11
$\begingroup$

In Higher topos theory and Higher algebra Lurie defines (see section 2.3.1 of HTT) a correspondence between two $\infty$-categories $C$ and $D$ as being an $\infty$-category $\mathcal{M}$ over $\Delta[1]$ whose fiber over $0$ and $1$ are respectively $C$ and $D$. Of course this is supposed to be equivalent to presheaves over $C^{op} \times D$, or to left adjoint functor $\widehat{D} \rightarrow \widehat{C}$. This is vaguely claimed as remark 2.3.1.4 in Higher topos theory, but I do not know if the details of this correspondence have been developped somewhere...

Is there a reference for this equivalence ? It is possible to deduce it without too much work from other results in the literature ?

I need this results for a technical lemma, and so far I don't see how to prove without some hard work on simplicial sets taking several pages... but I'm hopping I'm missing a nicer argument.

Remark: it is easy using Lurie's work to attach to $\mathcal{M} \rightarrow \Delta[1]$ a 'profunctor' $C^{op} \times D \rightarrow$ Space, In fact using the twisted arrow category construction (see section 5.2.1 of Higher algebra) one can even give a very explicit right fibration $T \rightarrow C^{op} \times D$ representing this profunctor, but going the other way seem harder...

$\endgroup$
3
  • 1
    $\begingroup$ In Example 2.4 of Barwick and Shah it's mentioned that at the time that note was written, a proof of this fact didn't seem to exist in the literature. I seem to vaguely recall that a proof was supposed to be forthcoming in work of Peter Haine. $\endgroup$ Commented Dec 22, 2019 at 18:15
  • 2
    $\begingroup$ This paper seems to do what you want: arxiv.org/abs/1807.08226 $\endgroup$ Commented Dec 22, 2019 at 23:30
  • 1
    $\begingroup$ If you just want to "go the other way" can't one start with $\hat{D}\to\hat{C}$, classified by a cartesian fibration over $[1]$ and then restrict to the full subcategory of objects corresponding to representable presheaves? Of course, one has to work harder to see that these two constructions constitute an equivalence of categories $\endgroup$ Commented Jan 7, 2020 at 20:43

1 Answer 1

12
$\begingroup$

Your question is answered by the following result, for which I will give a few references.


Theorem. For each pair of simplicial sets $A$ and $B$, there is a functor $$a_{A,B}^* \colon \mathbf{Cyl}(A,B) \longrightarrow \mathbf{sSet}/(A^\mathrm{op} \times B)$$ which is both a left and a right Quillen equivalence between the Joyal model structure on $\mathbf{Cyl}(A,B)$ and the covariant model structure on $\mathbf{sSet}/(A^\mathrm{op} \times B)$.


Definitions. (1) The category $\mathbf{Cyl}(A,B)$ of cylinders from $A$ to $B$ (this is Joyal's terminology) is the fibre over the pair $(A,B)$ of the functor $$\mathbf{sSet}/\Delta[1] \longrightarrow \mathbf{sSet} \times \mathbf{sSet}$$ that sends a map $X \to \Delta[1]$ to its fibres over $0$ and $1$. One can show that the category $\mathbf{Cyl}(A,B)$ admits a model structure (called the "Joyal model structure" above) created from Joyal's model structure for quasi-categories by the forgetful functor $$\mathbf{Cyl}(A,B) \longrightarrow \mathbf{sSet}/\Delta[1] \longrightarrow \mathbf{sSet}.$$ If $A$ and $B$ are quasi-categories, then the fibrant objects of this model structure are precisely the correspondences from $A$ to $B$ as you have defined them.

(2) The functor $a_{A,B}^*$ sends a cylinder $X \in \mathbf{Cyl}(A,B)$ to the pullback $\require{AMScd}$ \begin{CD} a_{A,B}^*(X) @>>> \mathrm{Tw}(X)\\ @V V V @VV V\\ A^\mathrm{op} \times B @>>i^\mathrm{op}\times j> X^\mathrm{op} \times X \end{CD} where $\mathrm{Tw}(X)$ denotes the twisted arrow of $X$ (whose $n$-simplices are maps $\Delta[n]^\mathrm{op} \star \Delta[n] \to X$), and where $i \colon A \to X$ and $j \colon B \to X$ denote the inclusions of the fibres of the structure map $X \to \Delta[1]$.


Proofs. In the case where $A$ and $B$ are quasi-categories, the theorem above is proved by Danny Stevenson in his recent preprint (see Theorem C therein)

Danny Stevenson. Model structures for correspondences and bifibrations. arXiv:1807.08226.

(Note that Stevenson uses the term "correspondence" for what I have called "cylinder".)

As I shall explain below, the general case of the theorem can be proved by a combination of results from my recent preprint

Alexander Campbell. Joyal's cylinder conjecture. arXiv:1911.02631.

and from Cisinski's recent book

Denis-Charles Cisinski. Higher categories and homotopical algebra, vol. 180 of Cambridge studies in advanced mathematics. Cambridge University Press, 2019.

First, note that the category $\mathbf{Cyl}(A,B)$ is equivalent to the category $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ of bisimplicial sets over the exterior product of $A^\mathrm{op}$ and $B$. It follows from Remark 3.23 and Theorem 5.4 of my preprint that, under this equivalence, the Joyal model structure on $\mathbf{Cyl}(A,B)$ corresponds to the bicovariant model structure on $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ defined in Section 5.5 of Cisinski's book.

Now, it is not difficult to see that, under this equivalence, the functor $a_{A,B}^*$ defined above corresponds to the functor $$\delta_{A^\mathrm{op},B}^* \colon \mathbf{ssSet}/(A^\mathrm{op} \boxtimes B) \longrightarrow \mathbf{sSet}/(A^\mathrm{op} \times B)$$ which sends a bisimplicial set over $A^\mathrm{op} \boxtimes B$ to its diagonal (note that the diagonal of $A^\mathrm{op} \boxtimes B$ is $A^\mathrm{op} \times B$).

Hence the theorem follows from Theorem 5.5.24 in Cisinski's book, where he proves that the functor $\delta_{A^\mathrm{op},B}^*$ is both a left and a right Quillen equivalence between the bicovariant model structure on $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ and the covariant model structure on $\mathbf{sSet}/(A^\mathrm{op} \times B)$.

$\endgroup$
1
  • 2
    $\begingroup$ Another reference is the paper "Fibrations of $\infty$-categories" (arXiv:1702.02681) by Ayala and Francis, section 4.1. $\endgroup$ Commented Dec 23, 2019 at 13:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .