4
$\begingroup$

Suppose I have a functor of quasi-categories $f: \mathcal{C} \to \mathcal{D}$. I want to show a criterion like: "$f$ is an equivalence of $\infty$-categories if the homotopy fiber of $f$ satisfies something". In my case, I have something close to being the identity, yet I don't know the best approach to show it is an equivalence if I don't want to mess up too much with combinatorial stuff.

If we were in the Quillen model structure, it would be almost straightforward: up to substituting $f$ with a Kan fibration, we would have a fiber sequence $|\textrm{hofib}(f) |\to |\mathcal{C}| \to |\mathcal{D}|$ and the criterion would be "if $\textrm{hofib}(f)$ is weakly contractible, then $f$ is a weak equivalence". Does this generalize to the Joyal model structure? I have the impression I am missing some model-categorical proof since the statement is general.

PS Since I believe my $f$ is a Joyal fibration, we can ignore the eventual difference between homotopy fibers in the two contexts (I am starting to doubt the two model categories both underlie the $\infty$-category of simplicial sets).

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Have you considered using Quillen's Theorem A, in Joyal's version for quasicategories? (For example, see Theorem 4.1.3.1 in Lurie's Higher Topos Theory.) $\endgroup$
    – Dmitri Pavlov
    Commented Aug 8, 2023 at 16:46
  • $\begingroup$ Not sure why a cofinal map should be a weak categorical equivalence. In HTT 4.1.1.3 it is proved that a cofinal map is also a weak homotopy equivalence, but not an equivalence of $\infty$-categories. What am I missing? However, that's exactly the kind of theorem I am looking for $\endgroup$
    – Andrea Marino
    Commented Aug 9, 2023 at 9:09
  • 1
    $\begingroup$ The necessary and sufficient criterion for categorical equivalences is similar: the map $f$ should be weakly equivalent to a cartesian (or cocartesian) fibration $g$ (i.e., $f$ is an ∞-categorical Street fibration), and the fibers of $g$ should be contractible. Of course, if $f$ is already a (co)cartesian fibration, we can simply take the fibers of $f$. $\endgroup$
    – Dmitri Pavlov
    Commented Aug 10, 2023 at 1:51
  • $\begingroup$ Would you mind posting this as an answer, maybe with a reference to the result you cited? Thanks!! $\endgroup$
    – Andrea Marino
    Commented Aug 10, 2023 at 16:46

1 Answer 1

Reset to default
3
$\begingroup$

An obvious necessary condition for $f$ to be a categorical equivalence is that $f$ is weakly equivalent to a (co)cartesian fibration of quasicategories, i.e., $f$ is an analogue of a Street fibration of quasicategories.

Since $f$ is already a Joyal fibration, a plausible course of action is to check right away whether $f$ is a (co)cartesian fibration, i.e., has (co)cartesian lifts for morphisms.

If the above necessary condition holds, then $f$ is a categorical equivalence if and only if its fibers (which are automatically homotopy fibers because $f$ is a Joyal fibration) over every vertex of $\cal D$ are contractible Kan complexes, e.g., by Lurie's straightening-unstraightening theorem (Higher Topos Theory, Theorem 3.2.0.1, combined with Proposition 2.4.2.4).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you, that's exactly what I needed. If you don't mind, I could come back in this comment section in the near future to ask for eventual technical difficulties. Best!! $\endgroup$
    – Andrea Marino
    Commented Aug 11, 2023 at 11:12
  • $\begingroup$ Would you please have a short remark on why categorical equivalence should be weakly equivalent to a (co)cartesian fiberation? $\endgroup$
    – Yang
    Commented Oct 22 at 13:37
  • 1
    $\begingroup$ @Yang: A categorical equivalence f:X→Y is weakly equivalent to the identity map id_Y:Y→Y via the commutative square with horizontal maps X→Y and Y→Y. An identity map is a (co)cartesian fibration because it satisfies the lifting property (tautologically). $\endgroup$
    – Dmitri Pavlov
    Commented Oct 22 at 15:13

You must log in to answer this question.

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