7
$\begingroup$

In a monoidal category $(C,\otimes)$ one can consider internal monoids and bimodules over these monoids (provided some additional requirements like existence of coequalizers hold). These monoids together with bimodules as 1-morphisms and bimodule morphisms as 2-morphisms should constitute a bicategory $\mathsf{Bimod}$.

Does anyone know a reference for this result? The only thing I could find so far is this page on nLab, but there is no proof given.

$\endgroup$
4
  • 2
    $\begingroup$ This work by Haugseng: arxiv.org/abs/1412.8459 might be relevant for you $\endgroup$
    – Adam Gal
    Aug 6, 2018 at 14:07
  • $\begingroup$ Specifically, Section 2 of Haugseng does what you're looking for in the further generality where $C$ is an $(\infty,1)$-category, using Segal spaces as the formalism for $(\infty,1)$-categories and $(\infty,2)$-categories. Since you just want the discrete setting and can use the easier notion of bicategory, it should be easier. But I don't know of the top of my head somewhere that really does the details. I thought it might be in Müger or Yamagami, but both of them just seem to treat this as obvious. It really should just be a tedious check, is there a specific step that tripped you up? $\endgroup$ Aug 6, 2018 at 15:00
  • $\begingroup$ Thanks for the help so far! I agree that this proof should a "simple" check of a long list of small statements. Nevertheless, when I came to defining the associativity natural isomorphism I really lost the overview, which made me wondering if there is conceptually clearer way to show all of this. And besides that I wanted to know if this has been done, or if it is just folklore (quite convincing, but nevertheles...). $\endgroup$ Aug 6, 2018 at 15:41
  • 1
    $\begingroup$ It probably is not the earlier reference, but all is proved carefully in this work of Joyal and Gambino (chapter 4), if I recall correctly. Also I remembered doing this myself in the context of pseudo double categories once and it was not that long, so it might worth integrating the proofs you want directly into your work if you don't find precise references. $\endgroup$ Aug 11, 2018 at 7:53

1 Answer 1

6
$\begingroup$

This example of a bicategory (in the case of monoids in Ab, i.e. rings) appears already in Benabou's original Introduction to bicategories (example 2.5), although he does not provide the detailed proof you seem to be looking for.

In searching the 2-categorical literature, you may want to look for constructions of the larger bicategory of profunctors (a.k.a. distributors or bimodules) whose objects are enriched categories, the bicategory of monoids and bimodules being its full sub-bicategory on the one-object categories. I believe the earliest references for this include Benabou's Les Distributeurs and Lawvere's Metric spaces, generalized logic, and closed categories, though on a quick glance I do not see a detailed proof in either of them either.

For actually checking axioms of this sort, it is often convenient to exhibit the operations as having a universal property. In the case of bicategories, or more generally pseudo double categories (the bicategory Bimod enlarges to a pseudo double category whose additional morphisms are monoid homomorphisms), such a universal property can be expressed in terms of virtual double categories (a.k.a. fc-multicategories). A construction of the virtual double category of monoids and bimodules can be found in section 5.3 of Leinster's Higher operads, higher categories, though he doesn't give a detailed proof of the conditions under which it is a pseudo double category. A sketch of the latter --- but, again, with details left to the reader --- can be found in Appendix A of Cruttwell-Shulman, A unified framework for generalized multicategories.

$\endgroup$
1
  • $\begingroup$ Unfortunately for the OP, Les Distributeurs is not generally available (although I have a pdf I can send if Bénabou is not forthcoming). $\endgroup$
    – David Roberts
    Aug 6, 2018 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.