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Stream: learning: questions

Topic: Exotic categorical completions

John Onstead (Jul 29 2025 at 11:47):

In general, a free completion of a category freely adjoins new objects—often thought of as “weaker” or “virtual” versions of the originals that "should have been" in the original category if not for some restricting condition—to simplify certain constructions (for example, the free colimit cocompletion is a topos with all that entails). We usually reserve the term “completion” for free (co)limit (co)completions, and so "ordinary" embeddings like $\mathrm{Grpd}\;\hookrightarrow\;\mathrm{Cat}$ are rarely considered completions. But while this might sound a little crazy, if we switched to the right perspective, might it be possible to consider this to be some form of "completion"?

John Onstead (Jul 29 2025 at 11:48):

Here's how I'm thinking. Every category can be found "within"- IE, as a 1-categorical subobject of- some groupoid (namely, its groupoidification), and so there's a sense they "should" be present in the category of groupoids. But they aren't "visible" due to the restriction that a subobject of a groupoid must have all inverses. From this, it almost seems like categories are like a "free subobject completion" of groupoids. This would consist of freely adding in precisely the subobjects that allow us to separate morphisms in groupoids from their inverses. Perhaps some procedure like freely adding in the walking morphism to $\mathrm{Grpd}$, freely adding the two inclusions from it to the walking isomorphism, and somehow generating all the rest of $\mathrm{Cat}$ based on that.

John Onstead (Jul 29 2025 at 11:49):

And so my question is: is there any valid, well defined construction like this proposed "free subobject completion", and is there any way the groupoid-category embedding might be interpreted as one?

Mike Shulman (Jul 29 2025 at 15:49):

The map from a category to its groupoid reflection is not always even faithful, so it doesn't make sense to think of it as a subobject.

John Baez (Jul 29 2025 at 16:14):

Let's see, what's the simplest example of that? A monoid that does not inject into its group completion should do the job. Okay, how about the monoid $\mathbb{B} = \{F,T\}$ with "or" as its monoid operation? Since this monoid is not cancellative, it cannot inject into its group completion.

Mike Shulman (Jul 29 2025 at 16:17):

And the one-object category corresponding to that monoid is the walking idempotent, whose groupoid reflection is the terminal category, since any invertible idempotent is an identity.

Mike Shulman (Jul 29 2025 at 16:18):

So more generally, no category containing a nontrivial idempotent can inject into its groupoid reflection.

John Baez (Jul 29 2025 at 16:27):

You can split 'em, but you can't invert 'em.

(Without turning them into the identity.)

John Onstead (Jul 29 2025 at 20:40):

Oh, I didn't know that a monoid/category won't always inject into the group/groupoid completion! I guess I thought that we could only ever add morphisms to the category by freely adding inverses, but it's interesting that if there's an idempotent adding this extra inverse will "collapse" it into the identity. What seems even more interesting here is that it seems you can characterize the identity morphisms as any endomorphism which is both idempotent and invertible (an automorphism).

Kevin Carlson (Jul 29 2025 at 21:24):

Yes, and that's quite elementary: if $e^2=e$ and $f=e^{-1}$ then $\mathrm{id}=f\circ e=f\circ e^2=e.$ (Note, this works even if $f$ is just a one-sided inverse, and implies also $f=e$!)

John Onstead (Jul 29 2025 at 21:28):

Kevin Carlson said:

Yes, and that's quite elementary: if $e^2=e$ and $f=e^{-1}$ then $\mathrm{id}=f\circ e=f\circ e^2=e.$ (Note, this works even if $f$ is just a one-sided inverse, and implies also $f=e$!)

Wow, that's quite cool!

John Onstead (Jul 29 2025 at 22:05):

With the above in mind, let me see if I can try to rephrase my initial question. Let $\mathrm{CanMon}$ be the category of cancellative monoids. Is there a way to view $\mathrm{CanMon}$ as being some form of "completion" of $\mathrm{Grp}$?

John Onstead (Jul 30 2025 at 11:29):

Or if we want to be more general about this- let $D$ be a mono-reflective subcategory of $C$. Of course, $D$ might be a mono-reflective subcategory of all sorts of other categories, so $C$ is nowhere near unique in this capacity. Perhaps there's some missing ingredient/information we can supply to make the determination unique. What might be this "extra structure" I should add into/onto $D$ in order to be able to uniquely reconstruct the objects of $C$ from those in $D$, if such a thing is possible at all?

John Onstead (Aug 05 2025 at 11:32):

In Garner and Mike Shulman's work "Enriched Categories as Free Cocompletion", it is proven that given a locally cocomplete bicategory $V$, the bicategory $V$-$\mathrm{Mod}$ of enriched categories and profunctors is the free cocompletion of $V$ under collages. While the topic was fascinating, I was left unsatisfied by two main things:

1. The requirement of being locally cocontinuous is quite a stringent requirement
2. The part where they generalize to equipments was very unwieldy, since they were treating equipments "piecewise" (as a pair of categories with a certain functor between them) instead of as a single cohesive object, which made the section more confusing for me

John Onstead (Aug 05 2025 at 11:33):

These issues can be resolved if we move to the setting of virtual equipments and VDCs, where equipments are single objects and cocompleteness is no longer needed since profunctors need not be composable. But of course, we have to be careful that the results actually transfer. Let $V$ be a VDC and define $\mathrm{Coll}(V)$ to be the VDC with all collages and an embedding $V \to \mathrm{Coll}(V)$ such that for any VD functor to any other VDC with collages $V \to D$, there is a unique extension to a collage-preserving (if necessary; the paper seems to imply collages are absolute thus making this requirement automatic) VD functor $\mathrm{Coll}(V) \to D$. The magic question is then: yes or no, is the following true: $\mathrm{Coll}(V) \cong V$-$\mathrm{Mod}$?

Nathanael Arkor (Aug 05 2025 at 11:57):

I gave a talk on the extension to virtual double categories at the Tallinn Category Theory Seminar last year: A recipe for enriched categories. I haven't yet found time to finish writing it up, though.

Nathanael Arkor (Aug 05 2025 at 11:58):

(Though I would disagree that the functorial perspective on equipments is any less "cohesive" than the double categorical one. As with any presentation, there are advantages and disadvantages.)

Nathanael Arkor (Aug 05 2025 at 11:59):

the paper seems to imply collages are absolute thus making this requirement automatic

Collages are not absolute if you care about the (virtual) double category of enriched categories, rather than the bicategory of enriched distributors. So the preservation condition is non-trivial.

Mike Shulman (Aug 05 2025 at 13:48):

I actually agree that the definition of an equipment as an identity-on-objects pseudofunctor is a bit unlovely. Richard and I started with that definition since it's the most traditional in the bicategory literature. But in actually stating and proving our theorems, we shifted to the "enhanced 2-category theory" perspective that an equipment is a bicategory enriched over the monoidal bicategory whose objects are fully faithful functors, which I find significantly more appealing (and useful).

Mike Shulman (Aug 05 2025 at 13:56):

The locally cocomplete version is nice because then you can regard this "cocompletion" as being under a class of weighted colimits in the ordinary (bicategorical) sense. But of course the virtual version is nice too for the extra generality. It sounds like Nathanael has worked out its universal property! Let me just add that the construction of this VDC goes back to Leinster.

John Onstead (Aug 05 2025 at 19:49):

Nathanael Arkor said:

I gave a talk on the extension to virtual double categories at the Tallinn Category Theory Seminar last year: A recipe for enriched categories.

Wow, that seems to be exactly what I'm looking for, thanks! So my intuition wasn't too far off. Though if there was one surprise, it's that the result of the category of monads being a free cocompletion doesn't seem to extend beyond "normal" VDCs.

John Onstead (Aug 05 2025 at 19:51):

Mike Shulman said:

I actually agree that the definition of an equipment as an identity-on-objects pseudofunctor is a bit unlovely. Richard and I started with that definition since it's the most traditional in the bicategory literature. But in actually stating and proving our theorems, we shifted to the "enhanced 2-category theory" perspective that an equipment is a bicategory enriched over the monoidal bicategory whose objects are fully faithful functors, which I find significantly more appealing (and useful).

I also like this "F-category" picture, it allows you to use the tools of enriched category theory in the equipment setting!

John Onstead (Aug 05 2025 at 19:56):

Overall, I think this makes sense. If we think of a monad as a single object enriched category, then a "polyad" is a more general enriched category. We can derive a polyad as a monad if the monad is a matrix of other objects- such as if it's a monad in a VDC of matrices, or equivalently, as Arkor showed, in a sort of VDC of "families". Then collages encode similar information to this polyad. Arkor explains this as being due to the fact an enriched category is a diagram in a VDC. I don't know what he means by this since it isn't elaborated on, but maybe it refers to how a polyad can also equivalently be expressed as a lax functor (or virtual double functor) from a "codiscrete category" into a VDC (with a monad the lax functor from the terminal category). Though I might be wrong about that.

Nathanael Arkor (Aug 05 2025 at 21:43):

John Onstead said:

Wow, that seems to be exactly what I'm looking for, thanks! So my intuition wasn't too far off. Though if there was one surprise, it's that the result of the category of monads being a free cocompletion doesn't seem to extend beyond "normal" VDCs.

An intuition for this is that, if $\mathbb V$ is a VDC that does not necessarily have loose identities, then we cannot necessarily define a functor $\mathbb V \to \mathbb V\text{-}\mathbb D\mathbf{ist}$ sending an object $V$ of $\mathbb V$ to a one-object $\mathbb V$-category with extent $V$ (because the endo-hom of the $\mathbb V$-category must be a loose morphism in $\mathbb V$, but $\mathbb V$ may not even have any loose morphisms at all).

Nathanael Arkor (Aug 05 2025 at 21:44):

John Onstead said:

I don't know what he means by this since it isn't elaborated on, but maybe it refers to how a polyad can also equivalently be expressed as a lax functor (or virtual double functor) from a "codiscrete category" into a VDC (with a monad the lax functor from the terminal category).

Yes, this is the intuition.

John Onstead (Aug 05 2025 at 23:00):

Nathanael Arkor said:

An intuition for this is that, if $\mathbb V$ is a VDC that does not necessarily have loose identities, then we cannot necessarily define a functor $\mathbb V \to \mathbb V\text{-}\mathbb D\mathbf{ist}$ sending an object $V$ of $\mathbb V$ to a one-object $\mathbb V$-category with extent $V$ (because the endo-hom of the $\mathbb V$-category must be a loose morphism in $\mathbb V$, but $\mathbb V$ may not even have any loose morphisms at all).

Ah, ok, that makes sense. You need to have loose identities in the first place in order to be able to associate an object with its loose identity via the embedding.

Matteo Capucci (he/him) (Aug 06 2025 at 12:35):

Nathanael Arkor said:

I gave a talk on the extension to virtual double categories at the Tallinn Category Theory Seminar last year: A recipe for enriched categories. I haven't yet found time to finish writing it up, though.

What a nice work Nathanael!