Kerodon

Proof. Using Proposition 2.4.4.15, we can identify the pushout $\operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet }$ with the simplicial path category of the simplicial set $S = \operatorname{N}_{\bullet }(Q_{-} ) \coprod _{ \{ q\} } \operatorname{N}_{\bullet }( Q_{+} )$. The tautological map $S \rightarrow \operatorname{N}_{\bullet }(Q)$ is a monomorphism of simplicial sets, and therefore induces an equivalence from $\operatorname{Path}[S]_{\bullet }$ to a simplicial subcategory $\operatorname{\mathcal{C}}\subseteq \operatorname{Path}[Q]_{\bullet }$ (Remark 2.4.4.12). It is clear that this subcategory contains both $\operatorname{Path}[ Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$. To complete the proof, it will suffice to show that if $\operatorname{\mathcal{D}}$ is any other simplicial subcategory of $\operatorname{Path}[Q]_{\bullet }$ which contains $\operatorname{Path}[Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$, then $\operatorname{\mathcal{D}}$ contains $\operatorname{\mathcal{C}}$. This is clear: the universal property of $\operatorname{\mathcal{C}}$ guarantees that there is a unique simplicial functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ which is the identity on both $\operatorname{Path}[ Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$. Invoking the universal property of $\operatorname{\mathcal{C}}$ again, we deduce that the composite functor $\operatorname{\mathcal{C}}\xrightarrow {F} \operatorname{\mathcal{D}}\hookrightarrow \operatorname{Path}[Q]_{\bullet }$ coincides with the inclusion map, so that $\operatorname{\mathcal{C}}\subseteq \operatorname{\mathcal{D}}$. $\square$