4
$\begingroup$

Not related to this old question of mine, but takes the question from a different perspective.

Let $\mathcal V$ be a monoidal model category (following the def of Hovey, for example). Then there is a bicategory $\text{Prof}(\mathcal V)$ of $\cal V$-valued profunctors, which has the following interesting property:

every hom-category has a model structure[1].

This is, I think, the paradigmatic example of a "locally model bicategory". There can be others: I'm interested in examples for this notion and related results.

I'm trying to get a list of properties to impose to the general notion (a "locally model 2-category", i.e. a 2-category "enriched" over model categories -this is not a true definition, as there is no sensible monoidal structure on model categories-).

Let's for example consider the following explicit question:

Let $\varphi\colon {\bf A}\looparrowright{\bf B}$ be a profunctor, and $\bf X$ be a category; then precomposition by $\varphi$ gives $$ \text{Prof}({\bf B},{\bf X}) \overset{-\diamond \varphi}\to \text{Prof}({\bf A},{\bf X}) $$ which has left and right adjoints $\text{Lan}_\varphi$, $\text{Ran}_\varphi$ defined by the co/ends $$ \text{Ran}_\varphi\psi(b,x) \cong \int_a \hom(\varphi(a,b), \psi(a,x)) $$ (Lan is similar). Does $-\diamond \varphi \dashv \text{Ran}_\varphi$ form a Quillen adjunction?

===

[1] in fact, many! Let's take $\mathcal V = \bf sSet$ and declare that I want to study the injective model structure on $\text{Prof}(\mathbf{sSet})(\mathbf A,\mathbf B)=[\mathbf A^\text{op}\times \mathbf B,\mathbf{sSet}]$.

$\endgroup$
5
  • $\begingroup$ I don't think that precomposition by a profunctor has a left adjoint in general; it is the left adjoint. Precomposition by a functor has both a left and a right adjoint, but that string of three adjoints arises from the single adjunctions associated to the two representable profunctors. $\endgroup$ Oct 1, 2016 at 20:16
  • $\begingroup$ You're right indeed. It is nevertheless possible to define $\rm Ran$ and $\rm Rift$, right? $\endgroup$
    – fosco
    Oct 1, 2016 at 21:32
  • $\begingroup$ Yes. That is, the bicategory of profunctors is closed. $\endgroup$ Oct 2, 2016 at 20:50
  • 1
    $\begingroup$ Also, it's not clear to me that the hom-categories of $\mathrm{Prof}(\mathcal{V})$ have model structures absent any further assumptions on $\mathcal{V}$. If $\mathcal{V}$ is cofibrantly generated, then you can try to define a projective model structure at least, but even then I don't know how to show that it actually works unless the hom-objects of your $\mathcal{V}$-categories are cofibrant, or if $\mathcal{V}$ satisfies the "monoid axiom". $\endgroup$ Oct 2, 2016 at 21:03
  • $\begingroup$ I'm not claiming I don't need further assumptions on $\cal V$! $\endgroup$
    – fosco
    Oct 2, 2016 at 22:03

1 Answer 1

3
$\begingroup$

Given two model categories $\mathcal{M},\mathcal{N}$, one does know what would have been a left Quillen functor out of what would have been the tensor product $\mathcal{M} \otimes \mathcal{N}$ into a third model category $\mathcal{K}$, and that is a left Quillen bifunctor $\mathcal{M} \times \mathcal{N} \to \mathcal{K}.$ This leads to a natural notion of a category (weakly) enriched in model categories, namely, a bicategory $\mathcal{C}$ such that each mapping category $\mathrm{Map}_{\mathcal{C}}(X,Y)$ carries a model structure and such that each composition operation $$ \mathrm{Map}_{\mathcal{C}}(X,Y) \times \mathrm{Map}_{\mathcal{C}}(Y,Z) \to \mathrm{Map}_{\mathcal{C}}(X,Z) $$ is a left Quillen bifunctor. In the case of profunctors this can be achieved, for example, if one endows $\mathrm{Fun}(\mathbf{A}^{\mathrm{op}} \times \mathbf{B},\mathcal{V})$ with the model structure which is injective in the $\mathbf{A}^{\mathrm{op}}$ coordinate and projective in the $\mathbf{B}$ coordinate (call it the injective-projective model structure). Note that if one defines enrichment in model categories in this way then pre-composition and post-composition with a cofibrant morphism is a left Quillen functor. In particular, pre-composition or post-composition with a cofibrant profunctor $\varphi:\mathbf{A} \looparrowright \mathbf{B}$ (with respect to the injective-projective model structure) induces a left Quillen functor $\text{Prof}({\bf B},{\bf X}) \overset{-\diamond \varphi}\to \text{Prof}({\bf A},{\bf X})$ (both equipped with the injective-projective model structure).

$\endgroup$
8
  • $\begingroup$ Why co-ending a cofibration in $\cal V$ should give a cofibration? $\endgroup$
    – fosco
    Oct 1, 2016 at 16:41
  • $\begingroup$ Ah! Now I see, maybe. It's a nontrivial statement but it's true if the categories are combinatorial. $\endgroup$
    – fosco
    Oct 1, 2016 at 16:44
  • $\begingroup$ Yes, you should assume $\mathcal{V}$ to be combinatorial. The main point is that the coend operation $Fun({\bf B},\mathcal{V}) \times Fun({\bf B^{op}},\mathcal{V}) \to \mathcal{V}$ is a left Quillen bifunctor (see Remark A.2.9.27 in HTT), where the first factor is endowed with the projective model structure and the second factor with the injective model structure. -> $\endgroup$ Oct 1, 2016 at 19:30
  • $\begingroup$ -> One can then deduce that the composition operation $Fun({\bf A^{op}} \times {\bf B},\mathcal{V}) \times Fun({\bf B^{op}} \times {\bf C} ,\mathcal{V}) \to Fun({\bf A^{op}} \times {\bf C}, \mathcal{V})$ is a left Quillen bifunctor as well (where both side carry the injective-projective model structure). $\endgroup$ Oct 1, 2016 at 19:33
  • 1
    $\begingroup$ This definition (plus the corresponding condition on the units) appears in section 11 of tac.mta.ca/tac/volumes/31/23/31-23abs.html . I wouldn't claim any originality for it -- as this question and answer shows, it's a fairly obvious thing to write down and has surely occured to plenty of people -- but we didn't know any published citation for it. $\endgroup$ Oct 2, 2016 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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