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Proposition Let $X$ be a simplicial set. Then $\operatorname{Ex}^{\infty }(X)$ is a Kan complex.

Proof. Let $X$ be a simplicial set and suppose we are given a morphism of simplicial sets $f_0: \Lambda ^{n}_{i} \rightarrow \operatorname{Ex}^{\infty }(X)$, for some $n > 0$ and $0 \leq i \leq n$. We wish to show that $f_0$ can be extended to an $n$-simplex of $\operatorname{Ex}^{\infty }(X)$. Since the simplicial set $\Lambda ^{n}_{i}$ has finitely many nondegenerate simplices, we can assume that $f_0$ factors as a composition $\Lambda ^{n}_{i} \xrightarrow { f'_0 } \operatorname{Ex}^{m}(X) \rightarrow \operatorname{Ex}^{\infty }(X)$, for some positive integer $m$. We will complete the proof by showing that $f'_0$ can be extended to an $n$-simplex of $\operatorname{Ex}^{m+1}(X)$: that is, that there exists a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_ i \ar [r]^-{f'_0} \ar [d] & \operatorname{Ex}^{m}(X) \ar [d]^{ \rho _{ \operatorname{Ex}^{m}(X) } } \\ \Delta ^{n} \ar [r]^-{f'} & \operatorname{Ex}^{m+1}(X). } \]

Replacing $X$ by $\operatorname{Ex}^{m-1}$, we can reduce to the case $m = 1$. In this case, $f'_0$ can be identified with a morphism of simplicial sets $g_0: \operatorname{Sd}( \Lambda ^{n}_{i} ) \rightarrow X$. Unwinding the definitions, we see that the problem of finding a simplex $f': \Delta ^{n} \rightarrow \operatorname{Ex}^{2}(X)$. with the desired property is equivalent to the problem of finding a morphism $g: \operatorname{Sd}( \operatorname{Sd}( \Delta ^ n) ) \rightarrow X$ whose restriction to $\operatorname{Sd}( \operatorname{Sd}( \Lambda ^{n}_{i} ) )$ is equal to the composition

\[ \operatorname{Sd}( \operatorname{Sd}( \Lambda ^{n}_{i} ) ) \xrightarrow { \operatorname{Sd}( \lambda _{ \Lambda ^{n}_ i} ) } \operatorname{Sd}( \Lambda ^{n}_{i} ) \xrightarrow {g_0} X. \]

Without loss of generality, we may assume that $X = \operatorname{Sd}( \Lambda ^{n}_{i} )$ and that $g_0$ is the identity map. Let $\operatorname{Chain}[n]$ be the collection of all nonempty subsets of $[n]$ (Notation and let $Q \subset \operatorname{Chain}[n]$ be the subset obtained by removing $[n]$ and $[n] \setminus \{ i\} $. Using Proposition, we can identify $\operatorname{Sd}( \Lambda ^{n}_{i})$, $\operatorname{Sd}( \operatorname{Sd}( \Lambda ^{n}_{i} ) )$, and $\operatorname{Sd}( \operatorname{Sd}( \Delta ^ n ) )$ with the nerves of the partially ordered sets $Q$, $\operatorname{Chain}[Q]$, and $\operatorname{Chain}[ \operatorname{Chain}[n] ]$, respectively. To complete the proof, it will suffice to show that there exists a nondecreasing function of partially ordered sets $g: \operatorname{Chain}[ \operatorname{Chain}[n] ]\rightarrow Q$ having the property that, for every element $(S_0 \subset S_1 \subset \cdots \subset S_ m)$ of $\operatorname{Chain}[Q]$, we have $g(S_0 \subset S_1 \subset \cdots \subset S_ m) = \{ \mathrm{Max}(S_0), \mathrm{Max}(S_1), \ldots , \mathrm{Max}(S_ m) \} \in Q$. This requirement is satisfied if $g$ is defined by the formula

\[ g(S_0 \subset S_1 \subset \cdots \subset S_ m) = \{ \mathrm{Max}'(S_0), \mathrm{Max}'(S_1), \ldots , \mathrm{Max}'(S_ m) \} , \]

where $\mathrm{Max}': \operatorname{Chain}[n] \rightarrow [n]$ is the (non-monotone) map of sets given by

\[ \mathrm{Max}'(S) = \begin{cases} i & \text{ if $S = [n]$ or $S = [n] \setminus \{ i\} $ } \\ \mathrm{Max}(S) & \text{ otherwise. } \end{cases} \]