Corollary 2.4.4.20. Let $Q$ be a partially ordered set, let $q \in Q$ be an element, and suppose that $Q = Q_{-} \cup Q_{+}$ for $Q_{-} = \{ q_{-} \in Q: q_{-} \leq q \} $ and $Q_{+} = \{ q_+ \in Q: q \leq q_{+} \} $ (this condition is automatically satisfied, for example, if $Q$ is linearly ordered). Then the simplicial functor
\[ \operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet } \rightarrow \operatorname{Path}[ Q ]_{\bullet } \]
has a unique left inverse $R: \operatorname{Path}[ Q ]_{\bullet } \rightarrow \operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet }$.
Proof.
By virtue of Corollary 2.4.4.18, we can identify the pushout $\operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet }$ with a simplicial subcategory $\operatorname{\mathcal{C}}\subseteq \operatorname{Path}[Q]_{\bullet }$; we wish to show that there is a unique simplicial functor $R: \operatorname{Path}[Q]_{\bullet } \rightarrow \operatorname{\mathcal{C}}$ satisfying $R|_{\operatorname{\mathcal{C}}} = \operatorname{id}_{\operatorname{\mathcal{C}}}$. Our assumption that $Q = Q_{-} \cup Q_{+}$ guarantees that $\operatorname{\mathcal{C}}$ contains every object of $\operatorname{Path}[Q]_{\bullet }$. To prove existence, we take the simplicial functor $R$ to be the identity on objects and given on morphisms by the maps
\[ \operatorname{Hom}_{ \operatorname{Path}[Q] }(a,b)_{\bullet } = \operatorname{N}_{\bullet }(P_{a,b} ) \rightarrow \operatorname{N}_{\bullet }( P'_{a,b} ) = \operatorname{Hom}_{\operatorname{\mathcal{C}}}(a,b)_{\bullet } \]
\[ (J \in P_{a,b}) \mapsto \begin{cases} J \cup \{ q\} & \text{ if $a \leq q \leq b$} \\ J & \text{ otherwise, } \end{cases} \]
where $P_{a,b}$ and the subset $P'_{a,b} \subseteq P_{a,b}$ are defined as in Remark 2.4.4.19.
To prove uniqueness, let $R': \operatorname{Path}[Q]_{\bullet } \rightarrow \operatorname{\mathcal{C}}$ be another simplicial functor satisfying $R'|_{\operatorname{\mathcal{C}}} = \operatorname{id}_{\operatorname{\mathcal{C}}}$; we wish to show that $R' = R$. It is clear that $R$ and $R'$ agree at the level of objects. For every pair of elements $a,b \in Q$, the simplicial functors $R$ and $R'$ induce maps $\theta , \theta ': \operatorname{Hom}_{ \operatorname{Path}[Q] }(a,b)_{\bullet } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(a,b)_{\bullet }$; we wish to show that $\theta = \theta '$. Since $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(a,b)_{\bullet }$ can be identified with the nerve of the partially ordered set $P'_{a,b}$, it will suffice to show that $\theta $ and $\theta '$ agree on vertices. For every finite linearly ordered subset $J \subseteq Q$ having least element $a$ and greatest element $b$, let $f_{J}: a \rightarrow b$ denote the corresponding morphism in the path category $\operatorname{Path}[Q]$; we wish to show that $\theta ( f_{J} ) = \theta '( f_{J} )$. Without loss of generality, we may assume that the morphism $f_{J}$ is indecomposable: that is, that we have $a \neq b$ and that $J = \{ a < b \} $. We may further assume that $a < q < b$ (otherwise, $f_{J}$ is a morphism in the category $\operatorname{\mathcal{C}}$ and we have $\theta (f_ J) = f_{J} = \theta '(f_ J)$). Set $J^{+} = \{ a < q < b \} $, so that $\theta ( f_ J ) = f_{J^{+} }$. Write $\theta '( f_ J ) = f_{K}$ where $K \subseteq Q$ is a finite linearly ordered subset having least element $a$ and greatest element $b$. Since $f_{J^{+}}$ is a morphism of $\operatorname{\mathcal{C}}$, we have $\theta '( f_{J^{+}} ) = f_{J^{+} }$. The inclusion $J \subseteq J^{+}$ then implies that $K \subseteq J^{+}$. On the other hand, $f_{K}$ is also a morphism of $\operatorname{\mathcal{C}}$, so we must have $q \in K$. It follows that $K = J^{+}$, so that $\theta ( f_ J) = f_{J^{+}} = f_{K} = \theta '( f_ J )$ as desired.
$\square$