two equations above and drawing them as diagrams: this suggests what the modification should do!

James Wood (Nov 23 2020 at 11:16):

Feels like this would be easier to draw on a cylinder (with end faces), but following along I seem to get something reasonable.

Jan Pax (Nov 23 2020 at 13:38):

I could only understand that these 2 equation are syntactically correct,meaning the domain and codomain are right. Unfortunately,I cannot do more. I'd be happy if you can guide me through this in some detail.

Morgan Rogers (he/him) (Nov 23 2020 at 14:39):

He was asking you to draw this...
modification.png.

To produce this diagram, I wrote the following code in LaTeX, but John was trying to get you to do this by hand (ie, on paper, for your own benefit, rather than to share with everyone!).

% \usepackage{tikz-cd}
\[\begin{tikzcd}
f(a) \ar[r, "f(h)"] \ar[d, "\beta_a", bend left] \ar[d, "\alpha_a"', bend right] & f(b) \ar[d, "\beta_b", bend left]  \ar[d, "\alpha_b"', bend right]  \\
g(a) \ar[r, "g(h)"'] & g(b)
\end{tikzcd}\]

Looking at this, what would you say the data of a modification from $\alpha$ to $\beta$ should be?

Jan Pax (Nov 23 2020 at 14:52):

Nice picture, thank you. I do not know the entire 3-cell though.

James Wood (Nov 23 2020 at 14:54):

Just to make sure I've got all the details, if we consider one of these naturality squares (say, the one about α), does it commuting mean the existence of a 2-cell isomorphism between the two halves?

James Wood (Nov 23 2020 at 14:57):

i.e, $α_b \circ f(h) ≅ g(h) \circ α_a$?

Morgan Rogers (he/him) (Nov 23 2020 at 15:00):

Indeed. I believe this data is included in natural transformations between 2-functors between 2-categories.

Jan Pax (Nov 23 2020 at 15:02):

Does this isomorphism correspond to the weak case ?

Morgan Rogers (he/him) (Nov 23 2020 at 15:03):

Jan Pax said:

Nice picture, thank you. I do not know the entire 3-cell though.

Well, we should expect a 3-cell to compare data like-for-like. From the diagram, you might be able to intuite that the basic data consists, for each 0-cell $a$ of $A$, of a 2-cell $\alpha_a \to \beta_a$. The extra data and conditions on this basic data should be the ones that make this "cylinder" diagram commute in all of the ways we can ask it to.

Morgan Rogers (he/him) (Nov 23 2020 at 15:05):

Jan Pax said:

Does this isomorphism correspond to the weak case ?

I want to say yes, so that in the strict case (of "strict natural transformations" between these functors), these isomorphisms will be identities. But you typically have to be careful about how much of the data you can force to be trivial/strict, so it's best to keep in the back of your mind that this exists.

John Baez (Nov 23 2020 at 17:06):

In the case of a strict natural transformation we have equations

$\alpha_b \circ f(h) = g(h) \circ \alpha_a$

and

$\beta_b \circ f(h) = g(h) \circ \beta_a$

John Baez (Nov 23 2020 at 17:08):

But still, I was asking Jan Pax to draw this picture on a piece of paper:

Morgan Rogers' picture

John Baez (Nov 23 2020 at 17:13):

because this picture is how you figure out what we need to define a "modification" between natural transformations. Think of this picture as a 3-dimensional cylinder with the the edges $\alpha_a$ and $\alpha_b$ curving towards you, out of the page and $\beta_a$ and $\beta_b$ curving down into the page.

John Baez (Nov 23 2020 at 17:14):

The square curving towards you commutes when we have a strict natural transformation (or is filled in by a 2-cell if we have a weak one), and so does the square curving down into the page.

John Baez (Nov 23 2020 at 17:16):

But the two disk-shaped ends of the cylinder are still open, not filled in. And so is the 3-dimensional body of the cylinder. Those are the things the modification could provide!

John Baez (Nov 23 2020 at 17:17):

Since we're only doing 2-categories today, the 3-dimensional body must be filled with an equation. But the disk-shaped ends of the cylinder must be filled with 2-morphisms.

John Baez (Nov 23 2020 at 17:19):

I think I've given enough hints. So here's my question again: what is a modification between strict natural transformations between strict functors between strict 2-categories?

Jan Pax (Nov 23 2020 at 17:33):

a natural transformation between the components of the natural transformations $\alpha_A$ and $\beta_A$ ?

sarahzrf (Nov 23 2020 at 23:22):

James Wood said:

Feels like this would be easier to draw on a cylinder (with end faces), but following along I seem to get something reasonable.

i got you!
https://twitter.com/sarah_zrf/status/1329092567573569539

i diagrammed out the definition of a modification in a form that i can at least *kinda* look at pieces: 1. Γ is a modification φ → ψ 2. Γ maps 0-cells A ∈ K to 2-cells Γ_A ∈ L 3. Γ maps 1-cells f ∈ K to "3-cells", commutative diagrams of 2-cells, in L https://twitter.com/sarah_zrf/status/1329092567573569539/photo/1

- profinite sarahzrf (@sarah_zrf)

John Baez (Nov 24 2020 at 00:00):

Jan Pax said:

a natural transformation between the components of the natural transformations $\alpha_A$ and $\beta_A$ ?

Yes, something like that... can you say exactly what 2-morphisms your modification will consist of, and what equations they should obey?

Btw, you can use \Rightarrow to draw 2-morphisms: $\Rightarrow$.

Jan Pax (Nov 24 2020 at 00:03):

just the most naive one: $\alpha_A\implies \beta_A$, but your question suggests that some more equations are to hold ?

John Baez (Nov 24 2020 at 00:05):

At least give the 2-morphism a name - say something like this:

A modification $M$ from $\alpha$ to $\beta$ gives a 2-morphism $M_a : \alpha_a \Rightarrow \beta_a$ in $B$ for each object $a \in A$.

Jan Pax (Nov 24 2020 at 00:06):

Your $\to$ should read $\implies$ right ?

John Baez (Nov 24 2020 at 00:06):

Yes.

Jan Pax (Nov 24 2020 at 00:07):

Have we finished or some new things are to be understood about 3Cat ?

Jan Pax (Nov 24 2020 at 00:07):

or can we move to the weak case ?

John Baez (Nov 24 2020 at 00:07):

What equations should the $M_a$ obey? I don't care if you give me "obvious" or "unobvious" equations, just tell me all the equations you can think of, that should hold.

John Baez (Nov 24 2020 at 00:09):

Let's finish doing this case, strict 2-categories.

Jan Pax (Nov 24 2020 at 00:09):

$\alpha_a\circ f(h)=g(h)\circ\alpha_b$ for all $h:a\to b$ ?

John Baez (Nov 24 2020 at 00:10):

That sounds good so far: one equation for each 1-morphism in $A$.

John Baez (Nov 24 2020 at 00:11):

So notice $m$ is giving one 2-cell for each 0-cell in $A$, and one 3-cell for each 1-cell in $A$.

John Baez (Nov 24 2020 at 00:11):

The 3-cells in a 2-category are equations between 2-cells.

John Baez (Nov 24 2020 at 00:12):

So what does $m$ give for each 2-cell in $A$?

Jan Pax (Nov 24 2020 at 00:13):

an equation of 2-cells between 2-cells ?

John Baez (Nov 24 2020 at 00:14):

I don't think you're following the pattern:

So notice $m$ is giving one 2-cell for each 0-cell in $A$, and one 3-cell for each 1-cell in $A$.

John Baez (Nov 24 2020 at 00:14):

Remember what I said:

John Baez (Nov 24 2020 at 00:15):

Now we see the general pattern. A j-cell in nCat sends any k-cell in some n-category A to some (k+j)-cell in some n-category B if k+j ≤ n....

Jan Pax (Nov 24 2020 at 00:16):

$(j+1)$-cell in nCat, right ?

Jan Pax (Nov 24 2020 at 00:42):

I would guess that we are beyond k+j ≤ n.... so some equations will come into the play.

John Baez (Nov 24 2020 at 16:37):

I should have said (k+j+1)-cell in nCat.

John Baez (Nov 24 2020 at 16:39):

Jan Pax said:

I would guess that we are beyond k+j ≤ n.... so some equations will come into the play.

Yes, but here's the pattern - with that mistake fixed:

John Baez said:

Now we see the general pattern. A (j+1)-cell in nCat sends any k-cell in some n-category A to some (k+j)-cell in some n-category B if k+j ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+j = n+1. And it does nothing at all to k-cells with k+j > n+1.

Jan Pax (Nov 24 2020 at 17:03):

I'm lost, can you still help more ? what equations $m$ involves/implies ?

John Baez (Nov 24 2020 at 17:07):

This was my question:

John Baez said:

So notice $m$ is giving one 2-cell for each 0-cell in $A$, and one 3-cell for each 1-cell in $A$.

The 3-cells in a 2-category are equations between 2-cells.

So what does $m$ give for each 2-cell in $A$?

So what's your answer now?

John Baez (Nov 24 2020 at 17:07):

I gave a hint:

Now we see the general pattern. A (j+1)-cell in nCat sends any k-cell in some n-category A to some (k+j)-cell in some n-category B if k+j ≤ n. It sends any k-cell in A to an equation between k-cells in B if k+j = n+1. And it does nothing at all to k-cells with k+j > n+1.

Jan Pax (Nov 24 2020 at 17:16):

I would say $j=2,n=2,k=2$ so nothing at all.

John Baez (Nov 24 2020 at 17:20):

Right, that's the answer I was hoping for!

Jan Pax (Nov 24 2020 at 17:42):

Would you like to move to the weak case ? What will change for it ?

John Baez (Nov 24 2020 at 18:40):

Sure - try the weak case! It's very helpful to consider strict functors between strict 2-categories, so nothing changes at that level... but then consider weak natural transformations between those, and weak modifications between those.

Can you guess how the weak natural transformations are different?

John Baez (Nov 24 2020 at 18:42):

So, suppose $A$ and $B$ are strict 2-categories and $f: A \to B$ and $g: A \to B$ are strict functors. When we had a strict natural transformation, for every morphism $h : a \to b$ in $A$ this equation would hold:

$\alpha_b \circ f(h) = g(h) \circ \alpha_a$

How should this change in the weak case? Again it's very helpful to draw a picture on some paper.

Jan Pax (Nov 24 2020 at 18:43):

instead of = will be $\cong$.

John Baez (Nov 24 2020 at 18:43):

Right. Now be more explicit: give things names, since we'll need them.

Jan Pax (Nov 24 2020 at 18:45):

let $m$ be the invertible 3-cell.

John Baez (Nov 24 2020 at 18:49):

Huh?

John Baez (Nov 24 2020 at 18:50):

You said there should be an $\cong$ instead of an $=$ somewhere. We have to give this 2-isomorphism a name, so we can begin to work with it.

John Baez (Nov 24 2020 at 18:50):

Say something like "let .... be the 2-isomorphism from .... to ....".

Jan Pax (Nov 24 2020 at 18:51):

I was trying to name it $m$

John Baez (Nov 24 2020 at 18:51):

But you said "3-cell".

John Baez (Nov 24 2020 at 18:51):

There's no 3-cell in this story yet.

Jan Pax (Nov 24 2020 at 18:52):

I thought that $\cong$ will be the 3-cell, but I'm certainly wrong.

John Baez (Nov 24 2020 at 18:53):

I think you didn't read my question carefully.

John Baez (Nov 24 2020 at 18:54):

John Baez said:

So, suppose $A$ and $B$ are strict 2-categories and $f: A \to B$ and $g: A \to B$ are strict functors. When we had a strict natural transformation, for every morphism $h : a \to b$ in $A$ this equation would hold:

$\alpha_b \circ f(h) = g(h) \circ \alpha_a$

How should this change in the weak case? Again it's very helpful to draw a picture on some paper.

Jan Pax (Nov 24 2020 at 18:54):

Let $m$ be an isomorphism from $\alpha_b\circ f(h)$ to $g(h)\circ \alpha_a$.

John Baez (Nov 24 2020 at 18:55):

What sort of cell is this? A 1-cell, 2-cell, 3-cell?

John Baez (Nov 24 2020 at 18:55):

And what does it depend on? Should we call it just $m$, or $m_{a,b}$, or... what?

Jan Pax (Nov 24 2020 at 18:56):

$m_{a,b,h}$ will do, but I'm not sure.

Jan Pax (Nov 24 2020 at 18:57):

I'd say 3-cell but you have denied this.

John Baez (Nov 24 2020 at 19:01):

Jan Pax said:

$m_{a,b,h}$ will do, but I'm not sure.

That's okay. Good!

But notice: $h: a \to b$. We usually say that a morphism determins its source and target. So if you know the morphism $h$, you know its source is $a$ and its target is $b$. So we can just write $m_h$; we don't need to say $m_{a,b,h}$. $h$ tells you what $a$ and $b$ are.

John Baez (Nov 24 2020 at 19:02):

To see if $m_h$ is a 3-cell we need to know about its source and target. You said

John Baez (Nov 24 2020 at 19:02):

Let $m$ be an isomorphism from $\alpha_b\circ f(h)$ to $g(h)\circ \alpha_a$.

What sort of cells are $\alpha_b\circ f(h)$ and $g(h)\circ \alpha_a$?

Jan Pax (Nov 24 2020 at 19:04):

1-cells ?

John Baez (Nov 24 2020 at 19:05):

Right. So what sort of cell can go between them?

John Baez (Nov 24 2020 at 19:05):

Again, if you draw a picture it all becomes easy.

Jan Pax (Nov 24 2020 at 19:06):

2-cells

Jan Pax (Nov 24 2020 at 19:07):

Pictures are easy only if someone else draws them for me.

John Baez (Nov 24 2020 at 19:09):

Okay, so what sort of cell is $m$?

Jan Pax (Nov 24 2020 at 19:10):

2-cell.

John Baez (Nov 24 2020 at 19:10):

Okay - not a 3-cell, a 2-cell.

John Baez (Nov 24 2020 at 19:11):

$m_h : \alpha_b\circ f(h) \Rightarrow g(h)\circ \alpha_a$

Jan Pax (Nov 24 2020 at 19:11):

Yes, this is what I meant from the beginning.

John Baez (Nov 24 2020 at 19:13):

Okay, good. But you said it was a 3-cell.

John Baez (Nov 24 2020 at 19:13):

Now, the next step is quite hard without a picture.

Jan Pax (Nov 24 2020 at 19:14):

I thought that as $\alpha$ is 2-cell then $\cong$ will shift up 1 degree.

John Baez (Nov 24 2020 at 19:14):

$\alpha$ is a natural isomorphism, which is a 2-cell in 2Cat, and $a$ is a 0-cell, so $\alpha_a$ is a 1-cell.

Jan Pax (Nov 24 2020 at 19:15):

OK. Guide me please gently through my next picture.

John Baez (Nov 24 2020 at 19:16):

Well, there are two things we could do. First, these 2-cells $m_h$ need to be obey some equations, in order for $\alpha$ to be a natural isomorphism. We could figure those out.

John Baez (Nov 24 2020 at 19:17):

Second, we could try to figure out what a modification looks like.

John Baez (Nov 24 2020 at 19:18):

The second is easier since Morgan already drew the necessary picture:

Morgan Rogers' picture

John Baez (Nov 24 2020 at 19:19):

Oh, by the way, we made a mistake. We should not call that 2-cell $m_h$.

Jan Pax (Nov 24 2020 at 19:20):

Now I'd put $\implies$ in between $\alpha_a$ and $\beta_a$

Jan Pax (Nov 24 2020 at 19:20):

Why ?

John Baez (Nov 24 2020 at 19:21):

We were figuring out what a natural transformation $\alpha$ is, in the weak case. We noticed it gives a 1-cell $\alpha_a$ for each 0-cell $a \in A$, and then we noticed it gives a 2-cell $\alpha_h$ for each 1-cell $h$ in $A$... but you called that 2-cell "$m$" and I went along with you.

John Baez (Nov 24 2020 at 19:22):

We should call it $\alpha_h$.

Jan Pax (Nov 24 2020 at 19:22):

Thing are clearer now. One letter less is always good.

John Baez (Nov 24 2020 at 19:22):

Okay. So where does $\alpha_h$ go into Morgan's picture?

John Baez (Nov 24 2020 at 19:23):

In category theory you make progress by drawing everything in one picture.

Jan Pax (Nov 24 2020 at 19:23):

Between the 2 $\alpha$'s there: $\alpha_a$ and $\alpha_b$ composed with $f(h)$ and $g(h)$ resp.

John Baez (Nov 24 2020 at 19:30):

Remember some stuff I already told you:

modification.png

Think of this picture as a 3-dimensional cylinder with the the edges $\alpha_a$ and $\alpha_b$ curving towards you, out of the page and $\beta_a$ and $\beta_b$ curving down into the page.

The square curving towards you commutes when we have a strict natural transformation (or is filled in by a 2-cell if we have a weak one), and so does the square curving down into the page.

John Baez (Nov 24 2020 at 19:32):

Which 2-morphism fits into the square curving towards you?

Jan Pax (Nov 24 2020 at 19:33):

$\alpha_h$

John Baez (Nov 24 2020 at 19:34):

Great! Which 2-morphism fits into the square curving down into the page?

Jan Pax (Nov 24 2020 at 19:35):

$\beta_h$

John Baez (Nov 24 2020 at 19:37):

Great! Which 2-morphism fits into the disk at the left end of the cylinder?

Jan Pax (Nov 24 2020 at 19:38):

Is the $\alpha_h$ really from $\alpha_a$ or rather from $g(h)\circ \alpha_a$ ?

John Baez (Nov 24 2020 at 19:39):

We decided that

$\alpha_h : \alpha_b\circ f(h) \Rightarrow g(h)\circ \alpha_a$

John Baez (Nov 24 2020 at 19:40):

This is why it fits into the square curving out towards you.

Jan Pax (Nov 24 2020 at 19:40):

OK.

John Baez (Nov 24 2020 at 19:40):

John Baez said:

Which 2-morphism fits into the disk at the left end of the cylinder?

Jan Pax (Nov 24 2020 at 19:41):

$\alpha_h$

John Baez (Nov 24 2020 at 19:42):

Huh?

Jan Pax (Nov 24 2020 at 19:42):

this is our 2-cell ?

John Baez (Nov 24 2020 at 19:42):

I don't know what "our 2-cell" is.

Jan Pax (Nov 24 2020 at 19:43):

Our isomorphism up to which it commutes the square.

John Baez (Nov 24 2020 at 19:44):

We agreed that $\alpha_h$ fits into the square curving towards us. Now I'm asking what 2-cell fits into the disk at the left end of the cylinder. You said $\alpha_h$. But it can't be in two places at once!

Jan Pax (Nov 24 2020 at 19:45):

Perhaps I do not follow what does it mean "fits into the disk at the left end of the cylinder"? This is not the same thing as in your first sentence ?

Jan Pax (Nov 24 2020 at 19:47):

I cannot find the difference.

John Baez (Nov 24 2020 at 19:49):

We're looking at a picture of a can. The can has a front, a back, a left side, and a right side.

tin-can-lying-its-side-isolated-white-background-tin-can-lying-its-side-isolated-white-background.jpg

John Baez (Nov 24 2020 at 19:50):

That's what Morgan drew:

modification.png

John Baez (Nov 24 2020 at 19:50):

We agreed the front was $\alpha_h$ and the back was $\beta_h$. These are curved squares.

Now I'm asking about the left side and the right side, which are round disks.

John Baez (Nov 24 2020 at 19:51):

This diagram is called a "tin can diagram", by the way.

Jan Pax (Nov 24 2020 at 19:51):

OK, some 2-cell then ?

John Baez (Nov 24 2020 at 19:54):

Right. And I'm asking you which 2-cell goes into the left side.

Jan Pax (Nov 24 2020 at 19:55):

(deleted)

Jan Pax (Nov 24 2020 at 19:59):

the composition $\beta^{-1}_h\circ \alpha_h$ ?

John Baez (Nov 24 2020 at 20:22):

It's one we haven't mentioned yet. What is its source and what is its target?

John Baez (Nov 24 2020 at 20:22):

Also, what's the source and target of $\beta_h^{-1} \circ \alpha_h$?

Jan Pax (Nov 24 2020 at 20:26):

$\alpha_a\implies \beta_a$

John Baez (Nov 24 2020 at 20:32):

You answered the first question, I guess. Yes, the source of the 2-morphism on the left of the tin can is $\alpha_a$, and its target is $\beta_a$.

John Baez (Nov 24 2020 at 20:33):

How about my second question?

John Baez (Nov 24 2020 at 20:33):

John Baez said:

Also, what's the source and target of $\beta_h^{-1} \circ \alpha_h$?

Jan Pax (Nov 24 2020 at 20:36):

$\alpha_a\circ g(h)$ is the source and $\beta_a\circ g(h)$ is the target. But I'm not sure at all.

John Baez (Nov 24 2020 at 20:38):

John Baez said:

We decided that

$\alpha_h : \alpha_b\circ f(h) \Rightarrow g(h)\circ \alpha_a$

John Baez (Nov 24 2020 at 20:39):

so the source of $\alpha_h$ is $\alpha_b \circ f(h)$, so the source of $\beta^{-1}_h \circ \alpha_h$ is also $\alpha_b \circ f(h)$.

John Baez (Nov 24 2020 at 20:41):

So the source of $\beta_h^{-1} \circ \alpha_h$ is different from the source of the 2-morphism on the left of the tin can.

So the 2-morphism on the left of the tin can can't possibly be $\beta_h^{-1} \circ \alpha_h$.

John Baez (Nov 24 2020 at 20:42):

Moral: when you guess what a morphism or 2-morphism might be, check its source and target and see if your guess makes sense.

Jan Pax (Nov 24 2020 at 20:42):

So what it is ?

John Baez (Nov 24 2020 at 20:42):

It's part of a modification from $\alpha$ to $\beta$. We talked about this earlier.

Jan Pax (Nov 24 2020 at 20:44):

I can see now. It's part of our old $m$ ?

John Baez (Nov 24 2020 at 20:44):

John Baez said:

But the two disk-shaped ends of the cylinder are still open, not filled in. And so is the 3-dimensional body of the cylinder. Those are the things the modification could provide!

Since we're only doing 2-categories today, the 3-dimensional body must be filled with an equation. But the disk-shaped ends of the cylinder must be filled with 2-morphisms.

John Baez (Nov 24 2020 at 20:45):

Jan Pax said:

I can see now. It's part of our old $m$ ?

Yes!

Jan Pax (Nov 24 2020 at 20:46):

is $m_a$ 2-cell ?

John Baez (Nov 24 2020 at 20:47):

The left disk is the 2-cell

$m_a: \alpha_a \Rightarrow \beta_a$

John Baez (Nov 24 2020 at 20:48):

where $m$ is our modification going from $\alpha$ to $\beta$.

Jan Pax (Nov 24 2020 at 20:48):

is it subject any equations ?

John Baez (Nov 24 2020 at 20:48):

Yes, and you're looking at that equation.

John Baez (Nov 24 2020 at 20:49):

modification.png

If you fill in all the 2-cells here, this is the equation - it's a commutative diagram of 2-cells.

John Baez (Nov 24 2020 at 20:50):

What's the 2-cell at the right end of the tin can? We haven't discussed that one yet.

Jan Pax (Nov 24 2020 at 20:50):

$m_b$

John Baez (Nov 24 2020 at 20:51):

Right!

John Baez (Nov 24 2020 at 20:52):

So there's some equation involving $m_a, m_b, \alpha_h$ and $\beta_h$ that must hold. And if you know enough about 2-categories you can take Morgan's picture and turn it into this equation!

Jan Pax (Nov 24 2020 at 20:56):

$m_b \ast \alpha_{f(h)} =\beta_{f(h)}\ast m_a$

John Baez (Nov 24 2020 at 20:58):

Something like that. But...

1) What's $\alpha_{f(h)}$? We've never talked about $\alpha_{f(h)}$. What's $f(h)$?

John Baez (Nov 24 2020 at 20:59):

(I have more questions later, but this is the first.)

Jan Pax (Nov 24 2020 at 21:00):

component of a natural transformation under functor $f$ ? I do not know.

Jan Pax (Nov 24 2020 at 21:01):

or $f(\alpha_a)$

John Baez (Nov 24 2020 at 21:21):

Okay: the answer is $f(h)$ doesn't mean anything, so $\alpha_{f(h)}$ doesn't mean anything. Neither does $f(\alpha_a)$. Remember, $f$ is not a function. So, we shouldn't be talking about these things.

John Baez (Nov 24 2020 at 21:22):

We have names for the 4 sides of our tin can, and I was asking you to write down an equation involving these 4 2-morphisms.

Jan Pax (Nov 24 2020 at 21:23):

I did my best, but I will know once you write them for me!

John Baez (Nov 24 2020 at 21:24):

You didn't do your best, because you wrote down an equation involving a nonexistent thing $\alpha_{f(h)}$ that we'd never talked about.

John Baez (Nov 24 2020 at 21:24):

I think you can do better.

John Baez (Nov 24 2020 at 21:27):

Your equation looked right to me at first, but then I noticed several problems with it, the first being this $\alpha_{f(h)}$ thing.

Jan Pax (Nov 24 2020 at 21:28):

(deleted)

Jan Pax (Nov 24 2020 at 21:31):

$g(h)\ast m_a=m_b\ast f(h)$ I would have used $\circ$ instead but the composition doesn't make sense.

John Baez (Nov 24 2020 at 21:31):

What's $f(h)$?

John Baez (Nov 24 2020 at 21:31):

John Baez said:

Okay: the answer is $f(h)$ doesn't mean anything, so $\alpha_{f(h)}$ doesn't mean anything. Neither does $f(\alpha_a)$. Remember, $f$ is not a function. So, we shouldn't be talking about these things.

John Baez (Nov 24 2020 at 21:32):

I hope you have a picture now of the tin can, with each of the 4 sides labelled by its 2-morphism.

Jan Pax (Nov 24 2020 at 21:32):

$f(h):f(a)\to f(b)$

Jan Pax (Nov 24 2020 at 21:33):

it is the naturality square

John Baez (Nov 24 2020 at 21:33):

Oh, okay! I was confused. Thanks.

John Baez (Nov 24 2020 at 21:34):

$f(h)$ does make sense. $f(\alpha_a)$ does not make sense, since $\alpha_a$ is a 1-morphism in $B$ and $f: A \to B$.

John Baez (Nov 24 2020 at 21:35):

Anyway, I said:

So there's some equation involving $m_a, m_b, \alpha_h$ and $\beta_h$ that must hold. And if you know enough about 2-categories you can take Morgan's picture and turn it into this equation!

So I'm looking for this equation, which is sitting there in Morgan's picture.

Jan Pax (Nov 24 2020 at 21:36):

I just wonder how do I connect $f(h)$ with $m_b$

John Baez (Nov 24 2020 at 21:36):

modification.png

John Baez (Nov 24 2020 at 21:37):

Jan Pax said:

I just wonder how do I connect $f(h)$ with $m_b$

That's a good question. You combine $f(h)$ and $m_b$ using whiskering.

Jan Pax (Nov 24 2020 at 21:38):

I have heard that notion from Makkai but I do not recall what it is. I would like you to explain this to me via a picture.

John Baez (Nov 24 2020 at 21:40):

Unfortunately the nLab article does not have a picture, but in any 2-category you can combine a 1-cell $f: x \to y$ and a 2-cell $\alpha: g \Rightarrow h$ where $g,h: y \to z$ and get a 2-cell $\alpha * f : g \circ f \to h \circ f$.

Jan Pax (Nov 24 2020 at 21:41):

So my $\ast$ was not such a bad choice ?

John Baez (Nov 24 2020 at 21:41):

If you draw all the stuff I just described you'll see a mouth with a whisker coming out of it...

John Baez (Nov 24 2020 at 21:41):

I decided to use your notation.

Jan Pax (Nov 24 2020 at 21:48):

So our equation becomes $m\ast f=m\ast g$ ?

Jan Pax (Nov 24 2020 at 21:50):

You should have written "and get a 2-cell $\alpha\ast f:g\circ f\implies h \circ f$" ?

John Baez (Nov 24 2020 at 22:41):

Jan Pax said:

You should have written "and get a 2-cell $\alpha\ast f:g\circ f\implies h \circ f$" ?

Yes, I should have written that.

John Baez (Nov 24 2020 at 22:42):

Jan Pax said:

So our equation becomes $m\ast f=m\ast g$ ?

$m$ by itself makes no sense here. Remember, we have 2-cells like $m_a$ and $m_b$.

John Baez (Nov 24 2020 at 22:43):

Hint:

John Baez said:

So there's some equation involving $m_a, m_b, \alpha_h$ and $\beta_h$ that must hold. And if you know enough about 2-categories you can take Morgan's picture and turn it into this equation!

Morgan Rogers (he/him) (Nov 25 2020 at 09:20):

@John Baez and @Jan Pax I really appreciate the patience you've displayed in hashing this out together. It's really nice for people to be able to come here and go away having genuinely learned something through interacting with other category theorists :tada:

Jan Pax (Nov 25 2020 at 12:01):

Can you reveal to me the final equation ?

Jan Pax (Nov 25 2020 at 12:04):

$m_a\ast \alpha_h=\beta_h\ast m_b$ But I'm sure this is not the correct equation.

John Baez (Nov 25 2020 at 18:45):

Jan Pax said:

Can you reveal to me the final equation ?

No, it's better to figure these things out yourself. Passive learning is not nearly as good.

John Baez (Nov 25 2020 at 18:48):

Jan Pax said:

$m_a\ast \alpha_h=\beta_h\ast m_b$ But I'm sure this is not the correct equation.

That equation has the right look to it. The problem is just: what does "$\ast$" mean? It can't be vertical composition of 2-morphisms, since that doesn't parse. It can't be horizontal composition of 2-morphisms, either, since that doesn't parse. So you need to learn a bit about whiskering.

John Baez (Nov 25 2020 at 18:50):

I don't have the energy to explain whiskering. I wish I knew a nice really elementary introduction to 2-categories that explains - it's very basic.

John Baez (Nov 25 2020 at 18:51):

All the introductions to 2-categories I know are too fancy.

Nathanael Arkor (Nov 25 2020 at 18:56):

I think it's helpful to view whiskering as the horizontal composition of some 2-cell with an identity 2-cell. That way simply knowing about vertical and horizontal composition is enough.

Jan Pax (Nov 25 2020 at 19:12):

What about making somehow identity from $\alpha_h$ ? I would guess that horizontal composition is what we need.

John Baez (Nov 25 2020 at 22:39):

I'm not sure what your first sentence means. However, whiskering is just horizontally composing a 2-morphism and an identity 1-morphism.

Jan Pax (Nov 26 2020 at 14:47):

Could you give me the equation required so that we can finish ? I cannot find it myself. Or is there more to say on nCat ?

John Baez (Nov 26 2020 at 19:17):

You need to learn about whiskering to be able to understand the required equation. Maybe someone can point to a nice introduction to 2-categories that explains whiskering in a simple way.

Jan Pax (Nov 26 2020 at 19:19):

Right, could you send me that equation just for my curiosity ? I believe I will learn something from it itself. Also, if you think that the equation will be hard to understand for me it will definitely not be possible that I'd find it out on my own.

John Baez (Nov 26 2020 at 19:53):

You already found the equation - it's visible in Morgan Rogers' picture:

Jan Pax said:

$m_a\ast \alpha_h=\beta_h\ast m_b$

The only problem is figuring out what $\ast$ means here.

Jan Pax (Nov 26 2020 at 19:53):

So what it does ?

John Baez (Nov 26 2020 at 19:55):

For that, as I already said, you need to understand whiskering.

Jan Pax (Nov 26 2020 at 19:55):

I understand the definition of whiskering.

John Baez (Nov 26 2020 at 19:56):

What is the definition of whiskering?

Jan Pax (Nov 26 2020 at 19:57):

identity of 1-cell horizontally composed with 2-cell

John Baez (Nov 26 2020 at 20:02):

Okay, good. So here:

$m_a\ast \alpha_h=\beta_h\ast m_b$

$\ast$ cannot be vertical composition: you can't vertically compose $m_a$ and $\alpha_h$ because the target of $\alpha_h$ does not equal the source of $m_\ast$. You need to whisker $m_a$ with something first!

So:

Question: what do you need to whisker $m_a$ by, to get a 2-morphism whose source equals the target of $\alpha_h$?

John Baez (Nov 26 2020 at 20:03):

Again, the picture shows the answer.

Jan Pax (Nov 26 2020 at 20:06):

I don't know.

John Baez (Nov 26 2020 at 20:07):

Okay, I'll let you think about it for a few weeks.

John Baez (Nov 26 2020 at 20:08):

These problems take a while sometime, but it's much better to figure out things yourself and build your mathematical muscles, rather than passively being told facts.

Jan Pax (Nov 26 2020 at 20:09):

I'll think about it for 3 days. Will you then reveal the answer for me if I do not succeed ?

John Baez (Nov 26 2020 at 20:11):

In two weeks.

John Baez (Nov 26 2020 at 20:11):

So, on December 10th.

John Baez (Nov 26 2020 at 20:11):

3 days may not be enough.

John Baez (Nov 26 2020 at 20:12):

There are lots of problems I've been working on for years...

Jan Pax (Nov 26 2020 at 20:14):

I'll do my best but I'm sure I won't succeed. I know something from your book Towards higher categories: that's not for beginners!

John Baez (Nov 26 2020 at 20:16):

It's not good to think "I'm sure I won't succeed". You should be sure that you will succeed, and work hard enough that you do succeed. This is a basic test of your understanding, not Fermat's Last Theorem, so I promise that you can succeed.

Jan Pax (Nov 26 2020 at 20:17):

Can I whisker with $f$ ?

John Baez (Nov 26 2020 at 20:17):

Let me know your final answer on December 10th.

Jan Pax (Nov 26 2020 at 20:19):

Can I whisker $m_a$ with $g(h)$ ?

John Baez (Nov 26 2020 at 21:59):

Actually since you're spending two weeks on this you might as well solve the whole problem. For this to make sense:

$m_a\ast \alpha_h=\beta_h\ast m_b$

you need to answer two questions. They are very similar:

Question 1: what do you need to whisker $m_a$ by, to get a 2-morphism whose source equals the target of $\alpha_h$?

Question 2: what do you need to whisker $m_b$ by, to get a 2-morphism whose target equals the source of $\beta_h$?

Jan Pax (Nov 26 2020 at 22:05):

This is to be some nice 1-cell for both Question 1 and Question 2. I'm gonna to look at it (and fail). There far too many symbols involved for me to orient myself. So $g(h)$ for Question 1 is wrong otherwise you'd agree ?

Jan Pax (Nov 27 2020 at 16:30):

The target of $\alpha_h$ is an 1-cell $f(a)\to g(b)$. There is no arrow from this to elsewhere in general except for $\text{id}_{g(b)}$. This gives $m_a\ast {g(b)}$. But I'm confused now whether $m_a\ast {g(b)}$ or $m_a\ast \text{id}_{g(b)}$ to be honest. I would guess the second possibility.

John Baez (Nov 28 2020 at 19:02):

1) What does $\ast$ mean? You invented this notation.

2) Tell me your notation for vertical composition of 2-morphisms and your notation for horizontal composition of 2-morphisms.

3) If you want to invent a notation for whiskering, tell me that too. You don't need a notation for whiskering because it's just horizontal composition with an identity 2-morphism. But some people find it convenient.

Jan Pax (Nov 28 2020 at 19:14):

I mean $m_a$ horizontally composed with $\text{id}_g$.

Jan Pax (Nov 28 2020 at 19:15):

The notation for vertical composition is $\circ$ and for horizontal $\ast$. Are you happy with this ?

John Baez (Nov 28 2020 at 19:15):

I'm happy with that.

Jan Pax (Nov 28 2020 at 19:16):

Right, what do you want make me now ?

John Baez (Nov 28 2020 at 19:17):

You wrote:

But I'm confused now whether $m_a\ast {g(b)}$ or $m_a\ast \text{id}_{g(b)}$ to be honest.

What do these two things mean? You say that $\ast$ means horizontally composition of 2-morphisms. If so, only one of these expressions make sense.

Jan Pax (Nov 28 2020 at 19:18):

Actually your question about the notation has made me believed that $m_a\ast \text{id}_g$ is the correct one expression. Please note that I have omitted $b$ in $g(b)$ for this version.

John Baez (Nov 28 2020 at 19:30):

Actually now that I look at it, $m_a \ast \mathrm{id}_{g(b)}$ does not make sense since $\mathrm{id}_{g(b)}$ is not a 2-morphism. $m_a \ast \mathrm{id}_g$ does not make sense. $m_a \ast g(b)$ does not make sense.

John Baez (Nov 28 2020 at 19:32):

Your chance of guessing the right answer to a question increases dramatically if you only write down guesses that make sense. This is a key principle in category theory. Very often in category theory there is only one guess that makes sense.

Jan Pax (Nov 28 2020 at 19:38):

I do not follow this: $g$ is a 1-cell so why $m_a\ast \text{id}_g$ does not make sense ? The second thing is the identity 2-cell on the 1-cell $g$. I wish to understand this point.

John Baez (Nov 28 2020 at 19:39):

There is no 1-cell $g$ in $B$.

John Baez (Nov 28 2020 at 19:40):

I made a mistake... here is my corrected opinion:

John Baez said:

Actually now that I look at it, $m_a \ast \mathrm{id}_{g(b)}$ does not make sense since $\mathrm{id}_{g(b)}$ is not a 2-morphism. $m_a \ast \mathrm{id}_g$ does not make sense. $m_a \ast g(b)$ does not make sense.

John Baez (Nov 28 2020 at 19:41):

You are using $\ast$ to mean horizontal composition of 2-morphisms, and we're working in $B$ here.

Jan Pax (Nov 28 2020 at 19:41):

Right, so for what $k$ is $g(b)$ a $k$-cell in $B$ ?

John Baez (Nov 28 2020 at 19:41):

Look at Morgan's picture. What does $g(b)$ look like?

John Baez (Nov 28 2020 at 19:42):

If you don't draw his picture and keep looking at it, you are doomed.

Jan Pax (Nov 28 2020 at 19:42):

1-cell in $B$ ?

John Baez (Nov 28 2020 at 19:42):

Let's talk more on December 10th. You are not looking at the picture.

John Baez (Nov 28 2020 at 19:44):

$k$-cells look $k$-dimensional.

Jan Pax (Nov 28 2020 at 19:44):

oh sorry I meant $0$-cell, an object

Jan Pax (Nov 28 2020 at 19:49):

What about $m_a\ast \text{id}_{g(h)}$ ?

John Baez (Nov 28 2020 at 21:04):

There are two questions whose answers I'd like on December 10th:

Question 1: what do you need to whisker $m_a$ by, to get a 2-morphism whose source equals the target of $\alpha_h$?

Question 2: what do you need to whisker $m_b$ by, to get a 2-morphism whose target equals the source of $\beta_h$?

You can think about them until then, and become 100% sure that you are correct.

Jan Pax (Nov 28 2020 at 22:23):

My best guess for now is this answer:for Question 1: $m_a\ast \text{id}_{g(h)}$ and for Question 2: $\text{id}_{f(h)} \ast m_b$. But your demand with "100% sure that I'm correct" is unattainable though.

John Baez (Nov 28 2020 at 22:25):

Okay, I'll settle for 90%. See you on December 10th!

Jan Pax (Nov 28 2020 at 22:31):

The source of $m_a\ast\text{id}_{g(h)}$ is $g(h)\circ \alpha_a$ which is the target of $\alpha_h$. I do not know what's wrong.

John Baez (Nov 28 2020 at 22:36):

I didn't say anything is wrong. I'm not saying it's right, either. I need a break from this, so we can continue on December 10th.

Jan Pax (Nov 28 2020 at 22:37):

But I think that if my whiskering were correct you wouldn't wait. This in turn makes me think that something is wrong. I've drawn some pictures which supports my belief.

John Baez (Nov 28 2020 at 22:38):

Good!

John Baez (Nov 28 2020 at 22:44):

I'm a stubborn guy so if I say I'll tell you the answer on December 10th, that means I'll tell you the answer on December 10th - unless of course I have an accident.

Jan Pax (Nov 28 2020 at 22:47):

I believe that you are doing the very best for me for free. I appreciate that very much and will obey all your instructions/explanations for whatever reasons you may have for this.

Jan Pax (Dec 10 2020 at 19:54):

Hi John,my time has come, and I have one more question besides if my answer in my -4th comment above is correct. Where do all $(j+1)$-cells in nCat live in contrast to what they satisfy by an action of a $k$-cell ?
I know this for $n=j=1$. What about various other possibilities taking into account things like $j>n$ or $j\leq n$ ?

John Baez (Dec 10 2020 at 22:26):

The j-cells in nCat live in nCat, of course. They are cells in a particular (n+1)-category called nCat. So, they are complicated and interesting things up to j = n+1, and then for j = n+2 they are just equations (which are still somewhat interesting), and for higher j they are completely boring and vacuous "equations between equations".

Jan Pax (Dec 10 2020 at 22:34):

Is (n+1)Cat defined by induction on enriching nCat over itself ?

John Baez (Dec 11 2020 at 00:17):

It's easy to define an (n+1)-category of strict n-categories inductively by enrichment: (n+1)Cat is the category of categories enriched over nCat.

John Baez (Dec 11 2020 at 00:17):

But the really interesting weak n-categories are harder to define.

Jan Pax (Dec 11 2020 at 15:20):

Right. Is my answer correct: :for Question 1: $m_a\ast \text{id}_{g(h)}$ and for Question 2: $\text{id}_{f(h)} \ast m_b$ ?

Jan Pax (Dec 16 2020 at 16:15):

John, you have promised your verdict about my answer to the 10th of December haven't you ?

John Baez (Dec 16 2020 at 19:51):

I'm sorry to take so long to reply!

Jan Pax said:

Is my answer correct: :for Question 1: $m_a\ast \text{id}_{g(h)}$ and for Question 2: $\text{id}_{f(h)} \ast m_b$ ?

Yes, these are right, if you're using a convention for $\ast$ where this horizontal composite:

horizontal composite

is called $\alpha \ast \beta$.

John Baez (Dec 16 2020 at 19:51):

I'll just warn you that lots of people call this $\beta \ast \alpha$, writing things backwards:

John Baez (Dec 16 2020 at 19:52):

https://ncatlab.org/nlab/show/horizontal+composition

John Baez (Dec 16 2020 at 19:52):

But I don't care what convention is used as long as we agree on a convention.

Jan Pax (Dec 16 2020 at 20:15):

I'm now happy. It would be a weird thing if I were wrong as I have drawn all the pictures on a paper. Any continuation is possible ? Some dialogue with exercises ? I'm pretty sure that something more is to be said.

John Baez (Dec 17 2020 at 05:42):

There's a lot more to say, but I'm getting really busy trying to finish a couple of papers before January. Sorry! I'm glad you figured it all out.

Jan Pax (Dec 17 2020 at 12:19):

Right, let's continue on February. You can write the names of those papers here so that I can have a look at them ?

John Baez (Dec 17 2020 at 20:29):

I'm not promising to continue in February. I'll be really busy teaching a course on Lie groups from January to the end of March, for example.

John Baez (Dec 17 2020 at 20:31):

Those papers I'm trying to finish are papers I'm writing: "Structured versus decorated cospans" with @Kenny Courser and @Christina Vasilakopoulou, and "Categories of nets" with @Jade Master, @Fabrizio Genovese and @Mike Shulman.

Jan Pax (Dec 17 2020 at 21:01):

Right, we may shift our discussion to May. I would appreciate if you could dedicate 2 hours in May to me yet.

John Baez (Dec 17 2020 at 21:21):

By the way, if you want to learn about 2-categories, this book is good:

https://arxiv.org/abs/2002.06055

Jan Pax (Dec 18 2020 at 15:37):

Thank you for your link. We haven't mentioned yet 3-cells except for their name. We could do it in May.

Jan Pax (Dec 20 2020 at 20:10):

How is called the general 2category which has as associator arbitrary natural transformation rather than a natural isomorphism as for bicategory ?

John Baez (Dec 20 2020 at 20:46):

Maybe "lax bicategory"? You can read about the special case where there's just one object here:

• nLab, Lax monoidal category.

A lax monoidal category is perhaps more commonly called a "skew monoidal category" or "skew monoidale".

Jan Pax (Jan 04 2021 at 23:04):

Is there a good book similar to your link https://arxiv.org/abs/2002.06055 also for the Kan extension concept?

John Baez (Jan 05 2021 at 02:52):

I don't know one.

Jan Pax (Jan 26 2021 at 17:39):

John, why is $Set^{Set}$ not locally small ?? Is it locally legitimate ?

John Baez (Jan 26 2021 at 18:26):

What does "locally legitimate" mean?

Jan Pax (Jan 26 2021 at 19:45):

if the hom-set Nat($F,G$) for all functors $F$ and $G$ is at most a proper class, not bigger.

Todd Trimble (Jan 26 2021 at 22:24):

Jan Pax said:

John, why is $Set^{Set}$ not locally small ?? Is it locally legitimate ?

A very good question! Unless I am missing something, I think the "not locally small" question is already nontrivial, and to be honest this stumped me. Finally I remembered that there's a paper by Freyd and Street which has an answer in it, which I'll extract from.

For any set $A$, let $S(A)$ be the set of all subsets $B \subseteq A$ whose inclusion has a left inverse $f$, i.e., a function $f: A \to B$ such that $f(b) = b$ for all $b \in B$. If $A$ is nonempty, then $S(A)$ is the set of nonempty subsets of $A$, and if $A$ is empty, then $S(A)$ consists of the single element $\emptyset \subseteq \emptyset$. Let $T(A)$ be the disjoint union $S(A) \sqcup \{\infty\}$. Make $T$ into a functor: if $f: X \to Y$ is a function, then $T(f): T(X) \to T(Y)$ takes $B \subseteq X$ in $S(X)$ to $f(B) \in T(Y)$ in case the composite $B \stackrel{i}{\hookrightarrow} X \stackrel{f}{\to} Y$ has a left inverse, else it takes $B \subseteq X$ to $\infty$ (and also it takes $\infty \in T(X)$ to $\infty \in T(Y)$. To see $T$ is a functor, one uses the fact that if $g f i$ has a left inverse $h$, then $f i$ also has a left inverse (namely, $h g$).

Now we define a function $\phi$ from the class of cardinalities (isomorphism classes of sets) to the collection of natural transformations $T \to T$, as follows: if $B$ is a set with cardinality $|B|$, then $\phi_{|B|}: T \to T$ has, for its component $\phi_{|B|}(X): TX \to TX$, the function which maps $(A \subseteq X) \in S(X)$ to $(A \subseteq X)$ just in case $A \cong B$, else it maps $(A \subseteq X)$ to $\infty$ (and also it takes $\infty \in TA$ to itself). It's not at all difficult to show that $\phi_{|B|}$ satisfies the naturality condition.

Clearly $\phi$ is well-defined. Now suppose $\phi_{|A|} = \phi_{|B|}$. Then at the component $B$, we have

$\phi_{|A|}(B)(B \subseteq B) = \phi_{|B|}(B)(B \subseteq B) = (B \subseteq B) \neq \infty$

whence, following the definitions, we must have $A \cong B$, i.e., $|A| = |B|$. This proves that $\phi$ is injective on cardinalities. Since cardinalities form a proper class, so must the collection of natural transformations $T \to T$.

It may be tempting to think this is an overly complicated example, and that perhaps something simpler to describe than $T$ works. For example, one might think the double powerset functor $X \mapsto PPX$ could work in place of $T$. Actually, it doesn't. Want to know how many natural transformations $PP \to PP$ there are? I believe the answer is $65536$!

Fawzi Hreiki (Jan 26 2021 at 22:43):

A natural transformation from $\text{Set} \xrightarrow{!} 1 \xrightarrow{1} \text{Set}$ to $\text{id}_{\text{Set}}$ is the same thing as a section of the projection functor from pointed sets to sets. It seems intuitive to me that there should be class many such sections but I'm unsure how to prove that.

Todd Trimble (Jan 26 2021 at 22:45):

That's a natural transformation from $\hom(\emptyset, -)$ to $\hom(1, -)$. Yoneda...

Fawzi Hreiki (Jan 26 2021 at 22:46):

Ah right. As usual.