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Background

There is a bicategory where the objects are categories, the 1-morphisms are profunctors, and the 2-morphisms are morphisms of profunctors. The non-obvious part of this assertion is that profunctors admit a "good" (i.e., coherently associative) composition. The way one sees this is to use the fact that the Yoneda embedding is the free cocompletion, from which we may identify profunctors from $\mathcal{C}$ to $\mathcal{D}$ with cocontinuous functors from $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$ to $\operatorname{Set}^{\mathcal{D}^{\operatorname{op}}}$. Over here, there is a strictly associative composition, so life is good.

All of this extends easily to the enriched setting. In particular, if we enrich over $\mathcal{V} = \operatorname{Cat}$, we can talk about strict 2-profunctors between strict 2-categories, and we get a nice bicategory of 2-categories, strict 2-profunctors, and morphisms between these.

Question

I would like to know if there is an analogous way to obtain a tricategory of bicategories and biprofunctors. You can "follow your nose" and write down what the composition should be, but checking all the axioms of a tricategory does not seem like the simplest approach. Of course, we can strictify to get equivalent strict 2-categories and 2-profunctors, but I doubt the composition will be the same; I don't think the "strict" colimits will in general agree with the "weak" colimits. Also, using the standard enriched theory only gives a bicategory; the natural transformations are lost. It seems that an appropriate statement of the form "the bicategorical Yoneda embedding is a free cocompletion" will handle things just as it does in the setting of ordinary profunctors, but I am not sufficiently comfortable with limits and colimits in bicategories to try to figure out if such a statement makes sense. Has anybody worked all of this out?

(The closest thing I have seen to such a result is Justin Greenough’s paper Monoidal 2-structure of Bimodule Categories, which more or less establishes this result in the case of sufficiently nice $k$-linear monoidal categories. But his proof seems to be very specific to that setting, whereas one would hope that a proof could be established along the same lines as the result for usual profunctors.)

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    $\begingroup$ My guess is that the answer to your question will be "no: the tricategory of bicategories and biprofunctors has not yet been constructed". $\endgroup$
    – André Henriques
    Aug 27, 2010 at 4:08
  • $\begingroup$ It might be helpful. S. Bozapalides. A note on 2-distributors. J. Ausral. Math. Soc.,23:Series A (1977), 242–246 $\endgroup$
    – Ma Ming
    Apr 11, 2014 at 18:09

4 Answers 4

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For those coming across this question more recently, there is now an answer to the original question. In fact, the tricategory of pseudoprofunctors has been defined twice, independently, via different approaches.

  • Lawler defines the tricategory $2\text{-}\mathbb P\mathrm{rof}$ of bicategories and pseudoprofunctors (there called "2-profunctors") in Definition 4.1.1 of his 2014 thesis Fibrations of predicates and bicategories of relations. The hom-bicategories are given by the cocontinuous pseudofunctors between cocomplete bicategories, which is justified as corresponding to pseudoprofunctors by the results of Section 3.2.3. Defining the hom-bicategories in this way guarantees the coherence conditions hold immediately.
  • Chikhladze defines the tricategory $2\text{-}\mathcal P\mathrm{rof}$ of bicategories and pseudoprofunctors (there called "biprofunctors") in Section 8 of his 2014 paper Lax formal theory of monads, monoidal approach to bicategorical structures and generalized operads. Here, the approach taken is to define a bicategorical $\mathrm{Mat}$ construction, along with a bicategorical $\mathrm{Mod}$ construction, so that $2\text{-}\mathcal P\mathrm{rof} = \mathrm{Mod}(\mathrm{Mat}(\mathrm{Cat}))$. That $2\text{-}\mathcal P\mathrm{rof}$ is a tricategory then follows from the coherence of simpler constructions, which are shown to coincide with pseudoprofunctors in the same way as in the 1-categorical setting.
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My recent paper with Richard Garner, Enriched categories as a free cocompletion (version published in Adv. Math.) comes within about $3\epsilon$ of constructing the tricategory of enriched bicategories and enriched profunctors. I'm kind of sad that we didn't get around to crossing the remaining distance.

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  • $\begingroup$ Dear Mike, very interesting paper. A side question: it is stated that a module(bimodule) of the terminal bicategory $\bullet$ is simply a category (page 18 of the above paper, I am taking $\mathcal{V}=Cat$); I thought it is a monad on a category (two maonad structure with a distributive law): even if the bicategory $\bullet$ is trivial, the action may have non-trivial unital/associative constraints. What is wrong with my statement? $\endgroup$
    – Ma Ming
    Apr 21, 2014 at 20:40
  • $\begingroup$ The coherence isomorphism $\sigma$ tells you that the action must be isomorphic to the identity. $\endgroup$
    – Mike Shulman
    Apr 22, 2014 at 20:17
  • $\begingroup$ Do you mean the unit constraint? Well, the action map and the identity can still be differed by an invertable 2-cell. By isomorphic, do you mean that the action is equivalent to the trivial action? If so, I do agree. $\endgroup$
    – Ma Ming
    Apr 22, 2014 at 20:54
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    $\begingroup$ The action map is isomorphic to the identity (by the invertible 2-cell), and yes, that means the module is equivalent to one with the trivial action. $\endgroup$
    – Mike Shulman
    Apr 23, 2014 at 21:10
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    $\begingroup$ Yes, you can re-express our construction of the tensor product as a codescent object by first taking a bunch of coproducts. $\endgroup$
    – Mike Shulman
    May 2, 2014 at 21:36
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Depending on what you actually need for your application, there might be something useful already known. Do you really need a tricategory, as opposed to some other model of weak 3-categories? Even if you had that, would you be able to do much with it?

There are several equivalent ways to think of profunctors, and some of them lend themselves to categorification quite easily. One is this:

A profunctor from $C$ to $D$ is the same a a colimit-preserving ordinary functor between the presheaf categories $PSh(C)$ and $PSh(D)$.

(See here for details: http://ncatlab.org/nlab/show/profunctor#FuncsOnPresheaves)

That's good, because a lot is known about categorifications of categories of presheaves and of morphisms between them.

For instance it is straightforward to set this up over bicategories: take objects bicategories, and hom-bicategories to be the full sub-bicategory on bicolimit-preserving bifunctors between their bipresheaf categories.

In case that your application is such that you only need higher categories whose higher morphisms are all invertible, one can go much further and consider the (oo,1)-category of (oo,1)-profunctors. By the above, this is simply the gadget whose objects are small (oo,1)-categories and whose hom-oo-groupoids are the full sub-oo-groupoids on the (oo,1)-colimit preserving (oo,1)-functors between the corresponding (oo,1)-presheaf categories.

More generally, one can generalize here (oo,1)-presheaf categories and all in addition all their reflective sub-(oo,1)-categories. That structure of (oo,1)-profunctors has proven to play a major role as kind of categorification of the category of vector spaces: one thinks of a presentable (oo,1)-category as vector space, of colimits as being sums of vectors, as colimit preserving (oo,1)-functors as linear maps.

More on this is here: http://ncatlab.org/nlab/show/Pr(infinity,1)Cat .

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    $\begingroup$ Unfortunately, I really do need profunctors in the sense of maps $\mathcal{C} \times \mathcal{D}^{\operatorname{op}} \to \operatorname{Cat}$. As you say, if we could show that "my" profunctors are the same as "your" profunctors (bicolimit-preserving bifunctors between bipresheaf categories), we'd be done. The $(\infty, 1)$-categorical version would be sufficient for my purposes, but I do not believe anybody has yet worked out how to compose "my" profunctors in this setting (although I know that there are people actively thinking about these things). $\endgroup$
    – Evan Jenkins
    Aug 27, 2010 at 22:06
  • $\begingroup$ Do you need "your" profunctors to compose in a "coend" type way? Would it be enough to show that functors $C\times D^{op}\to Cat$ are equivalent to cocontinuous functors $Cat^{C^{op}} \to Cat^{D^{op}}$, and then use composition of the latter functors to induce a composition operation on the former? $\endgroup$
    – Mike Shulman
    Sep 3, 2010 at 7:11
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If you're still interested, I've worked this out on my personal web at nLab here, with supporting material linked to from this page.

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