Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: (n,r)-category

Leopold Schlicht (Nov 22 2021 at 15:59):

The nLab taught me:

An $(n,r)$-category is a higher category such that, essentially:
- all $k$-morphisms for $k>n$ are trivial.
- all $k$-morphisms for $k>r$ are reversible.

I want to convert this into a precise definition using the model of quasicategories.

• What is a $k$-morphism?
• When is a $k$-morphism trivial? (Guess: if it is equivalent to an identity $k$-morphism. But that just provokes the questions "what is an identity $k$-morphism?" and "when are two $k$-morphisms equivalent?", which I am not able to answer myself.)
• When is a $k$-morphism reversible?

John Baez (Nov 22 2021 at 16:01):

I think you should try these exercises yourself!

John Baez (Nov 22 2021 at 16:02):

But one thing I just told you is that in a quasicategory all $k$-morphisms are reversible for $k > 1$. So quasicategories are not going to help you understand the difference between reversible and nonreversible 2-morphisms, 3-morphisms, etc.

John Baez (Nov 22 2021 at 16:03):

You can, however, figure out why in a quasicategory all $k$-morphisms are reversible for $k > 1$.

Leopold Schlicht (Nov 22 2021 at 16:07):

John Baez said:

But one thing I just told you is that in a quasicategory all $k$-morphisms are reversible for $k > 1$.

I read this all the time. But bizarrely nobody says what $k$-morphisms are and what "reversible" means. I have the feeling category theory is a very beginner unfriendly subject, and people hide ideas and definitions.

I can do that exercise, but how can I ever check my solution?

John Baez (Nov 22 2021 at 16:08):

Just tell me.

You say that category theory is a beginner unfriendly subject. That might be true, but you're not doing category theory. You are now starting to do $(\infty,1)$-category theory. This cutting-edge math, not for beginners. If you don't know category theory and 2-category theory reasonably well, it's going to be hard.

John Baez (Nov 22 2021 at 16:12):

But if you read this:

• John Baez, An introduction to n-categories, 1997.

you'll see I talk about 2-morphisms in the simplicial approach. I say what they are, and I show how to reverse one, and why.

John Baez (Nov 22 2021 at 16:14):

I can do that exercise, but how can I ever check my solution?

In math you check things by seeing if they cohere: by seeing if they make sense in light of everything you know. Also, you talk to people about them.

Leopold Schlicht (Nov 25 2021 at 13:40):

John Baez said:

But if you read this:

• John Baez, An introduction to n-categories, 1997.

you'll see I talk about 2-morphisms in the simplicial approach. I say what they are, and I show how to reverse one, and why.

Great read! (I finally took the time to read most of it.)

John Baez (Nov 26 2021 at 03:07):

Thanks!

Leopold Schlicht (Dec 03 2021 at 15:53):

John Baez said:

I think you should try these exercises yourself!

Before talking about general $n$-morphisms, here is a guess for how 3-morphisms could be defined:

Let $A$ and $B$ be objects, and $f, g\colon A\to B$ 1-morphisms. I use the convention that a 2-morphism $\alpha\colon f\to g$ is a 2-simplex $\alpha$ such that

• $d_0\alpha = \mathrm{id}_B$ ($\mathrm{id}_B$ is $s_0B$),
• $d_1\alpha = g$, and
• $d_2\alpha = f$
  -- see (1) in the picture.

Now let $\alpha, \beta\colon f\to g$ be two 2-morphisms. We want to define what a "3-morphism from $\alpha$ to $\beta$" is. I propose the following: a 3-morphism $\Gamma\colon \alpha \to \beta$ is a 3-simplex $\Gamma$ such that

• $d_0 \Gamma = s_0\mathrm{id_B}$ (which is the same as $s_1 \mathrm{id}_B$ by the way),
• $d_1\Gamma=s_1 g$,
• $d_2\Gamma=\beta$, and
• $d_3\Gamma=\alpha$
  -- see (2) in the picture.

picture.jpg

But I have problems when it comes to defining the identity 3-morphism $\mathrm{id}_\alpha\colon\alpha\to \alpha$. I had the idea to define $\mathrm{id}_\alpha:=s_2\alpha$. But then the simplicial identities imply:

• $d_0 \mathrm{id}_\alpha = s_1 g$,
• $d_1 \mathrm{id}_\alpha = s_1 f$,
• $d_2 \mathrm{id}_\alpha = \alpha$,
• $d_3 \mathrm{id}_\alpha = \alpha$.

The last two equations make sense (they are compatible with the definition of "3-morphism from $\alpha$ to $\alpha$" I gave above). But the first two don't. Can one fix that somehow?

(By the way, here's a conceptual question: In a quasicategory composition is defined as a relation and not as an operation. That is, two 1-morphisms $f\colon A\to B$ and $g\colon B\to C$ don't have a specified composite $g\circ f\colon A\to C$. Instead, there is a relation of the form "a 1-morphism $h\colon A\to C$ is a composite of $f$ and $g$". This makes sense because such a composite $h$ is uniquely determined up to homotopy -- and we only care about a 1-morphism up to homotopy anyway. But why, then, is there a specified identity $\mathrm{id}_A:= s_0A$ instead of a relation of the form "$f\colon A \to A$ is a identity of $A$"?)

Reid Barton (Dec 03 2021 at 16:25):

Leopold Schlicht said:

• $d_0 \mathrm{id}_\alpha = s_1 g$,
• $d_1 \mathrm{id}_\alpha = s_1 f$,

I think everything looks right except that these two should be

• $d_0 (s_2 \alpha) = s_1 (d_0 \alpha) = s_1 \mathrm{id}_B$ ($= s_0 \mathrm{id}_B$),
• $d_1 (s_2 \alpha) = s_1 (d_1 \alpha) = s_1 g$.

Leopold Schlicht (Dec 03 2021 at 16:38):

Ah, right, thanks! I now see why I did that wrong: for 1-simplices $\sigma$ I am used to thinking that $d_0\sigma$ is the codomain of $\sigma$ and $d_1\sigma$ is the domain of $\sigma$. I shouldn't do that with 2-simplices. :grinning_face_with_smiling_eyes:

Leopold Schlicht (Dec 03 2021 at 16:43):

The next step would be to define composition of 3-morphisms. Probably composition of 3-morphisms is witnessed by 4-simplices. But how do you visualize these? (I can't draw 4-dimensional things.)

John Baez (Dec 03 2021 at 19:45):

At some point one has to stop drawing things. But it's not hard to draw a 4-simplex.

John Baez (Dec 03 2021 at 19:47):

And you can use this picture to figure out the famous "pentagon identity" for the associators when you're composing four 1-morphisms in a row in a bicategory.

Leopold Schlicht (Dec 27 2021 at 19:15):

@John Baez I don't like this way of drawing 4-simplices, because I want to write down the name of each $n$-simplex for $n=2,3$ in its interior, but in that picture this gets very messy.

Then let's forget about pictures. I want to inductively define what an $n$-morphism between two $(n-1)$-morphisms is.

Leopold Schlicht (Dec 27 2021 at 19:15):

Recall:

The identity 1-morphism $\mathrm{id_A}$ is defined to be $s_0A$.
I already said that a 2-morphism from $f$ to $g$ (where $f,g\colon A\to B$) is a 2-simplex $\alpha$ such that

• $d_0\alpha=\mathrm{id}_B$,
• $d_1\alpha=g$,
• $d_2\alpha=f$.

Furthermore: the identity 2-morphism $\mathrm{id_f}$ is defined to be $s_1f$.
A 3-morphism from $\alpha$ to $\beta$ (where $\alpha, \beta\colon f\to g$) is a 3-simplex $\Gamma$ such that

• $d_0\Gamma = \mathrm{id}_{\mathrm{id}_B}$,
• $d_1\Gamma = \mathrm{id}_g$,
• $d_2\Gamma = \beta$,
• $d_3\Gamma = \alpha$.

The identity 3-morphism $\mathrm{id_\alpha}$ is defined to be $s_2\alpha$.

Leopold Schlicht (Dec 27 2021 at 19:15):

This suggests the following definition of 4-morphism:

A 4-morphism from $\Gamma$ to $\Lambda$ is a 4-simplex $\eta$ such that

• $d_0\eta=\mathrm{id}_{\mathrm{id}_{\mathrm{id}_B}}$,
• $d_1\eta= \mathrm{id}_{\mathrm{id}_g}$,
• $d_2\eta= \mathrm{id}_\beta$,
• $d_3\eta= \Lambda$,
• $d_4\eta= \Gamma$.

Then it makes sense to define the identity 4-morphism $\mathrm{id}_\Gamma$ as $s_3\Gamma$.

Is that correct?

Leopold Schlicht (Dec 27 2021 at 19:15):

Is that the right pattern to inductively define "$n$-morphism $f$ between $(n-1)$-morphisms $A$, $B$":

• $d_nf=A$,
• $d_{n-1}f=B$,
• $d_{n-2}f=\mathrm{id}_{\mathrm{cod}(B)}$,
• $\dots$,
• $d_0f=\mathrm{id}_{\dots_{\mathrm{id}_{\mathrm{cod}(\dots(\mathrm{cod}(B)))}}}$?

Then it makes sense to define the identity $n$-morphism $\mathrm{id}_A$ as $s_{n-1}A$.

Leopold Schlicht (Dec 27 2021 at 19:16):

How to define composition of $n$-morphisms? When is an $n$-morphism reversible? When is an $n$-morphism trivial? (If it is equivalent to an identity $n$-morphism? But what does "equivalent" mean?)

John Baez (Dec 27 2021 at 21:51):

This is an interesting line of investigation, but alas, I don't have the energy for it myself - I'm busy with other projects. I hope someone else will enjoy working on this.

Leopold Schlicht (Dec 29 2021 at 13:54):

Here's an idea how to define composition of 2-morphisms $\alpha\colon f\to g$ and $\beta\colon g\to h$: say that $\gamma\colon f\to h$ is a composition of $\alpha$ and $\beta$ if there is a 3-simplex $\sigma$ with $d_1\sigma=\beta$, $d_2\sigma=\gamma$, and $d_3\sigma=\alpha$.

Leopold Schlicht (Jan 17 2022 at 19:29):

Probably the notion of a morphism space or mapping space provides an answer to my questions. However, I haven't encountered them until now, because they are discussed beginning on page 667 in Kerodon.