A simplicial category is a category enriched over the monoidal category of simplicial sets (morphism sets are now simplicial sets), and the collection of all such categories forms a category itself (modulo set theoretic issues). It is asserted on p. 23 of "Higher topos theory" that this latter category has all small colimits---why? How are such colimits computed?

I can certainly understand coproducts (take a disjoint union of everything in sight), but how does one construct coequalizers? In fact, I think I don't understand the latter even in the case of ordinary categories (to which one may perhaps reduce because a simplicial category is just a simplicial object in the category of categories).

For instance, consider the category $C_0$ that has a single object $*$ with no nonidentity morphisms, and consider the category $C_1$ which has four objects and two nonidentity morphisms $a \rightarrow b$ and $c \rightarrow d$. Consider the two functors $C_0 \rightarrow C_1$ that send $*$ to $b$ and $c$, respectively. What is the resulting colimit category in this case and why?

  • 3
    $\begingroup$ This seems like a question that gets asked a lot. Categories and simplicially enriched categories (they behave similarly here) are presentable categories, so are complete and cocomplete. $\endgroup$ Dec 22, 2016 at 18:23
  • $\begingroup$ This seems circular: part of the definition of presentable (Def. A.1.1.2 in HTT) is the existence of small colimits, so invoking this more general property does not justify the existence of colimits. Concretely, without invoking more abstract nonsense, how does one describe the colimit in the example I give in the last paragraph? $\endgroup$ Dec 22, 2016 at 18:43
  • 1
    $\begingroup$ There's a few equivalent characterizations of (locally) presentable. For instance, as a colimit preserving localization of a presheaf category with respect to a set of maps (which can be constructed explicitly using a small-object type argument, for instance). That's what I had in mind. $\endgroup$ Dec 22, 2016 at 18:46
  • 2
    $\begingroup$ @O-RenIshii One of the main motivations for using quasicategories instead of fibrant simplicial categories as a model for (∞,1)-categories is that computing colimits is much much easier in the former model. Hence, I really doubt there is a nice formula for colimits in simplicial categories. Also: a simplicial category in the sense of Lurie in not a simplicial object in categories (this is one of the reasons I dislike the name). $\endgroup$ Dec 22, 2016 at 19:34
  • 3
    $\begingroup$ Simplicial categories are monadic over simplicial graphs, thus coequalizers can be computed using the construction in Theorem 9.3.9 of Toposes, Triples, and Theories. $\endgroup$ Dec 23, 2016 at 2:07

1 Answer 1


To expand on Dmitri Pavlov's answer, the recipe for colimits, as in any monadic category, will be the following.

  1. Take the colimit of the underlying (simplicial) graphs.
  2. Apply the free functor.
  3. Mod out by all relations that existed in the (simplicial) categories you're taking the colimit of.

Colimits in $\mathsf{Cat}$ are already notoriously bad (depending on the indexing category). For example, the coequalizer of the two inclusions $[0]^\to_\to [1]$ is $\mathbf{B}\mathbb{N}$, the category with one object and an $\mathbb{N}$'s worth of endomorphisms. However, as you point out, coproducts are not so bad. Filtered colimits are likewise computed as on the underlying graphs. There is a model structure on $\mathsf{Cat}$ which gives you a notion of "homotopy colimit", and only those colimits which are actually homotopy colimits should be considered "good". Similarly, the Bergner model structure on simplicial categories can tell you which colimits of simplicial categories are good.

  • 1
    $\begingroup$ The relation of colimits with homotopy colimits is certainly interesting. I'd just like to comment on the words "notoriously bad"! In fact this seems to be a useful description of $\mathbb N$ . In the corresponding groupoid situation, we move from the groupoid $\mathcal I$ with exactly one arrow $0 \to 1$ to the group $\mathbb Z$ of integers, which is a "good" explanation for the fundamental group of the circle, $\endgroup$ Dec 25, 2016 at 10:05

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.