# Kerodon

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### 3.3.6 The $\operatorname{Ex}^{\infty }$ Functor

Let $X$ be a simplicial set. In ยง3.1.7, we proved that one can always choose an embedding $j: X \hookrightarrow Q$, where $Q$ is a Kan complex and $j$ is a weak homotopy equivalence (Corollary 3.1.7.2). In [MR90047], Kan gave an explicit construction of such an embedding, based on iteration of the construction $X \mapsto \operatorname{Ex}(X)$.

Construction 3.3.6.1 (The $\operatorname{Ex}^{\infty }$ Functor). For every nonnegative integer $n$, we let $\operatorname{Ex}^{n}$ denote the $n$-fold iteration of the functor $\operatorname{Ex}: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set_{\Delta }}$ of Construction 3.3.2.5, given inductively by the formula

$\operatorname{Ex}^{n}(X) = \begin{cases} X & \text{ if } n = 0 \\ \operatorname{Ex}( \operatorname{Ex}^{n-1}(X) ) & \text{ if } n > 0. \end{cases}$

For every simplicial set $X$, we let $\operatorname{Ex}^{\infty }(X)$ denote the colimit of the diagram

$X \xrightarrow { \rho _{X} } \operatorname{Ex}(X) \xrightarrow { \rho _{ \operatorname{Ex}(X) } } \operatorname{Ex}^2(X) \xrightarrow { \rho _{ \operatorname{Ex}^2(X)} } \operatorname{Ex}^3(X) \rightarrow \cdots ,$

where each $\rho _{ \operatorname{Ex}^{n}(X) }$ denotes the comparison map of Construction 3.3.4.3, and we let $\rho ^{\infty }_{X}: X \rightarrow \operatorname{Ex}^{\infty }(X)$ denote the tautological map. The construction $X \mapsto \operatorname{Ex}^{\infty }(X)$ determines a functor $\operatorname{Ex}^{\infty }$ from the category of simplicial sets to itself, and the construction $X \mapsto \rho _{X}^{\infty }$ determines a natural transformation of functors $\operatorname{id}_{ \operatorname{Set_{\Delta }}} \rightarrow \operatorname{Ex}^{\infty }$.

Our goal in this section is to record the main properties of Construction 3.3.6.1. In particular, for every simplicial set $X$, we show that $\operatorname{Ex}^{\infty }(X)$ is a Kan complex (Proposition 3.3.6.9) and that the comparison map $\rho _{X}^{\infty }: X \rightarrow \operatorname{Ex}^{\infty }(X)$ is a weak homotopy equivalence (Proposition 3.3.6.7).

Proposition 3.3.6.2. Let $X$ be a simplicial set. Then the comparison map $\rho _{X}^{\infty }: X \rightarrow \operatorname{Ex}^{\infty }(X)$ is a monomorphism of simplicial sets. Moreover, it is bijective on vertices.

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

$\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{m}, X ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{Sd}(\Delta ^{m}), X)$

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 $m=0$. $\square$

Example 3.3.6.3. Let $X$ be a discrete simplicial set (Definition 1.1.5.10). Invoking Example 3.3.4.6 repeatedly, we deduce that the transition maps in the diagram

$X \xrightarrow { \rho _{X} } \operatorname{Ex}(X) \xrightarrow { \rho _{ \operatorname{Ex}(X) } } \operatorname{Ex}^2(X) \xrightarrow { \rho _{ \operatorname{Ex}^2(X)} } \operatorname{Ex}^3(X) \rightarrow \cdots ,$

are isomorphisms. It follows that the comparison map $\rho _{X}^{\infty }: X \rightarrow \operatorname{Ex}^{\infty }(X)$ is also an isomorphism.

Proposition 3.3.6.4. The functor $X \mapsto \operatorname{Ex}^{\infty }(X)$ preserves filtered colimits and finite limits.

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. $\square$

Corollary 3.3.6.5. Let $f: X \rightarrow S$ be a morphism of simplicial sets. Let $s$ be a vertex of $S$, which we will identify (via Proposition 3.3.6.2) with its image in $\operatorname{Ex}^{\infty }(S)$. Then the canonical map $\operatorname{Ex}^{\infty }( X_ s ) \rightarrow \operatorname{Ex}^{\infty }(X)_{s}$ is an isomorphism of simplicial sets. Here $X_ s = \{ s\} \times _{S} X$ denotes the fiber of $f$ over the vertex $s$, and $\operatorname{Ex}^{\infty }(X)_{s} = \{ s\} \times _{ \operatorname{Ex}^{\infty }(S) } \operatorname{Ex}^{\infty }(X)$ is defined similarly.

Proposition 3.3.6.6. Let $f: X \rightarrow S$ be a morphism of simplicial sets. If $f$ is a Kan fibration, then the induced map $\operatorname{Ex}^{\infty }(f): \operatorname{Ex}^{\infty }(X) \rightarrow \operatorname{Ex}^{\infty }(S)$ is also a Kan fibration.

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. $\square$

Proposition 3.3.6.7. Let $X$ be a simplicial set. Then the comparison map $\rho _{X}^{\infty }: X \rightarrow \operatorname{Ex}^{\infty }(X)$ is a weak homotopy equivalence.

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

$X \xrightarrow { \rho _{X} } \operatorname{Ex}(X) \xrightarrow { \rho _{ \operatorname{Ex}(X) } } \cdots \xrightarrow { \rho _{ \operatorname{Ex}^{n-1}(X)} } \operatorname{Ex}^ n(X)$

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. $\square$

Corollary 3.3.6.8. Let $f: X \rightarrow Y$ be a morphism of simplicial sets. Then $f$ is a weak homotopy equivalence if and only if $\operatorname{Ex}^{\infty }(f)$ is a weak homotopy equivalence.

Proof. We have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ X \ar [d]^{ \rho ^{\infty }_ X} \ar [r]^-{f} & Y \ar [d]^{ \rho ^{\infty }_{Y} } \\ \operatorname{Ex}^{\infty }(X) \ar [r]^-{ \operatorname{Ex}^{\infty }(f) } & \operatorname{Ex}^{\infty }(Y), }$

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). $\square$

Proposition 3.3.6.9. 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 $\operatorname{Ex}^{m-1}(X)$, 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

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

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 3.3.2.1) and let $Q \subset \operatorname{Chain}[n]$ be the subset obtained by removing $[n]$ and $[n] \setminus \{ i\}$. Using Proposition 3.3.3.16, 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

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

Corollary 3.3.6.10. Let $X$ be a Kan complex. Then the comparison map $\rho _{X}^{\infty }: X \rightarrow \operatorname{Ex}^{\infty }(X)$ is a homotopy equivalence.

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. $\square$