20
$\begingroup$

There are many models for $(\infty,1)$-categories: simplicial categories, Segal categories, complete Segal spaces, and quasi-categories.

Doubtlessly the model most used to do higher category theory in is the model of quasi-categories, due to the work of Lurie (Higher Topos Theory, Kerodon), who calls them $\infty$-categories.

Just browsing through these books, I noticed that however, simplicial categories are used too, for instance to construct examples of quasi-categories, as in the construction of the quasi-category of spaces: there is a simplicial category of Kan complexes, and to get the quasi-category of Kan complexes we take the homotopy coherent nerve (also called simplicial nerve in HTT I think) of that.

Question: If simplicial categories are a more practical model for actually constructing examples, why are quasi-categories used for most of the theory? That is, what advantages do quasi-categories have over simplicial categories that outweigh the complications of constructing quasi-categories directly?

$\endgroup$

2 Answers 2

28
$\begingroup$

As a preface, I think that this question should be viewed as analogous to "what are the advantages of ZFC over type theory" or vice versa. We're talking about foundations -- in principle, it doesn't matter what foundations you use; you end up with an equivalent model-independent theory of $\infty$-categories.

The paradigm is "use simplicial categories for examples; use quasicategories for general theorems". There are a lot of constructions which are simpler in quasicategories. I tend to think that a lot of the difference is visible at the model category level: the Joyal model structure on $sSet$ is just much nicer to work with than the Bergner model structure on $sCat$ (of course, the latter is still theoretically very important, at the very least for the purposes of importing examples which start life as simplicial categories). Some of these differences are:

  1. The Joyal model structure is defined on a presheaf category.

  2. The Joyal model structure is cartesian, making it much easier to talk about functor categories.

  3. In the Joyal model structure, every object is cofibrant.

There's a synergy between (1) and (2) -- if $X$ is a quasicategory and $A$ is any simplicial set, then the mapping simplicial set $Map(A,X)$ gives a correct model for the functor category from $A$ to $X$ -- you don't need to do any kind of cofibrant replacement of $A$. This is nicely explained in Justin Hilburn's answer.

(2) is quite convenient. For example, in general frameworks like Riehl and Verity's $\infty$-cosmoi, a lot of headaches are avoided by assuming something like (2).

Here are a few examples of some things which are easier in quasicategories -- I'd be curious to hear other examples folks might mention!

  • The join functor is very nice.

  • Consequently (in combination with the nice mapping spaces), limits and colimits can be defined pretty cleanly.

An example of a theorem proven in HTT using quasicategories which I imagine would be hard to prove (maybe even to formulate) directly in simplicial categories is the theorem that an $\infty$-category with products and pullbacks has all limits. The proof uses the fact that the nerve of the poset $\omega$ is equivalent to a 1-skeletal (non-fibrant) simplicial set, and relies on knowing how to compute co/limits indexed by non-fibrant simplicial sets like this.

  • The theory of cofinality is very nice, arising from the (left adodyne, left fibration) weak factorization system on the underlying category -- I imagine it would be quite complicated with simplicial categories.

Roughly at this point in the theory, though, one starts to have enough categorical infrastructure available that it becomes more possible to think "model-independently", and the differences start to matter less.

Here's a few more:

  • When you take the maximal sub-$\infty$-groupoid of a quasicategory, it is literally a Kan complex, ready and waiting for you to do simpicial homotopy with. This is especially nice when you take the maximal sub-$\infty$-groupoid of a mapping object -- which doesn't quite make the model structure simplicial, but it's kind of "close".

  • The theory of fibrations is pretty nice in quasicategories -- just like in ordinary categories, left, right, cartesian, and cocartesian fibrations are "slightly-too-strict" notions which are very useful and have nice properties like literally being stable under pullback. I don't know what the theory of such fibrations looks like in simplicial categories.

  • The fact that $sSet$ is locally cartesian closed is sneakily useful. Even though $Cat_\infty$ is not locally cartesian closed, there's a pretty good supply of exponentiable functors, and it's not uncommon to define various quasicategories using the right adjoint to pullback of simplicial sets. (Rule of thumb: in HTT, when Lurie starts describing a simplicial set by describing its maps in from simplices over a base, 90% of the time he's secretly describing the local internal hom of simplicial sets.)

$\endgroup$
1
  • 1
    $\begingroup$ +1 - the fact that it's a presheaf category also sometimes makes (strict) colimit computations easy. Like, a pushout in sSet is so much easier than in sCat. $\endgroup$ Jan 20 at 8:33
6
$\begingroup$

In order to compute the correct mapping space $\text{Hom}(x,y)$ in a model category you have to replace $x$ with a cofibrant object and $y$ with a fibrant object.

In the Joyal model structure on simplicial sets everything is cofibrant and quasi-categories are fibrant. In the Bergner model structure on simplicial categories most naturally occurring objects are not fibrant or cofibrant.

$\endgroup$
2
  • 2
    $\begingroup$ I think there is a good supply of naturally occurring fibrant simplicial categories, i.e. Kan-enriched categories. The Kan-enriched category of Kan complexes, for one! $\endgroup$
    – Zhen Lin
    Jan 18 at 14:41
  • $\begingroup$ Point taken. I guess it is always cofibrant replacement that has caused me the most problems. $\endgroup$ Jan 18 at 15:17

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.