17
$\begingroup$

I'm interested in precisely what information an infinity category encodes. For example, consider the infinity category of spaces. I like to think about this as the homotopy category of spaces equipped with some extra structure, coming from the simplicial set model, which allows us to compute homotopy limits and so on. My (rather vague) question is:

  1. How unique is this structure? In other words, given a homotopy category, is there a canonical associated infinity category?
  2. Relatedly, do infinity categories encode any information that is non-homotopical (by this, I mean information that cannot be obtained from the homotopy category) in nature? If so, what?
$\endgroup$
3
  • $\begingroup$ There are uncharitable interpretations of your questions where the answer is (1) the homotopy category does not determine the $(\infty, 1)$-category and (2) the $(\infty, 1)$-category contains information that cannot be obtained from the homotopy category. But this is because the homotopy category is a very impoverished structure. $\endgroup$
    – Zhen Lin
    Aug 18, 2020 at 13:08
  • $\begingroup$ It may be necessary to be a bit more explicit about what extra structure you consider the homotopy category to possess. For example, if $M$ is a model category, and we write $H(-)$ for the ordinary category that is usually called the homotopy category, then $H(M)$ don't have enough information to compute homotopy co/limits. On the other hand, if we keep track of $H(\mathrm{Fun}(I,M))$ for all small categories $I$, we retain enough information to compute homotopy co/limits. (I'm simplifying matters, of course.) $\endgroup$
    – Brian Shin
    Aug 18, 2020 at 13:15
  • $\begingroup$ Well it of course depends on what you call extra structure. An $\infty$-category $C$ is the data of its homotopy category $Ho(C)$ together with the localization functor $C\to Ho(C)$. That is, in a sense, extra structure. If you define "structure" by something like faihfulness, or monadicity, then these are precise statements to which an answer can be given (probably a negative one), but you have to clarify what you mean $\endgroup$ Aug 18, 2020 at 17:19

1 Answer 1

24
$\begingroup$

I think it's useful to consider a much lower-dimensional analogue of your question, which is (at least for me) much easier to reason about intuitively, but still gets some of the message across.

Let's compare $0$-categories (i.e., sets) and $1$-categories (i.e., categories) based on what they're able to encode.

  • a $0$-category is just a class of objects. Two objects of a $(0,1)$-category are equivalent precisely if they are equal (this is the $0$-categorical truncation of equivalence), and nothing more can really be said about the objects.
  • a $1$-category is a $0$-category (weakly) enriched in $(0,0)$-categories (i.e., sets), which allows us to be more delicate about how one object relates to another; in particular, morphisms allow us to describe the structure of the objects, and $1$-categorical language thus addresses properties of objects regarding their structure. More precisely, two objects of a $1$-category are equivalent precisely if they are isomorphic (i.e., they have the same structure), and $1$-categorical constructions (such as co/limits) are defined up to isomorphism.

Given a $1$-category $\def\cC{\mathcal C}$ $\cC$, we can define its homotopy $0$-category $\def\Ho{\operatorname{Ho}}$ $\Ho\cC$ as the $0$-category whose objects are isomorphism classes of objects of $\cC$. This serves as an effective presentation of $\cC$ with a $0$-category in the sense that objects of $\cC$ are isomorphic precisely if the corresponding objects in $\Ho\cC$ are equal.

However, we can also see that this is difficult to reverse-engineer, even canonically, as several non-equivalent $1$-categories can have the same homotopy $0$-category. The quickest way to see this is to note that a $0$-category $X$ can be thought of as a $1$-category with only identity morphisms, and in this case $\Ho X=X$; in particular, given any $1$-category $\cC$, its homotopy $0$-category $\Ho\cC$ is also a presentation of the $0$-category $X := \Ho\cC$ viewed as a $1$-category. Which of $\cC$ and $X$ would be a more suitable choice of a "canonical $1$-category" associated to $\Ho\cC$?

Moreover, as the comments mention, it's nearly impossible to perform $1$-categorical constructions in the homotopy $0$-category: the only diagrams $F:J\to\Ho\cC$ that have limits are constant diagrams. In fact, even if we were computing the limit of a functor $F:J\to\cC$ where all the objects in the diagram were isomorphic to each other (that is, the induced map $F:\operatorname{Ob}J\to\Ho\cC$ is a constant map) so that the limit in the homotopy $0$-category exists, the limit in $\Ho\cC$ need not be related at all to the limit in $\cC$. For example, the Cartesian product $X\times X$ is generally not isomorphic to $X$, but the limit in the corresponding map $\{\bullet\,\,\,\bullet\}\to\Ho\cC$ (which is a constant map) will always be the isomorphism class of $X$.


The story is similar for $(\infty,1)$-categories. As these can be thought of as categories weakly enriched in spaces (or $\infty$-groupoids), we can be even more delicate about how we compare objects. Just as categories concern themselves over the structure of the objects, $(\infty,1)$-categories are concerned with homotopy coherent structure of objects. For instance:

  • consider the topological spaces $\Bbb R$, $(0,1)$, and $\{0\}$. If we look at them $0$-categorically (in the $0$-category $\mathbf{Top}_0$ of topological spaces), then they are all completely different, as they consist of different elements. If we look at them $1$-categorically (in the $1$-category $\mathbf{Top}$ of topological spaces and continuous maps), then $\Bbb R$ and $(0,1)$ are the same because they have the same topological structure, but they are different from $\{0\}$ because they can't be put in bijection. Finally, if we look at them $(\infty,1)$-categorically, then all three objects are the same, as they can be contracted to a point.
  • similarly, consider the categories $\mathbf{FinSet}$ of finite sets and its full subcategory $\mathbf{FinOrd}$ on finite ordinals. They are non-isomorphic as categories because the former has a proper class of objects while the latter has a set and thus can't be put in bijection; however, they are equivalent as categories because we can contract the objects of $\mathbf{FinSet}$ together by bijections together (by their cardinalities) and find that $\mathbf{FinOrd}$ is the skeleton of $\mathbf{FinSet}$

We can certainly associate to an $(\infty,1)$-category $\def\sC{\mathscr C}$ $\sC$ a homotopy category $\Ho\sC$, where objects of $\Ho\sC$ are isomorphic precisely if they are equivalent in $\sC$, but we see the same problem when trying to reverse-engineer this. Just as before, a category $\cC$ can be thought of as an $(\infty,1)$-category where all higher cells are trivial, and in this case $\Ho\cC=\cC$, so given an $(\infty,1)$-category $\sC$, its homotopy category is also a presentation of the category $\cC := \Ho\sC$ viewed as an $(\infty,1)$-category.

Moreover, computing limits in $\Ho\sC$ will not say anything about how to compute limits in $\sC$. For example, consider the $(2,1)$-category $\mathbf{Cat}$ of (small) categories, functors, and natural isomorphisms, viewed as an $(\infty,1)$-category. Then, its homotopy category $\Ho\mathbf{Cat}$ actually fails to have pullbacks, which is shown here. The distinction between homotopy limits in general and limits in the corresponding homotopy category is also stressed here, where they emphasise that even if the limit in $\Ho\sC$ exists, it need not correspond to the limit in $\sC$.


In certain cases, you can present an $(\infty,1)$-category with a $1$-category equipped with extra structure so that you can work with $1$-categorical language to discuss the structure of the $(\infty,1)$-category it presents, and you may even be able to recover the $(\infty,1)$-category canonically. For example, if $\sC$ is a locally presentable $(\infty,1)$-category, then you can present it with a combinatorial simplicial model category $\cC$. Then, limits in $\sC$ correspond to homotopy limits in $\cC$, and they even have the same homotopy categories. Moreover, you can recover $\sC$ by (for example) taking the homotopy coherent nerve of the simplicially enriched subcategory of $\cC$ on the cofibrant fibrant objects, so in this sense there is a canonical way of going backwards as well.

$\endgroup$
1
  • $\begingroup$ This is a really great explanation - that clears most of what I was wondering about. Thank you! $\endgroup$ Aug 19, 2020 at 15:01

You must log in to answer this question.

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