Suppose we are given a category enriched over semi-simplicial sets, i.e. for the simplices in this category we have well-defined boundary maps, but no degeneracy maps. Suppose also that in this category all inner Kan conditions are satisfied, i.e. we have $\mathrm{Kan}(n,k)$ for all $n$ and $1\le k \le n-1$, but not necessarily $\mathrm{Kan}(n,0)$ and $\mathrm{Kan}(n,n)$.

Is there any "canonical" way to make this category become enriched over simplicial sets, i.e. to find "canonical up to equivalence" degeneracy maps, so that the category will become an $(\infty,1)$-category?

There are several papers on making a Kan-simplicial set out of Kan-semi-simplicial one, including the classical paper by Rouke and Sanderson and a more recent by McClure, but I haven't found anything concerning quasi-categories.


The short answer is no. If you take the semi-simplicial set $X$ with one $0$-simplex and no higher simplices, then it has fillers for all inner horns (because it has no inner horns) but clearly does not admit degeneracies.

The question then becomes: what in addition to inner horn fillers could we assume to make this true (without restricting attention to Kan complexes)? One possible guess would be: add lifts along the inclusion of $X$ into $E[1]$, i.e. (the underlying semi-simplicial set of) the nerve of the groupoid freely generated by one isomorphism. I don't know whether this is sufficient, but this is a notable lifting property that follows from fillers for inner horns in the simplicial setting, but doesn't in the semi-simplicial one.


Kolya! Thanks an $\infty$ many times for your question, and for pointing (me) to McClure's paper! :)

You might be interested to take a look at this paper (of mine), where a definition of the notion of a "cat" is presented: the globular (see, e.g., John Baez's An Introduction to n-Categories) multi-simplicial sets satisfying the (multi-simplicial version of the) Kan condition presented by McClure in Definition 5.2 of his paper ( - I think "higher Kan complexes" are a good name for them - ) are special kinds of cats, just like Kan complexes are special kinds of $\infty$-categories (in the sense of Boardman-Vogt, Joyal, Lurie, ...).

I did not read McClure's proof yet, but I'm "pretty confident" that what you're asking for should be possible. Also, an analogue of McClure's Theorem 5.3 should be true: it should be possible to canonically build a cat from any given "semi-cat", even if the "horizontal composition" in the semi-cat isn't strict. But as I said, I did not prove this, it's just a guess.

Another question that one could come up with is whether the analogue of Moore's proposition ( - whose formulation should be obvious - ) in this "higher" situation holds.

For several months now I tried to find out whether this definition of a "cat" has already been studied, and if so, why it can't be found in the literature ( - at least I did not find it). If anybody knows of further papers in this direction, for example answering my previous question, please drop a reference in the comments section below ( - if that is allowed).

  • $\begingroup$ @Kolya Are you in Bonn, too? $\endgroup$ – Daniel Gerigk May 10 '14 at 2:03
  • 1
    $\begingroup$ I wrote (somewhat sloppy) that "what you're asking for should be possible". Perhaps this is misleading. As Karol pointed out, it is not possible to obtain a simplicial set (which then is an $\infty$-category) from a semi-simplicial set satisfying the inner horn conditions just by adding degeneracy maps. So the immediate analogue of McClure's Theorem 5.3 does not hold, of course. What I had in mind was "building" a cat, i.e. not only adjoining degeneracy maps, but also degenerate simplices. $\endgroup$ – Daniel Gerigk May 10 '14 at 16:13
  • $\begingroup$ Thanks for the answer and the comment. No, I wrote my PhD in Bonn, presently I'm in Goettingen, just haven't updated my profile. $\endgroup$ – Kolya Ivankov May 11 '14 at 8:05

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.