# (Co)completness of the 2-category of 2-Rings"

Let $2Ring$ denote the 2-category of cocomplete categories with monoidal structures that preserve colimits in each argument. The morphisms are cocontinuous and strong monoidal functors (which are categorified ring homomorphisms). This notion is described in detail in Martin Brandenburg's thesis Tensor categorical foundations of algebraic geometry, arXiv:1410.1716 (pdf) and many authors use a slightly more/less structured notion as a categorification of the category of rings.

Is $2Ring$ is (co)complete? I have in mind strict 2-(co)limits but would be interested in other flavors of 2-categorical (co)limits as well. Additionally, I would also like to consider the same question but replacing $2Ring$ with $Comm2Ring$. $Comm2Ring$ is the same as $2Ring$ except with symmetric monoidal structures.

What if we impose that $2Ring$ is the 2-category of locally presentable categories with monoidal structures which preserve colimits in each argument? (This is the approach taken in Chrivasitu and Johnson-Freyd's The fundamental pro-groupoid of an affine 2-scheme, arXiv:1105.3104v4 (pdf).)

• Do you mean symmetric monoidal here? Also, you haven't specified what functors you want to consider; I think a reasonable choice is cocontinuous and strong monoidal. – Qiaochu Yuan Aug 24 '16 at 23:09
• I'd like to consider plain monoidal along with symmetric monoidal. Those functors are exactly what I wanted. I'll add this into the question. – user84563 Aug 24 '16 at 23:12
• The coproduct of two commutative rings has underlying abelian group the tensor product of the underlying abelian groups. Similarly, one expects that the coproduct of two cocomplete symmetric monoidal categories has as underlying cocomplete category the tensor product of the underlying cocomplete categories. The problem is, however, that it seems to be unknown if every pair of two cocomplete categories admits a tensor product (defined by classifying functors which are cocontinuous in both variables), although this is the case if one of them is locally presentable. – HeinrichD Sep 12 '16 at 8:12
• Similarly, the coequalizer of two cocontinuous symmetric monoidal functors will be a localization of the domain, but (reflective) localizations only really work well in the locally presentable setting, as is explained in Brandenburg's thesis, Section 5.8. It seems to be quite reasonable that the 2-category of locally presentable symmetric monoidal categories (by definition, this means that the tensor product is cocontinuous in both variables) is 2-cocomplete and that locally presentability is crucial here. – HeinrichD Sep 12 '16 at 8:16

I would argue that the "correct" version of (co)limits is the "pseudo" or "bi" or "weak" or "strong" or "homotopy" version. (As far as I can tell, all of these words mean the same thing.) The bicategories of cocomplete or of locally presentable categories are closed under limits: indeed, you can show by hand that the limit just as categories of cocomplete categories along cocontinuous functors is itself cocomplete, and Bird's thesis verifies that if all constituents are locally presentable, then so is the limit. Lurie has shown the same statement in the $\infty$-categorical world.

2-rings, in either Brandenburg's sense or ours, are therefore the symmetric monoidal objects in a complete bicategory. Just as in the 1-categorical case, I believe it is straightforward (if tedious) to show that the limit of underlying categories carries a canonical symmetric monoidal structure, making it the limit of 2-rings. So this handles the "limit" version of your question.

The colimit version I believe also works, although you cannot just take the colimit of underlying categories. I haven't thought as much about the bicategory of all cocomplete categories, but the bicategory of locally presentable categories is known, again by Bird's thesis, to contain all colimits. (They are computed by computing instead the limit along right adjoints to the cocontinuous functors you have in mind.) Then you can present a colimit of 2-rings (in our sense) much as you would present a colimit of commutative algebras: any time you see a coproduct, for example, write down a free commutative algebra; any time you see a quotient, do it symmetric monoidally; etc. for the other types of colimits in the bicategorical world. I haven't done the exercise fully myself, but again at least in the locally presentable case I'm sure you can do it (in terms of sketches if necessary).

I don't recall off the top of my head whether colimits of arbitrary cocomplete categories exist. I could imagine that you would run into size problems, but I don't know.

I should say, there is also a fair amount of work on the bicategorical version of monads, and (co)limits, in all senses (including lax and oplax), of their algebras. I don't recall the references, but my memory is that there are no surprises from the 1-categorical case.

• "...are therefore the symmetric monoidal objects in a complete bicategory." So in your sense, this would be the bicategory of presentable categories with the cartesian product? – user84563 Aug 25 '16 at 1:46
• Also do you have references for presenting the colimit of 2-rings and the limits of monoid objects carrying a canonical monoidal structure? – user84563 Aug 25 '16 at 4:46
• @user84563 Ah, herm, you're right, I need to think more. The monoidal structure is the tensor product, so abstract nonsense of monads might not be good enough. Results from Lurie's work are plenty strong enough --- you say the right words about algebras for an operad in a closed monoidal higher category ---, but I don't know of a paper that carefully matches Lurie-style notions of (co)limit with the older version used in the bicategory literature. There might be one, and I think everyone trusts that these do match, I just don't know a paper. – Theo Johnson-Freyd Aug 25 '16 at 20:27
• The long and the short of it is that the answer to your question (at least in the locally presentable version --- I have less intuition in the realm of all cocomplete categories) is Platonically yes, but I don't have papers to point to. I'm confident that I could prove directly that all limits and colimits of 2-rings (in our sense) exist, if I ever needed to. I'm also confident that this is covered by (known) abstract nonsense of $\infty$-operads, modulo (possibly not yet written down?) translations between the bicategory and $\infty$-category worlds. – Theo Johnson-Freyd Aug 25 '16 at 20:30
• Symmertic or braided or plain doesn't matter, since you should just appeal (either formally, by citing Lurie et al, or informally by unpacking the arguments) to facts about limits and colimits of algebras for operads. – Theo Johnson-Freyd Aug 25 '16 at 20:31