Proposition 3.3.5.8. Let $X$ be a simplicial set, and let $\rho _{X}: X \rightarrow \operatorname{Ex}(X)$ be the comparison map of Construction 3.3.4.3. Then the morphisms $\rho _{ \operatorname{Ex}(X) }, \operatorname{Ex}( \rho _{X} ): \operatorname{Ex}(X) \rightarrow \operatorname{Ex}( \operatorname{Ex}(X) )$ are homotopic.
Proof. Let $Q$ be a partially ordered set. Using Example 3.3.3.18, we can identify the subdivisions $\operatorname{Sd}( \operatorname{N}_{\bullet }(Q) )$ and $\operatorname{Sd}( \operatorname{Sd}( \operatorname{N}_{\bullet }(Q) ) )$ with the nerves of partially ordered sets $\operatorname{Chain}[Q]$ and $\operatorname{Chain}[\operatorname{Chain}[Q]]$, respectively. Under this identification, a morphism of simplicial sets
corresponds to a nondecreasing functions $\operatorname{Chain}[\operatorname{Chain}[Q]] \rightarrow \operatorname{Chain}[Q]$, whose value on a linearly ordered subset $\vec{S} = (S_0 \subset S_1 \subset \cdots \subset S_ n)$ of $\operatorname{Chain}[Q]$ is given by
Note that we always have an inclusion $\{ \mathrm{Max}(S_0), \ldots , \mathrm{Max}(S_ n) \} \subseteq S_{n}$. It follows that there is a unique map of simplicial sets
satisfying $h_{Q} |_{ \{ 0\} \times \operatorname{Sd}( \operatorname{Sd}( \operatorname{N}_{\bullet }(Q) ) )} = \operatorname{Sd}( \lambda _{ \operatorname{N}_{\bullet }(Q) } )$ and $h_{Q} |_{ \{ 1\} \times \operatorname{Sd}( \operatorname{Sd}( \operatorname{N}_{\bullet }(Q) ) ) } = \lambda _{ \operatorname{Sd}( \operatorname{N}_{\bullet }(Q) )}$, depending functorially on $Q$.
Let $\sigma $ be an $n$-simplex of the simplicial set $\operatorname{Ex}(X)$, which we identify with a map $\sigma : \operatorname{Sd}( \Delta ^ n ) \rightarrow X$. We let $f(\sigma )$ denote the composite map
which we will identify with an $n$-simplex of the simplicial set $\operatorname{Fun}( \Delta ^1, \operatorname{Ex}( \operatorname{Ex}(X) ) )$. The construction $\sigma \mapsto f(\sigma )$ then determines a morphism of simplicial sets $f: \operatorname{Ex}(X) \rightarrow \operatorname{Fun}( \Delta ^1, \operatorname{Ex}( \operatorname{Ex}(X) ) )$, which we can identify with a map $\Delta ^1 \times \operatorname{Ex}(X) \rightarrow \operatorname{Ex}( \operatorname{Ex}(X) )$. By construction, this map is a homotopy from $\rho _{ \operatorname{Ex}(X) }$ to $\operatorname{Ex}( \rho _{X} )$. $\square$