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Stream: learning: questions

Topic: Confused with Riehl's Decategorification example

Alex Kreitzberg (Sep 24 2024 at 19:43):

Emily Riehl exemplifies decategorization decategorification by restricting $Set$ to $\text{Fin}_{iso}$, finite Sets and their bijections, and then defining the cardinality Functor $|-| : \text{Fin}_{iso} \rightarrow \mathbb N$ treating $\mathbb N$ as a discrete category. Essentially mapping the category of isomorphic sets to their skeleton.

She explains natural transformations, like $A \times B \cong B \times A$, get mapped to equations between natural numbers, like $ab = ba$

But if $\mathbb N$ is a discrete category, it doesn't have (categorical) products right? So in what sense are the products from Set getting carried over?

Emily's example for context:
1be8b34d-18a0-4720-8886-ae11eb94f9c7.png

John Baez (Sep 24 2024 at 19:50):

It's decategorification, not "decategorization".

But if $\mathbb{N}$ is a discrete category, it doesn't have products right?

Right. $\mathbb{N}$ does not have products, which is why it's good to realize it's the decategorification of a category that does.

I'd rather think of it as a mere set than a discrete category, in what follows. Thinking of it as a discrete category is not helpful here.

So in what sense are the products from Set getting carried over?

This is how decategorification works:

Any functor $F : C \to D$ between essentially small categories gives a function

$\mathrm{Decat}(F) : \mathrm{Decat}(C) \to \mathrm{Decat}(D)$

where $\mathrm{Decat}(C)$ is the set of isomorphism classes of objects of $C$ and similarly for $D$. Here's how it works:

$\mathrm{Decat}(F)([c]) = [d]$

where $[c]$ is the isomorphism class of the object $c \in C$, and similarly for $d$.

I think from this you can figure out how decategorifying the product functor

$\times : \mathsf{FinSet} \times \mathsf{FinSet} \to \mathsf{FinSet}$

we get the usual multiplication of natural numbers, which is

$\mathrm{Decat}(\times) : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$

John Baez (Sep 24 2024 at 19:55):

Similarly addition comes from coproducts, and so on.

John Baez (Sep 24 2024 at 19:57):

You will need to ponder $\mathrm{Decat}(\mathsf{FinSet} \times \mathsf{FinSet})$ and simplify it a bit, to see what's going on here. More generally: given categories $C$ and $D$, what's $\mathrm{Decat}(C \times D)$? Make the obvious guess, then prove it!

David Michael Roberts (Sep 24 2024 at 21:39):

Emily is working with the groupoid of sets and bijections, which also doesn't have categorical products — but it has a monoidal product given as the cartesian product of sets.

David Michael Roberts (Sep 24 2024 at 21:40):

Likewise the monoid N with multiplication is a discrete category with a monoidal product...

John Baez (Sep 24 2024 at 22:17):

Emily should (in my humble opinion) be working with the category of sets and functions, so that 'products' have a nice universal property. Anyway, that's the story I was telling above.

David Michael Roberts (Sep 24 2024 at 22:22):

I was more addressing @Alex Kreitzberg but it wasn't clear from your post if you were thinking about the groupoid or the category of finite sets.

Alex Kreitzberg (Sep 24 2024 at 22:26):

I did assume he was talking about finite sets and his presentation is making sense (I'm still putting the pieces together).

When that was clear I was planning on circling back to clarifying what you pointed out, so thank you! Here decatigorification was stepping a category down to a discrete monoidal groupoid.

Alex Kreitzberg (Sep 24 2024 at 22:29):

(if I got that last sentence right, this does seem more fussy than just saying "a set with product")

John Baez (Sep 24 2024 at 22:45):

David Michael Roberts said:

I was more addressing Alex Kreitzberg but it wasn't clear from your post if you were thinking about the groupoid or the category of finite sets.

Whoops, sorry - for me $\mathsf{FinSet}$ is the category of finite sets and functions, while something like $\mathrm{core}(\mathsf{FinSet})$ or $\mathsf{S}$ is the groupoid of finite sets and bijections. ($\mathsf{S}$ because it's a good way of thinking of all symmetric groups $S_n$ together - @Joe Moeller and @Todd Trimble have been engaged in a long detailed study of this groupoid and its offspring.)

John Baez (Sep 24 2024 at 22:50):

Alex Kreitzberg said:

Here decategorification was stepping a category down to a discrete monoidal groupoid.

'Discrete monoidal groupoid' is how people say 'monoid' when they've studied too much category theory too fast and haven't recovered yet. :upside_down: Some of these people also say 'object of the category of sets' instead of 'set'.

(if I got that last sentence right, this does seem more fussy than just saying "a set with product")

A monoid is a set with an associative product and identity element, and that is essentially the same as a 'discrete monoidal category' or 'discrete monoidal groupoid'.

When you decategorify a monoidal category you get a monoid.

'Set with product' is a bit ambiguous because we're left wondering about whether it obeys the associative law and has an identity. If you don't require these, say [[magma]]. If you do, say [[monoid]].

Alex Kreitzberg (Sep 25 2024 at 01:02):

Lemma For all categories C and D, $\text{Decat}(C \times D) \cong \text{Decat}(C)\times \text{Decat(D)}$

Proof:

First, for all $c,c' \in C$ and $d, d' \in D$ an object $(c, d) \in C \times D$ is isomorphic to another object $(c', d') \in C \times D$ exactly when we can find isomorphisms $f : c \rightarrow c'$ and $g : d \rightarrow d'$ because then and only then can we have the isomorphism $(f, g) : (c, d) \rightarrow (c', d')$

So $(c, d) \cong (c', d')$ if and only if $c \cong c'$ and $d \cong d'$

Now define a function $q : C \times D \rightarrow Decat(C)\times Decat(D)$ by
$q(c, d) = ([c], [d])$

The following chain of equivalences is useful:

$(c, d) \cong (c', d') \Leftrightarrow c \cong c' \text{ and } d \cong d' \Leftrightarrow [c] = [c'] \text{ and } [d] = [d'] \Leftrightarrow ([c], [d]) = ([c'], [d']) \Leftrightarrow q(c, d) = q(c', d')$

The top down direction of the logic chain implies there exists a unique $\bar q : Decat(C\times D) \rightarrow Decat(C) \times Decat(D)$ so that $\bar q ([(x, y)]) = q(x, y)$ for all $x \in C, y \in D$.

We know $\bar q$ is surjective because $q$ is surjective. We also now $\bar q$ is injective since $q$ is injective on isomorphism classes. That's proved by following the previous chain of equivalences from the bottom up.

This proves our Lemma.

So now what happens when we decatigorify $\text{FinSet}$?

Define $0_s = \{\}$ and for all $n \in \mathbb N$ define $n_s = \{1, 2, \ldots, n\}$. Every finite set is isomorphic to one of these.

So now the following:

$n_s \times m_s = \{ (1, 1), \ldots, (n, m) \}\cong nm_s$

Becomes under $\text{Decat}$, and precomposition by $\bar q^{-1}$:

$[n_s] \times_{\mathbb N} [m_s] = [nm_s]$

But $[n_s] = n$ so this is just
$n \times m = nm$

Which is just multiplication of natural numbers.

John Baez (Sep 25 2024 at 01:12):

:tada: GREAT! :tada:

John Baez (Sep 25 2024 at 02:21):

By the way, if you know about coproducts, we also have

$\mathrm{Decat}(C + D) \cong \mathrm{Decat}(C) + \mathrm{Decat}(D)$

so it's all quite nice.

Alex Kreitzberg (Sep 25 2024 at 02:27):

So is $Decat$ just an ordinary functor that preserves limits and colimits?

Alex Kreitzberg (Sep 25 2024 at 02:32):

Making it a left and right adjoint functor?

John Baez (Sep 25 2024 at 02:33):

Hmm, I have no idea if it preserves limits and colimits - that sounds too good to be true, I just said products and coproducts. I'll think about it.

I guess you're thinking of $\mathrm{Decat}$ as going from the category of categories to the category of sets.

It also goes from the 2-category of categories, functors and natural isomorphisms to the category of sets regarded as a discrete 2-category. This is a very useful fact: e.g. it implies that equivalent categories have isomorphic decategorifications. But this changes the meaning of your question about whether it preserves limits and colimits.

John Baez (Sep 25 2024 at 02:35):

No, it doesn't preserve equalizers - at least not if we think of it going from the category of categories and functors to the category of sets and functions.

Alex Kreitzberg (Sep 25 2024 at 02:37):

Interesting, I guess I was just trying to ask something about $Decat$ to understand it better. While using it, it seemed like I wasn't using all the structure it might have.

Alex Kreitzberg (Sep 25 2024 at 02:38):

So your added context is what I was looking for

John Baez (Sep 25 2024 at 02:44):

Puzzle: find a category $C$ with two objects and only two non-identity morphisms, and a functor $F: C \to C$, such that

• the decategorification of the equalizer of $F$ and $1_C: C \to C$

is not isomorphic to

• the equalizer of $\mathrm{Decat}(F)$ and $\mathrm{Decat}(1_C)$

Here I'm thinking of $\mathrm{Decat}$ as a functor from the 1-category $\mathsf{Cat}$ to the 1-category $\mathsf{Set}$, which is not really ideal.

John Baez (Sep 25 2024 at 02:45):

(For anyone who wants a puzzle.)

Alex Kreitzberg (Sep 25 2024 at 19:39):

puzzle solution:

Define a category $C$ with two objects and an isomorphism between them. Define $F$ to swap the objects, and the components of the isomorphism.

Now $1_C$ and $F$ disagree on every input, so their equalizer is the empty category, which decatigorifies to the empty set.

Both $Decat(F)$ and $Decat(1_C)$ are maps from a singleton set to itself, which makes them both the identity function. So their equalizer is just the singleton set.

The singleton set isn't isomorphic to the empty set.

John Baez (Sep 25 2024 at 21:00):

Great! I believe this is the simplest counterexample to show $\mathsf{Decat}$ doesn't preserve limits, at least when we think of it as going from the mere 1-category of categories to $\mathsf{Set}$.

John Baez (Sep 25 2024 at 21:02):

If we work with the 2-category of categories, functors and natural isomorphisms, and replace the concept of 'limit' by the subtler 2-categorical concept of limit, things may work out better.

Alex Kreitzberg (Sep 26 2024 at 07:51):

I think I confused the story.

In the original example $Decat$ also didn't preserve limits, because $\times_{\mathbb N}$ wasn't a categorical product, it's a monoidal product. So $Decat$ is a monoidal functor.

But $\mathbb N$ is a ring, which I imagine categorifies to a fancier definition than a monoidal category. So $Decat$ is a bit better then that.

So maybe "does it preserve limits?" doesn't make sense? For the same reason it doesn't make sense to say $\times_{\mathbb N}$ is a product, the are no arrows in $\mathbb N$.

I'll think about how $Decat$ is a Functor between 2-categories.

Alex Kreitzberg (Sep 26 2024 at 07:58):

Or wait, maybe I'm level slipping.

Maybe we can say the products between categories gets sent to a product between sets, instead of the product within the category $Set$ being sent to Natural number's multiplication?

I needed to get this out before I went to sleep, tomorrow I'll try to draw a picture of everything I guess :joy:

John Baez (Sep 26 2024 at 18:06):

Maybe reread what I wrote?

Alex Kreitzberg (Sep 26 2024 at 18:12):

I'm sorry which part?

John Baez (Sep 26 2024 at 18:34):

Well, you're saying some curious things like this:

Alex Kreitzberg said:

In the original example $\mathrm{Decat}$ also didn't preserve limits, because $\times_{\mathbb N}$ wasn't a categorical product, it's a monoidal product. So $\mathrm{Decat}$ is a monoidal functor.

Alex Kreitzberg (Sep 26 2024 at 18:35):

In the first argument, a categorical product between categories was sent to a categorical product of sets. At that level products are preserved but not limits, at least when talking about 1-categories.

But I thought we were trying to understand how natural transformations between Set endofunctors stepped down to equations between natural numbers.

I'm getting confused with the relationship between the two.

Maybe I'm getting confused with Reihl's presentation again? Because I'm substituting in her Functor from $Set_{iso}$ to $\mathbb{N}$, with $Decat$ but they really aren't the same?

Alex Kreitzberg (Sep 26 2024 at 18:37):

No they can't be the same

Alex Kreitzberg (Sep 26 2024 at 18:39):

Okay, I'm going to try to understand the way $Decat$ is a Functor between 2-categories

Alex Kreitzberg (Sep 26 2024 at 18:55):

I shouldn't post anything when I'm hallucinating at 1 in the morning :upside_down:

Alex Kreitzberg (Oct 02 2024 at 17:32):

Okay, we have a 2-Category of Categories, Functors and natural isomorphisms

Decat steps this down to a 2-Category of Sets, functions, and function equality.

We've already explored a little bit how this works on objects and Functors. So now we should show natural isomorphisms map to equalities under $Decat$.

Suppose we have two Functors $F:\mathcal C\rightarrow \mathcal D$ and $G:\mathcal C\rightarrow \mathcal D$

And a natural isomorphism between them $F\cong G$. Then for each $c\in \mathcal C$ their outputs are isomorphic $F(c)\cong G(c)$.

I don't think I showed this earlier, but isomorphisms are preserved by functors, therefore isomorphic inputs map to isomorphic outputs; Hence, Decat taking Functors to functions between isomorphism classes is well defined.

In any case $F(c)\cong G(c)$ implies $f([c])=g([c])$ for all $c\in \mathcal C$, so $f=g$, where $f=Decat(F)$ and $g=Decat(G)$

Finally, this shows that if you know $A\times (B+C)\cong (A\times B)+(A\times C)$ noting both sides are the images of Functors $Set\times Set\times Set \rightarrow Set$

And have all of the product and union results about Decat, then

$$\begin{align} Decat(A\times (B+C) &\cong (A\times B)+(A\times C) )\\ Decat(A\times(B+C)) &= Decat((A\times B)+(A\times C))\\ Decat(A)\times Decat(B+C) &= Decat(A\times B) + Decat(A\times C)\\ a(b + c) &= ab + bc \end{align}$$

where $Decat(A)=a$, $Decat(B)=b$ and $Decat(C)=c$.

Which is the desired relationship Riehl was alluding to.

Note I think I'm abusing the notation above a bit by applying $Decat$ to a set, like in $Decat(A)$, because $Decat$ works on Functors and categories, not objects of a specific category, but I think if $Decat(A)$ is read as the equivalence class of $A$ then the above argument works.

An aside, I did need to reread what Baez wrote, because I read the 2-category as the category of categories, functors, and natural transformations. After getting my head sorted, even in this case I think it's possible to decatigorify to preorders and monotonic functions.

We draw an inequality between isomorphism classes if there is a monomorphism or section between representing objects. Then I think a natural (monomorphism? section?) from F to G could map to monotonic functions satisfying f≤g

Alex Kreitzberg (Oct 02 2024 at 18:25):

Edit:

I accidentally deleted everything at one point I hope it's sorted now.

I'd appreciate feedback on the argument, I'm worried I'm level slipping somewhere

Alex Kreitzberg (Oct 02 2024 at 18:48):

Can a server administrator restore the most recent message? Jean's suggestion helped!

Jean-Baptiste Vienney (Oct 02 2024 at 21:34):

If you click on «EDITED », you will see the history of the different versions of your message. Not in LaTeX though apparently.

Alex Kreitzberg (Oct 02 2024 at 21:40):

Thank you!

John Baez (Oct 02 2024 at 23:02):

An aside, I did need to reread what Baez wrote, because I read the 2-category as the category of categories, functors, and natural transformations.

Aha - yes, $\mathrm{Decat}$ does not extend to a 2-functor from that 2-category to $\mathsf{Set}$, because two functors $F, G: C \to D$ with a natural transformation between them can do different things to isomorphism classes of objects in $C$.

John Baez (Oct 02 2024 at 23:07):

After getting my head sorted, even in this case I think it's possible to decategorify to preorders and monotonic functions.

How does that work?

You can collapse any category $C$ to a preorder $\mathrm{Pre}(C)$ whose elements are objects of $C$ and $c \le c'$ iff there's a morphism $f: c \to c'$.

Any functor $F: C \to D$ then gives a monotone map, let's call it

$\mathrm{Pre}(F): \mathrm{Pre}(C) \to \mathrm{Pre}(D)$,

in an evident way. We can show $\mathrm{Pre}$ thus defined is a functor from the category of categories to the category of preorders.

But if there's a natural transformation $\alpha : F \to G$ we don't usually have $\mathrm{Pre}(F) = \mathrm{Pre}(G)$. What we instead get is

$\mathrm{Pre}(F) \le \mathrm{Pre}(G)$

Here I'm using the fact that given two monotone maps between preorders, say $f, g: P \to Q$, we can define $f \le g$ to mean

$\forall p \in P \quad f(p) \le g(p)$

Using this we see that there's not just a set of monotone maps between preorders, but a preorder of them.

And using this we get a 2-category $\mathbf{Preord}$ of preorders, monotone maps, and $\le$ relations between monotone maps!

John Baez (Oct 02 2024 at 23:10):

And if I'm not screwing up, we can extend $\mathrm{Pre}$ as defined so far to a 2-functor

$\mathrm{Pre} : \mathbf{Cat} \to \mathbf{Preord}$

John Baez (Oct 02 2024 at 23:13):

Is that what you were talking about?

Alex Kreitzberg (Oct 02 2024 at 23:22):

I thought of that initially, but I was worried the choice of $c\rightarrow c'$ if and only if $c \le c'$, was somehow too permissive.

Because then in $Set$ you have two way inequalities between every nonempty set. And one from the empty set to every other set.

So you can only tell whether a set is empty or not right?

So I was wondering if there was a principled way to choose a functor across all categories, which for the case of $Set$ mapped to Natural numbers with the usual ordering.

That's why I was trying to restrict the categories to their subcategories of monomorphisms or sections (I hope that makes sense).

Alex Kreitzberg (Oct 02 2024 at 23:36):

(that is, I'm not sure the sections or monomorphisms by themselves form categories, I think the monomorphisms might...)

If $f,g$ are monomorphisms, then $f\, g \, x = f\, g\, y$ implies $g\, x = g\, y$ implies $x = y$ so the composition $f\, g$ is a monomorphism

Alex Kreitzberg (Oct 02 2024 at 23:40):

And the identity is a monomorphism, so I think the restriction of a category to those is still a category.

And then make the inequality $f \leq g$ if and only if $f \rightarrow g$, after making the restriction to monomorphisms.

Alex Kreitzberg (Oct 02 2024 at 23:44):

I think that'll take $Set$ to $\mathbb N$ with the usual ordering after decatigorification, but I haven't worked it out yet.

John Baez (Oct 03 2024 at 00:20):

Alex Kreitzberg said:

I thought of that initially, but I was worried the choice of $c\rightarrow c'$ if and only if $c \le c'$, was somehow too permissive.

I think you're imposing your taste on the math instead of letting the math tell you what it wants. Mathematics really loves the 2-functor $\mathrm{Pre}$ that I described. I could explain why. But let's solve your problem.

Because then in $\mathsf{Set}$ you have two way inequalities between every nonempty set. And one from the empty set to every other set. [...] That's why I was trying to restrict the categories to their subcategories of monomorphisms or sections (I hope that makes sense).

So you're saying that the category of sets and functions give you a very boring preorder if you apply the 2-functor $\mathrm{Pre}$ I described. True. For But if you want a more interesting preorder, just apply $\mathrm{Pre}$ to the category of sets and injective functions.

And notice this is something you can do quite generally. Given a category $C$, you can define a new category $\mathrm{Mono}(C)$ with the same objects but only monomorphisms as morphisms. Then you can apply $\mathrm{Pre}$ to this.

In other words: instead of discarding a beautiful tool because you don't like what it does in one particular example, notice that there's another tool available, and you can use that one first.

Alex Kreitzberg (Oct 03 2024 at 01:16):

Okay, very cool. I'll try to quiet my mind and listen to what math is trying to tell me.

It's cool (and kinda funny) that $Pre$ is directly useful for the problem I was trying to understand :joy:. That's good marketing!

John Baez (Oct 03 2024 at 01:53):

I can't resist saying a bit more. The category $\mathsf{Set}$ has a baby sister called $\mathsf{Boole}$ with just two objects, $F$ and $T$. There's a functor $\mathsf{Set} \to \mathsf{Boole}$ that sends empty sets to $F$ and nonempty ones to $T$. This functor is symmetric monoidal in two ways: it sends coproduct to "or" and product to "and".

In enriched category theory, a $\mathsf{Set}$-enriched category is just a category, since it has a set of morphisms from any object to any other. A $\mathsf{Boole}$-enriched category is a preorder, since it has a truth value of morphisms from any object to any other.

The functor $\mathsf{Set} \to \mathsf{Boole}$ I mentioned can be used to do base change. In this example, base change turns any $\mathsf{Set}$-enriched category to a $\mathsf{Boole}$-enriched category, by turning the hom-sets into hom-truth values. And this base change process is just what I've been calling $\mathrm{Pre} : \mathbf{Cat} \to \mathbf{Preord}$.

John Baez (Oct 03 2024 at 01:57):

This is why $\mathrm{Pre}$ is so beautiful: this process for turning categories into preorders emerges automatically from the simpler process of turning sets into truth values.

Alex Kreitzberg (Oct 04 2024 at 21:31):

I'm going to respond to what you posted after thinking carefully about it, but before moving on from $Decat$ to $\textrm{Pre}$, I wanted to clarify something.

There are a lot of products in our example and it's making me fidgety. Specifically we have three, we have $\times_{Cat}$ (I don't know what category this is an arrow in), $\times_{Set} : Set \times_{Cat} Set \rightarrow Set$, and $\times_\mathbb N : \mathbb N \times_{Set} \mathbb N \rightarrow \mathbb N$

Now $Decat$ takes $\times_{Set} : Set \times_{Cat} Set \rightarrow Set$ to $\times_\mathbb N : \mathbb N \times_{Set} \mathbb N \rightarrow \mathbb N$

So $Decat(Set \times_{Cat} Set) = \mathbb N \times_{Set} \mathbb N$ taking the categorical product to the set product. So it preserves products when we pay attention to what $Decat$ does on objects.

$Decat$ also operates on the Functor $\times_{Set}$, explicitly $Decat(\times_{Set}) = \times_\mathbb N$

I interpreted that last equality to mean $Decat$ was monoidal, because it took the tensor product $\times_{Set}$ to the monoid product $\times_\mathbb N$, is "monoidal functor" the right terminology here?

Does that all look correct?

Alex Kreitzberg (Oct 04 2024 at 21:40):

And if that is correct, waaaay back when you wrote $Decat(C + D) \cong Decat(C) + Decat(D)$ Was that + on the left on categories, and the + on the right on Sets?

Alex Kreitzberg (Oct 04 2024 at 21:45):

Doesn't $+_{Set} : Set \times_{Cat} Set \rightarrow Set$ map to $+_{\mathbb N} : \mathbb N \times \mathbb N \rightarrow \mathbb N$?

John Baez (Oct 04 2024 at 21:46):

Alex Kreitzberg said:

There are a lot of products in our example and it's making me fidgety. Specifically we have three, we have $\times_{Cat}$ (I don't know what category this is an arrow in),

You introduced this notation, and it sounds like you're asking us to guess exactly what you mean by it - which is extra fun because you say you don't know! :upside_down:

I can't tell whether this increases our chance of giving you an answer you consider correct, or decreases it. But I will rise to this challenge:

You may recall we discussed $\text{Decat}$ as a 2-functor

$\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$

where $\mathbf{Cat}_g$ is the 2-category of categories, functors and natural isomorphisms (this is Steve Lack's notation, I'm not wedded to it), and $\mathbf{Set}$ is the 2-category of sets, functions, and identity 2-morphisms between functions.

I asked you to show

$\text{Decat}(C \times D) \simeq \text{Decat}(C) \times \text{Decat}(D)$

for any categories $C, D$, and you happily did it.

And here's the point: the $\times$ at left is the product in $\mathbf{Cat}_g$, while the $\times$ at right is the product in $\mathbf{Set}$.

So I'm guessing that your new symbol $\times_{Cat}$ means the product in $\mathbf{Cat}_g$, while your $\times_{Set}$ means the product in $\mathbf{Set}$.

Alex Kreitzberg (Oct 04 2024 at 21:51):

That is how I understood $\times_{Cat}$ I didn't know what to put after the colon, $\times_{Cat} : (Cat, Cat) \rightarrow Cat$ feels weirdly recursive somehow. That's all that parenthetical was intended to convey.

John Baez (Oct 04 2024 at 21:52):

This seems to make your various formulas be correct, e.g.

$\times_{Set} : Set \times_{Cat} Set \rightarrow Set$

says that the product in $\mathbf{Set}$ is a 2-functor from the category $\mathbf{Set} \times \mathbf{Set}$, defined as a product in $\mathbf{Cat}_g$, to the category $\mathsf{Set}$.

John Baez (Oct 04 2024 at 21:55):

Alex Kreitzberg said:

And if that is correct, waaaay back when you wrote $Decat(C + D) \cong Decat(C) + Decat(D)$ Was that + on the left on categories, and the + on the right on Sets?

Yes indeed. If I were feeling like a smart-aleck I'd say "what else could it possibly be? - that's the only thing that parses!"

$C$ and $D$ are categories, i.e. objects of $\mathbf{Cat}_g$, while $\mathrm{Decat}(C)$ and $\mathsf{Decat}(D)$ are sets, i.e. objects of $\mathbf{Set}$. We always take products in the category or 2-category where our objects actually live.

John Baez (Oct 04 2024 at 21:57):

By the way, I should assert this claim:

The 2-functor $\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$ preserves products and coproducts.

John Baez (Oct 04 2024 at 21:58):

This claim implies

$\text{Decat}(C \times D) \simeq \text{Decat}(C) \times \text{Decat}(D)$

and

$\text{Decat}(C + D) \simeq \text{Decat}(C) + \text{Decat}(D)$

but also more (as you'll see from the definition of 'co/product preserving functor').

John Baez (Oct 04 2024 at 22:00):

Alex Kreitzberg said:

Doesn't $+_{Set} : Set \times_{Cat} Set \rightarrow Set$ map to $+_{\mathbb N} : \mathbb N \times \mathbb N \rightarrow \mathbb N$?

Yes indeed! Here $+_{\mathbb{N}}$ is ordinary addition of natural numbers, which is not a coproduct. So we say that the addition of natural numbers comes from decategorifying the coproduct of finite sets. Similarly multiplication of natural numbers comes from decategorifying the product of finite sets!

John Baez (Oct 04 2024 at 22:05):

I love how the various levels of multiplication - multiplication of natural numbers, product of sets and product of categories - interact with each other, with each relying on the next one. This is an example of the [[microcosm principle]].

And it goes on forever: for example, the product in $\mathsf{Cat}_g$ gives a 2-functor

$\times: \mathsf{Cat}_g \times \mathsf{Cat}_g \to \mathsf{Cat}_g$

where the times symbol in the expression $\mathsf{Cat}_g \times \mathsf{Cat}_g$ is the product in the 3-category of 2-categories! (You can use a 2-category or even a 1-category of 2-categories if you prefer, but the mathematics has a certain desire to keep escalating.)

Alex Kreitzberg (Oct 04 2024 at 22:09):

Okay, so we never use that $Decat$ preserves coproducts to prove the natural number equations come from decatigorifying the natural isomorphisms then (because all of them are of the form $F : Set \times Set \times Set ... \times Set \rightarrow Set$).

Instead I'd need to give an argument like

$\{1, 2, ..., n\} +_{Set} \{1, 2, ..., m\} = \{1, 2, ..., n + m\}$ or $[n] +_{Set} [m] = [n + m]$

Decatigorifies to

err... $n + m = n + m$, I dunno how to make this not look silly XD.

I think I understand all the parts now. And you're right the interactions are interesting.

Just one last question, can I say that $Decat$ is monoidal because it sends a tensor product to a monoid product? It seems like all the monoidal category natural isomorphisms will decatigorify to monoid multiplications - is it therefore a monoidal functor, or is that not quite right terminologically?

Alex Kreitzberg (Oct 04 2024 at 22:12):

I'm trying to understand David's earlier comment about monoidal products

John Baez (Oct 04 2024 at 22:18):

Alex Kreitzberg said:

Okay, so we never use that $Decat$ preserves coproducts to prove the natural number equations come from decatigorifying the natural isomorphisms then.

Right, there's an interesting asymmetry! You might think products and coproducts are perfectly symmetrical, we use products to describe binary operations, even coproducts! E.g.
coproduct of sets defines a functor

$+ : \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set}$

not

$+ : \mathsf{Set} + \mathsf{Set} \to \mathsf{Set}$

We can try to repair this asymmetry in various ways, but that's another story for another evening.

Alex Kreitzberg (Oct 04 2024 at 22:21):

Oh I wasn't planning on fixing that asymmetry, that's hilarious that you can XD, food for thought!

John Baez (Oct 04 2024 at 22:26):

Alex Kreitzberg said:

Just one last question....

I bet you're going to break that promise someday. :wink:

can I say that $Decat$ is monoidal because it sends a tensor product to a monoid?

Yikes! I get what you mean, but let's say it right:

1. The 2-functor $\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$ preserves products.
2. Any 2-category with products is an example of a [[monoidal 2-category]], and it's a theorem that any product preserving 2-functor is a monoidal 2-functor.
3. So, $\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$ is a monoidal 2-functor. But this statement is wimpy and weak compared to 1.
4. A monoidal 2-functor maps monoidal objects (usually called [[pseudomonoids]]) to monoidal objects.
5. So, $\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$ maps monoidal objects in $\mathbf{Cat}_g$ to monoidal objects in $\mathbf{Set}$
6. A monoidal object in $\mathbf{Cat}_g$ is a [[monoidal category]]. A monoidal object in $\mathbf{Set}$ is a [[monoid]].
7. So, $\text{Decat} : \mathbf{Cat}_g \to \mathbf{Set}$ maps monoidal categories to monoids!

And this is how multiplication of natural numbers emerges from product of finite sets! Now you can really see how all the levels are interacting.

Alex Kreitzberg (Oct 04 2024 at 22:37):

Woooow, okay cool. Believe it or not, that is all making sense (though it's clear to really get it I need to walk down each level manually)

I'll leave it there, I'll try to ask a good question about $\mathrm{Pre}$ later.

You have a good evening, thank you for your time.

John Baez (Oct 04 2024 at 22:46):

Great! Yes, it's easy to get lost among these levels, especially since they all look so similar... but that's actually one reason this subject is fun: it all makes perfect sense eventually and you feel like a 'master of many levels'.

Daniel Teixeira (Oct 07 2024 at 13:51):

Another perspective on this is that in $\mathsf{FinSet}_{iso}$ we have a morphism $A\times B \to B\times A$ which must be carried by $F:\mathsf{FinSet}_{iso}\to\mathbb N$ to a morphism in $\mathbb N$ (since it's a functor). But here we only have identities; so we must assert that $F(A\times B) = F(B\times A)$ by lack of a better option. (This might be why Emily doesn't work with $\mathsf{FinSet}$ directly, as a functor $\mathsf{FinSet}\to\mathbb N$ would be constant.)

Daniel Teixeira (Oct 07 2024 at 13:52):

This is something that systematically happens in decategorification: what was an $n$-morphism at one level must become an identity (equality) one level down, due to the lack of non-identities.

Alex Kreitzberg (Oct 07 2024 at 15:16):

That makes me think of $\mathrm{Pre}(\mathrm{Iso}(\mathrm{Set}_{\mathrm{Fin}})) = \mathbb N$ using Baez's preorder functor.

This is a good moment to call out, in Riehl's original example, she defined a cardinality Functor $|\cdot| : \mathrm{Set}_{\mathrm{iso}} \rightarrow \mathbb N$ which is a morphism in $\mathrm{Cat}$, but $\mathrm{Pre}$ and $\mathrm{Decat}$ are both morphisms in $\mathrm{2Cat}$.

If the goal is to understand Decatigorification beyond just $\mathrm{Set}$, then it makes more sense to talk about $\mathrm{Decat}$ and $\mathrm{Pre}$, so I get the motivation behind Baez's shift in focus.

Though I suppose Riehl was trying to avoid talking about 2-categories, since her goal here was to show off how nifty natural transformations are.

Alex Kreitzberg (Oct 07 2024 at 16:46):

Actually do we have $\textrm{Pre} \circ \textrm{Iso} \cong \textrm{Decat}$?

Hmm

Alex Kreitzberg (Oct 07 2024 at 16:50):

I guess this doesn't fit Daniel's point perfectly because they're saying the target doesn't have inequalities, only equalities, and so the natural transformations are forced to fit.

Alex Kreitzberg (Oct 07 2024 at 17:01):

Here I'm defining $\textrm{Iso}$ to restrict a category to its isomorphisms.

John Baez (Oct 07 2024 at 17:10):

Great puzzle there. What you're calling $\text{Iso}$ is now often called the core of a category and denoted $\text{core}$ --- this convention seems to have been invented by people at the nLab, but it seems to be catching on. So I'll use that.

Puzzle. Suppose we apply the 2-functor $\text{core}$ which takes a category and throws out all morphisms except isomorphisms, and then apply the 2-functor $\text{Pre}$ which collapses a category to a preorder where $x \le y$ iff there was a morphism from $x$ to $y$. How is this related to applying the 2-functor $\text{Decat}$, which sends a category to its set of isomorphism classes?

Alex Kreitzberg (Oct 10 2024 at 19:45):

I've been trying to keep myself from being confused by bicategories and pseudofunctors

All the Functors we have, $\mathrm{core}$, $\mathrm{Pre}$, and $\mathrm{Decat}$ are strict. I did a quick proof that each one was a "Functor" in the naive sense.

I did this because our maps disagree on the category with one isomorphism
$\mathrm{Decat}(a \leftrightarrows b) = \{ \cdot \} \neq \{ a \leq b, b \leq a\} = \mathrm{Pre} \circ \mathrm{core}(a \leftrightarrows b)$

However, the mapped categories are equivalent.

I don't think I can make a canonical choice of functors exhibiting this equivalence for every category, so I don't think $\mathrm{Pre} \circ \mathrm{core}$ and $\mathrm{Decat}$ have a natural equivalence.

I think I could get these "equal" if I used the skeleton pseudofunctor $\mathrm{sk}$ on the output of $\mathrm{Pre}$ making the preorder into a total order, so $a \leq b, b \leq a \Rightarrow a = b$

note: I don't really fully understand pseudofunctors but I get the motivation.

The relationships here seem innocuous but I'm worried that if I trip I'll fall off a cliff.

Why is it safe to work with strict 2-Functors and strict 2-categories generally? Is it true I need Pseudofunctors to get these Functors to be "the same", and therefore I need bicategories? Am I overthinking this?

John Baez (Oct 10 2024 at 20:11):

Every bicategory is equivalent to a strict 2-category - this is a souped-up version of Mac Lane's strictification theorem saying every monoidal category is equivalent to a strict monoidal category.

However, the equivalence here is not a strict 2-functor.

There's a lot more to say about this, but you've convinced me that Pre $\circ$ core and Decat are equivalent if we set up the machinery intelligently (e.g., not more strict than it should be, etc.).