Non-degenerate simplexes in a Kan complex

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 $h: \Delta^n\to X$.

• 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 $A\subset \{1, \ldots, n-1\}$ of cardinality $a$, let $\Lambda(A)$ be the horn $\Lambda^{a+2}_0 \hookrightarrow \Delta^{a+2} = \Delta^{\{0\}\cup A\cup \{n, n+1\}}\hookrightarrow \Delta^{n+1}$. Now observe that $i$ is the composition $i_{n-1}\circ\cdots\circ i_1 \circ i_{0}$, where $i_k$ is the "horn-filling inclusion" that fills $\{\Lambda(A)\mid |A|=k\}$.