Non-degenerate simplexes in a Kan complex

Here I write an answer in the form I like (which I hope to be useful for others):

It's essentially the same as Naruki Masuda's answer above, but I don't like things like $Delta^{{0,1,ldots,n-1,n+1}}$, which I would write as the image of the map $d^n: Delta^ntoDelta^{n+1}$.

As suggested in Tom Goodwillie's comment, I'll prove that if $f$ is a non-degenerate $n$-simplex in a Kan complex $X$ for $n>0$, then there exists a non-degenerate $(n+1)$-simplex $g$ such that $d_{n+1}g = f$.

Let $f: Delta^n=Delta^{{0, ldots, n}}to X$ be a non-degenerate simplex. Consider $f' = s_{n-1}d_n f: Delta^{{0, ldots, n-1, n+1}}to X$, whose restriction to the first $n$ vertices agrees with that of $f$. These glue together to define $bar f: Delta^{{0, ldots, n}}cup_{Delta^{{0, ldots, n-1}}}Delta^{{0, ldots, n-1, n+1}}to X$. Now I claim the following:

1. $bar f$ extends to a simplex $g: Delta^{{0, ldots, n+1}}to X$

2. The simplex $g$ is non-degenerate.

First, assume 1. and let us prove 2. Assume the contrary and suppose $g=s_i h$ for some $h: Delta^nto X$.

• If $i= n$, then this implies $f=d_{n+1}s_n h= h = d_n s_n h =f'$, but this is impossible since $f$ is non-degenerate and $f'$ is degenerate.
• If $i<n$, then $f=d_{n+1} g = d_{n+1}s_i h = s_i d_n h$, so again this contradicts to the assumption that $f$ is non-degenerate.

Therefore $g$ must be non-degenerate.

Now let us prove 1. It suffices to prove that the inclusion $i: Delta^{{0, ldots, n}}cup_{Delta^{{0, ldots, n-1}}}Delta^{{0, ldots, n-1, n+1}}to Delta^{n+1}$ is an anodyne extension. For any $Asubset {1, ldots, n-1}$ of cardinality $a$, let $Lambda(A)$ be the horn $Lambda^{a+2}_0 hookrightarrow Delta^{a+2} = Delta^{{0}cup Acup {n, n+1}}hookrightarrow Delta^{n+1}$. Now observe that $i$ is the composition $i_{n-1}circcdotscirc i_1 circ i_{0}$, where $i_k$ is the "horn-filling inclusion" that fills ${Lambda(A)mid |A|=k}$.