A category is a skeleton if, roughly speaking, no two distinct objects within the category are isomorphic. To every category is associated a skeleton, and two categories are categorically "equivalent" if and only if their skeletons are isomorphic. A fuller definition can be found here.

Consider the subcategory of $\bf{Cat}$ which takes as objects those categories which are skeletons, and morphisms the functors between them; I will call this $\bf{{Cat}_{Skel}}$. Note that this is not the skeleton of $\bf{Cat}$ itself, but a subcategory of $\bf{Cat}$ in which the objects are skeletal categories.

Within this subcategory of skeletal categories, there are a number of objects which are, themselves, isomorphic. So we can take the skeleton of this category, hence obtaining a new category, which I will call $\bf{Skel({Cat}_{Skel})}$. (I don't care which skeleton you take; pick one.)

This category is noteworthy in that it contains one object for each equivalence class of categories in $\textbf{Cat}$, making it perhaps more useful than looking at $\bf{Skel({Cat})}$ itself, which only contains one object for each isomorphism class of categories. My questions are:

  1. Does $\bf{Skel({Cat}_{Skel})}$ have a name?
  2. Has this category been studied in any detail, and if so, can someone please reference me towards any research that's been done on its structure?
  3. Is there an essentially equivalent construction which might be defined more simply than the way I've laid it out here?
  4. Are there any useful areas of study in which this category naturally arises?

Lastly, I've glossed over the usual foundational issues which arise when considering $\bf{Cat}$, mostly because I don't care whether you use Grothendieck universes, or a class-set theory, or only look at small categories, or some other way of solving the problem. Feel free to use any foundational approach that you want which makes $\bf{Skel({Cat}_{Skel})}$ to be consistent.

  • 6
    $\begingroup$ +1: this is a nicely worded and interesting question $\endgroup$ Feb 9, 2013 at 0:24
  • 6
    $\begingroup$ This is the same as the homotopy category associated to the canonical model structure on Cat. See, for example, sbseminar.wordpress.com/2012/11/16/… $\endgroup$ Feb 9, 2013 at 0:24
  • $\begingroup$ Mike, can you explain your description of Skel(Cat_Skel)? Why does it contain equivalence classes of categories? $\endgroup$ Feb 9, 2013 at 0:59
  • $\begingroup$ A skeletal category has "morally" as many non-trivial isomorphisms as its original category --- we are removing only isomorphic copies of objects; isomorphisms on objects have to be untouched. I am by no means an expert in higher-dimensional categories, but I would call your construction a 2-skeleton of a 2-category, where by a "2-skeleton" I mean a 2-category whose equivalent objects are equal. $\endgroup$ Feb 9, 2013 at 1:04
  • 4
    $\begingroup$ I like this question. I would bicker, however, with the first sentence. In general, a skeleton may contain lots of nontrivial isomorphisms. They just all must be automorphisms. A category with no nontrivial isomoprhisms is "gaunt" ncatlab.org/nlab/show/gaunt+category. $\endgroup$ Feb 9, 2013 at 4:09

1 Answer 1


I'm not entirely sure what you're looking for in an answer, but maybe I'll flesh out my comment.

It looks like what you're describing is equivalent to the homotopy category associated to the model structure on Cat where the weak equivalences are equivalences of categories. (I can say "the" because there is only one such, as pointed out in the comments. The cofibrations are functors injective on objects, and the fibrations are "isofibrations".)

I would say that in this context your category has been much studied. In particular, it is interesting to ask questions about homotopy limits and colimits in this category because many useful constructions arise in this way. (Homotopy (co)limits with this model structure are the same as "2-(co)limits" which is the name appearing in most of the literature, especially older literature.)

An example application of this language is the following theorem: The subcategory of presentable (resp. accessible) categories is closed under homotopy limits.

Using this one can prove that most of your favorite things are presentable (resp. accessible). For example, the category of modules over a monad arises via a homotopy limit construction, and this takes care of most things of interest.

Here's a neat application of this (which is the ordinary category version of a result that can be found, for example, in Lurie's HTT,

Say you want to localize a category $\mathcal{C}$ with respect to some collection of morphisms, $S$. Usually $S$ will not be given as a set, but if $\mathcal{C}$ is presentable you're usually okay if $S$ is generated by a set. Well, it turns out that if $F: \mathcal{C} \rightarrow \mathcal{D}$ is a colimit preserving functor between presentable categories, and $S$ is a (strongly saturated) collection of morphisms in $\mathcal{D}$ that is generated by a set, then $f^{-1}S$ is a (strongly saturated) collection of morphisms generated by a set. The argument goes by way of showing that the subcategory of the category of morphisms generated by $f^{-1}S$ is presentable, using a homotopy pullback square.

Adapting this to the model category or $\infty$-category setting, one sees immediately that localizing with respect to homology theories is totally okay and follows formally from this type of argument. (Basically, after fiddling around with cells to prove the category of spectra is presentable, you don't have to fiddle any more to get localizations. This is in contrast to the usual argument found in Bousfield's paper. You've moved the cardinality bookkeeping into a general argument about homotopy limits of presentable categories.)

Anyway, apologies for the very idiosyncratic application of this language; these things have been on my mind recently. I'm sure there are much more elementary reasons why one would care about using the model category structure on Cat.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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