Kerodon

Proof. It will suffice to show that each of the comparison maps $\rho _{ \operatorname{Ex}^{n}(X) }: \operatorname{Ex}^{n}(X) \rightarrow \operatorname{Ex}^{n+1}(X)$ is a monomorphism which is bijective on vertices. Replacing $X$ by $\operatorname{Ex}^{n}(X)$, we can reduce to the case $n = 0$. Fix $m \geq 0$. On $m$-simplices, the comparison map $\rho _{X}$ is given by the map of sets

induced by precomposition with the last vertex map $\lambda _{ \Delta ^{m} }: \operatorname{Sd}(\Delta ^ m) \rightarrow \Delta ^{m}$. To complete the proof, it suffices to observe that $\lambda _{ \Delta ^{m}}$ is an epimorphism of simplicial sets (in fact, it admits a section $\Delta ^{m} \rightarrow \operatorname{Sd}( \Delta ^{m} ) \simeq \operatorname{N}_{\bullet }( \operatorname{Chain}[m] )$, given by the chain of subsets $\{ 0\} \subset \{ 0,1 \} \subset \cdots \subset \{ 0,1, \ldots , m \} $), and an isomorphism in the special case $m=0$. $\square$

Proof. It will suffice to show that, for every nonnegative integer $n$, the functor $X \mapsto \operatorname{Ex}^{n}(X)$ preserves filtered colimits and finite limits. Proceeding by induction on $n$, we can reduce to the case $n = 1$. We now observe that the functor $\operatorname{Ex}$ preserves all limits of simplicial sets (either by construction, or because it admits a left adjoint $X \mapsto \operatorname{Sd}(X)$), and preserves filtered colimits by virtue of Remark 3.3.2.7. $\square$

Proof. Combine Proposition 3.3.6.4 with Example 3.3.6.3. $\square$

Proof. Since the collection of Kan fibrations is stable under the formation of filtered colimits (Remark 3.1.1.8), it will suffice to show that each of the maps $\operatorname{Ex}^{n}(f): \operatorname{Ex}^{n}(X) \rightarrow \operatorname{Ex}^{n}(S)$ is a Kan fibration. Proceeding by induction on $n$, we can reduce to the case $n=1$, which follows from Corollary 3.3.5.4. $\square$

Proof. By virtue of Proposition 3.2.8.3, it will suffice to show that for each $n \geq 0$, the composite map

is a weak homotopy equivalence. Proceeding by induction on $n$, we are reduced to showing that each of the comparison maps $\rho _{ \operatorname{Ex}^{n-1}(X) }: \operatorname{Ex}^{n-1}(X) \rightarrow \operatorname{Ex}^{n}(X)$ is a weak homotopy equivalence, which is a special case of Theorem 3.3.5.1. $\square$

Proof. We have a commutative diagram

where the vertical maps are weak homotopy equivalences (Proposition 3.3.6.7). The desired result now follows from the two-out-of-three property (Remark 3.1.6.16). $\square$

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

Replacing $X$ by $\operatorname{Ex}^{m-1}(X)$, 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

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 3.3.2.1) and let $Q \subset \operatorname{Chain}[n]$ be the subset obtained by removing $[n]$ and $[n] \setminus \{ i\} $. Using Proposition 3.3.3.17, 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

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

Proof. Since $\operatorname{Ex}^{\infty }(X)$ is also a Kan complex (Proposition 3.3.6.9), it will suffice to show that $\rho _{X}^{\infty }$ is a weak homotopy equivalence (Proposition 3.1.6.13), which follows from Proposition 3.3.6.7. $\square$