If $X$ is a Kan complex, then it is an easy consequence of the existence of the Quillen model structure or of the basic theory of anodyne extensions, that $X^{K}$ is also Kan. However, I am interested in a "direct" proof of this fact, appealing to neither of the above (which is apparently how these sorts of things were proved before Gabriel-Zisman).

Here is the argument I would like to complete:

Let $X$ be a Kan complex, and let $K$ be any simplicial set. It suffices to show that $Hom(K\times \Delta^{n},X) \rightarrow Hom(K \times \Lambda_{k}^{n},X)$ is surjective for all horn inclusions. Since $K\simeq \text{colim}_{m} sk_{m}(K)$ is a filtered colimit, the claim follows if $$Hom(sk_{m}(K)\times \Delta^{n},X) \rightarrow Hom(sk_{m}(K) \times \Lambda_{k}^{n},X)$$ is surjective for all $m \ge 0$. Proceed by induction. The base case is easy, but the induction step is what's giving me trouble. By the induction hypothesis, we can choose a solution

$$ h: sk_{m-1}K \times \Delta^{n} \rightarrow X$$

to the lifting problem

$\require{AMScd}$ $\begin{CD} sk_{m-1}K \times \Lambda^{n}_{k} @>>> sk_{m}K \times \Lambda^{n}_{k} @>f>> X \\ @VVV @VVV \\ sk_{m-1}K \times \Delta^{n} @>>> sk_{m}K \times \Delta^{n} \\ \end{CD}$

However, as the square above is not a pushout, we can't immediately get a solution to the lifting problem of interest. Now my thought is that maybe there's some way to directly construct the desired lift $g:sk_{m}K \times \Delta^{n} \rightarrow X$ from $f$ and $g$, since all we need to do is specify the images of the nondegenerate simplices of $sk_{m}K \times \Delta^{n}$ which are in neither $sk_{m-1}K \times \Delta^{n}$ nor $sk_{m}K \times \Lambda_{k}^{n}$. However, this is where I run into a wall, as I don't know a simple way to describe these.

Ideally, I would like to see this induction step completed. Does anyone have a solution to this or suggestions for possible solutions? It would be just as well if someone knows a different but equally "direct" argument, not appealing to the machinery mentioned above.

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    $\begingroup$ I would try this - $K$ is a $\varinjlim$ of a diagram of simplices, so the problem should be reducible to lifting along $\Delta^m\times\Lambda^n_k\hookrightarrow\Delta^m\times\Delta^n$; the latter are in turn explicit gluings of simplices, so this lifting problem should be reducible to gradual applications of the Kan property of $X$. Equivalently, one might try to show that $X^{\Delta^m}$ is a Kan complex by explicitly describing $n$-simplices of it as certain explicit compatible families of $m+n$-simplices of $X$ $\endgroup$ Commented Aug 6, 2018 at 5:16

1 Answer 1


For me, the 'usual' way is as suggested by მამუკა ჯიბლაძე. This is based on a systematic analysis of the simplices of $\Delta^m\times \Delta^n$ and of those missing in the corresponding subcomplex involving the k-horn of $\delta^n$. This sort of thing is classically well known and is based on collapsing arguments from `combinatorial' topology. It is handled in great detail in Peter May's Simplicial Objects in Algebraic Topology, starting on page 16.

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    $\begingroup$ There is also an alternative, slightly more conceptual, argument in Appendix H of Joyal's The Theory of Quasi-categories and Its Applications. $\endgroup$ Commented Aug 6, 2018 at 7:42
  • $\begingroup$ This is just the sort of argument I was looking for, thanks! $\endgroup$ Commented Aug 6, 2018 at 11:22
  • $\begingroup$ As Karol's comment makes clear this is a very useful piece of machinery. I have been trying to develop non-additive shuffle products in other settings and the anlysis of just such fillling arguments is very important in that context. $\endgroup$
    – Tim Porter
    Commented Aug 6, 2018 at 19:31
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    $\begingroup$ To be fair, I don't think it is accurate to say that this approach avoids "the theory of anodyne extensions". In both cases, we prove that the pushout product of $\partial \Delta[m] \to \Delta[m]$ and $\Lambda^{k}[n] \to \Delta[n]$ is anodyne which I is pretty much "the fundamental lemma of the theory of anodyne extensions". $\endgroup$ Commented Aug 7, 2018 at 7:35
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    $\begingroup$ Arguments of May and Joyal are combinatorial and very explicit while the argument of Gabriel and Zisman is categorical but still quite explicit. In the grand scheme of things they amount to the same. Of course, for some specialized purposes the combinatorial argument will be more appropriate. $\endgroup$ Commented Aug 7, 2018 at 7:35

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