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Proposition 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 $m=0$. $\square$