The n-Category Café

Well, the way Tim defined $\mathcal{R}C$ is invalid, because it refers to an adjunction that doesn’t exist. Suppose, for instance, that $C$ is a large discrete category. The empty functor $C^{op} \to Set$ is small, but its conjugate is the terminal functor $C \to Set$, which is not small. So conjugation doesn’t define a functor $\mathcal{P}C \to \mathcal{P}^\dagger C$.

So the issue is that the conjugate of a small presheaf needn’t be small (and the conjugate of a non-small presheaf needn’t even be defined). It therefore seems to be necessary to define reflexivity as follows: a presheaf $X$ is reflexive if $X$ is small, $\check{X}$ is small, and the unit map $X \to \hat{\check{X}}$ is iso.

The argument for “complete $\implies$ reflexively complete” in our paper (not yet finished) runs as follows. On any category, any small presheaf is a small colimit of representables, so the conjugate of any small presheaf is a small limit of representables. Hence on a complete category, the conjugate of any small presheaf is representable — and in particular, any reflexive presheaf is representable.

Hmm, that makes me somewhat suspicious of the whole business of defining reflexive completion of non-small categories using small presheaves.

If you have other ideas for how to do it, I’d be very interested to hear them.

We were more or less following Isbell in defining it that way. (I say “more or less” because he used a size condition on presheaves which, as far as we know, is not quite the same as smallness.)

But it’s a more intuitive idea than it seemed to me at first. Maybe that’s just because I’ve got used to it, or maybe it’s because of the following analogy with Fourier transforms.

Let’s consider functions (measurable, say) $$f: \mathbb{R}^n \to \mathbb{C}.$$ Not every such function has a Fourier transform. In order to define the Fourier transform $\hat{f}$ of $f$, we need to ask for $f$ to be integrable. Even so, $$\hat{f}: \mathbb{R}^n \to \mathbb{C}$$ needn’t be integrable. But if $\hat{f}$ is integrable, we can define the dual Fourier transform $$\check{\hat{f}}: \mathbb{R}^n \to \mathbb{C}.$$ (By “dual Fourier transform” I mean the usual formula for Fourier transform but with a change of sign in the exponential; it’s sometimes called the inverse Fourier transform, even though it doesn’t always give an inverse.)

This leads us to consider those functions $f$ on $\mathbb{R}^n$ such that $f$ is integrable, $\hat{f}$ is integrable, and $f = \check{\hat{f}}$. And that’s like the definition of the reflexive completion of $C$: it consists of those Set-valued functors $F$ on $C$ such that $F$ is small, $\hat{F}$ is small, and the unit $F \to \check{\hat{F}}$ is an isomorphism.

Now one can argue that this framework for Fourier transforms is unsatisfactory, what with continually having to worry about the space of functions we’re working with. But it does at least mean that the framework for reflexive completion has some respectable precedent.

Here’s one. An arbitrary presheaf $F:C\to Set$ has a conjugate $\hat{F}:C^{op}\to SET$ that takes values in a larger universe. Thus, we could define a $Set$-valued presheaf $F$ to be reflexive if $\hat{F}$ is also $Set$-valued and $F \to \check{\hat{F}}$ is iso. One could think of this as also like the Fourier example, with the $SET$-valued presheaves playing the role of the “distributions” — every function has a distributional Fourier transform, and we can then just restrict attention to those functions whose distributional Fourier transform happens to be another function.

Oh, that’s $\mathbf{B}(Set(3,3))$ – the endomorphism monoid of the set with 3 elements, regarded as a category. That’s me being a little too cute – although it’s interesting to see that a monoid can have a pretty big reflexive completion, given that you’ve shown a group can’t.

I wasn’t sure from your slides, but I guess you’re saying that when $A \to C$ is a snug embedding, the canonical functor $C \to \mathcal{R}(A)$ is itself an embedding (by which I mean fully faithful), right? Assuming that, the argument looks good to me!

Thanks for the clarification.

I don’t think the point about monoids is too cute! As you say, it’s interesting that reflexively completing a group barely enlarges it, but reflexively completing a certain 27-element monoid turns it into something infinite.

The slides don’t say it, but our paper (in draft) will: when $G: A \to C$ is a snug embedding, there’s a unique snug embedding $C \to \mathcal{R}(A)$ making the evident triangle commute.

If you view $\mathcal{R}(A)$ as a subcategory of $\hat{A}$, this embedding $C \to \mathcal{R}(A)$ is nothing more than the nerve-type functor $C \to \hat{A}$ induced by $G$. Since $G$ is dense, the functor $C \to \mathcal{R}(A)$ is full and faithful. The “nice property” on slide 12 then implies that it’s snug.

One might even want to state this (putative) result in the following way. Let $\mu$ be the smallest measurable cardinal, if it exists, and let $Set_\mu$ denote the sets smaller than $\mu$, or else all of $Set$ (ZFC implies measurable cardinals are inaccessible, but not ZF, so I’m not sure if I doing anything suspicious here). Then do the enriched version of your results using $Set_\mu$. It should be the case that there is a ‘small’ category (i.e. an internal category in $Set_\mu$ when that category is a topos) such that its reflexive completion is not ‘essentially small’ (i.e. it is not the externalisation of an internal category in $Set_\mu$).

What the terms in scare quotes mean (or should mean) when $Set_\mu$ is not a topos I don’t know. It probably would be handled by fibred categories or similar. Note that in ZF with the axiom of determinacy $\omega_1$, the first uncountable ordinal (denoted $\aleph_1$ when Choice is present), is measurable. So $Set_\mu = Set_{\omega_1}$ is the category of countable sets. The content of the (putative) result is then, morally, that there is a countable category whose reflexive completion is not equivalent to a countable category. This is a quasi-sensible-sounding statement, but as it requires the axiom of determinacy to prove $\omega_1$ measurable, it has to be a bit trickier than that (and apparently the consistency strength of AD is very high).

Interesting!

I had some related but different thoughts.

All I want to assume about $Set$ is that it satisfies the ETCS axioms. Now, we know that given an ETCS category, it’s possible to build another ETCS category $\mathcal{E}$ that fails the axiom of choice. If the axiom of choice fails badly enough then the full subcategory of $\mathcal{E}$ consisting of the single object $3 = 1 + 1 + 1$ should be codense. It should also be dense. In that case, we’ve got a finite $Set$-enriched category whose reflexive completion isn’t even small.

There may be a much easier way of implementing this basic idea, without thinking about ETCS at all. Maybe there’s some easily-constructed large topos $\mathcal{E}$ such that the full subcategory consisting of the single object $1 + 1 + 1$ is finite, dense and codense. That would then be an example of a finite category whose reflexive completion isn’t even small.

Away from set theory, I believe I know an example of a small category whose reflexive completion isn’t small. If I understand correctly, Isbell’s Theorem 2.6 states that the one-object full subcategory $[0, 1]^2$ of (compact Hausdorff spaces) is codense, as well as dense. So by the usual argument, the reflexive completion of this small subcategory is not essentially small.

How about this: the category of Hilbert spaces is countably acceessible and self-dual, via an object-fixing duality. So the reflexive completion of the small category of countably accessible Hilbert spaces (or just of the 1-object category $\mathbb{B}(End_{Hilb}(\ell_2(\mathbb{Z}))$) is the large category of Hilbert spaces.

Here’s another example with a similar feel.

The category $FinVect_k$ of finite-dimensional vector spaces over a field $k$ is equivalent to its opposite via an object-fixing duality. Since $k^2$ (or any vector space of dimension $\geq 2$) is dense in $FinVect_k$, it is also codense. So if $\mathbf{B}$ is the inclusion of a subcategory of $FinVect_k$ containing an object of dimension at least 2, then the inclusion $\mathbf{B} \to FinVect_k$ is snug, and again is probably the reflexive completion.

It’s interesting that by bumping up the dimension, we can recover the same reflexive completion as in the enriched case. I wonder what the $Set$-enriched reflexive completion of $\mathbf{B}(End_k(k))$ is?

We can try to play a similar game with finite (or finitely-generated) abelian groups. Using the classification theorem, it’s clear that the cyclic (or even: cyclic and indecomposable) groups form a cogenerator among the finite/finitely generated ones. But I suspect this cogenerator may not be codense.

Re the vector space example, $FinVect_k$ is of course essentially small, so it doesn’t help us get an example of a small category whose reflexive completion is large.

However, if we take $k$ to be a finite field and choose a finite full subcategory $\mathbf{B}$ of $FinVect_k$ containing at least one space of dimension at least $2$, then as you say, the embedding $\mathbf{B} \hookrightarrow FinVect_k$ is snug. Hence the reflexive completion of the finite category $\mathbf{B}$ (as an ordinary category) is infinite.