Lemma 7.4.4.1. Let $n \geq 2$, let $X: \Lambda ^{n}_{0} \rightarrow \operatorname{\mathcal{QC}}$ be a diagram, and let $W$ be a collection of morphisms of the $\infty $-category $X(0)$ which satisfies the following pair of conditions:
- $(1)$
Let $1 \leq i \leq n$, and let $X(0 < i): X(0) \rightarrow X(i)$ be the functor obtained by evaluating $X$ on the edge $\operatorname{N}_{\bullet }( \{ 0 < i \} ) \subseteq \Lambda ^{n}_{0}$. Then $X(0 < i )$ carries each element of $W$ to an isomorphism in the $\infty $-category $X(i)$.
- $(2)$
The functor $X(0 < 1): X(0) \rightarrow X(1)$ exhibits $X(1)$ as a localization of $X(0)$ with respect to $W$.
Then $X$ can be extended to an $n$-simplex $\Delta ^{n} \rightarrow \operatorname{\mathcal{QC}}$.
Proof.
Set $\operatorname{\mathcal{C}}= X(0)$, $\operatorname{\mathcal{D}}= X(1)$, and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be the functor $X(0 < 1)$. Using the isomorphism $\Lambda ^{n}_{0} \simeq ( \operatorname{\partial \Delta }^{n-1} )^{\triangleleft }$, we can identify $X$ with a diagram $\sigma _0: \operatorname{\partial \Delta }^{n-1} \rightarrow \operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }$. To complete the proof, it will suffice to show that $\sigma _0$ can be extended to an $(n-1)$-simplex of $\operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }$. Let us identify the objects of the $\infty $-category $\operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }$ with pairs $( \operatorname{\mathcal{E}}, G )$, where $\operatorname{\mathcal{E}}$ is a small $\infty $-category and $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ is a functor. Let $\operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }^{W}$ denote the full subcategory of $\operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }$ spanned by those pairs $(\operatorname{\mathcal{E}},G)$, where the functor $G$ carries each element of $W$ to an isomorphism in $\operatorname{\mathcal{E}}$. It follows from assumption $(1)$ that the diagram $\sigma _0$ factors through the subcategory $\operatorname{\mathcal{QC}}_{ \operatorname{\mathcal{C}}/ }^{W} \subseteq \operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/}$. To prove the existence of $\sigma $, it will suffice (by virtue of Corollary 4.6.7.13) to show that $\sigma _0( 0 ) = (\operatorname{\mathcal{D}}, F)$ is an initial object of the $\infty $-category $\operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/ }^{W}$. Fix another object $(\operatorname{\mathcal{E}}, G) \in \operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/ }^{W}$; we wish to show that the morphism space $\operatorname{Hom}_{\operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/}^{W}}( (\operatorname{\mathcal{D}}, F), (\operatorname{\mathcal{E}}, G) ) = \operatorname{Hom}_{ \operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/} }( (\operatorname{\mathcal{D}}, F), (\operatorname{\mathcal{E}}, G) )$ is a contractible Kan complex. Using Corollary 4.6.9.18 and Remark 5.5.4.6, we can identify $\operatorname{Hom}_{ \operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/} }( (\operatorname{\mathcal{D}}, F), (\operatorname{\mathcal{E}}, G) )$ with the homotopy fiber of the map of Kan complexes
\[ \operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})^{\simeq } \xrightarrow { \circ F } \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})^{\simeq } \]
over the vertex $G \in \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})^{\simeq }$. Assumption $(2)$ guarantees that this map is a homotopy equivalence onto the summand of $\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})^{\simeq }$ spanned by those functors $\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ which carry each element of $W$ to an isomorphism in $\operatorname{\mathcal{E}}$. It will therefore suffice to show that this summand contains the functor $G$, which follows from the definition of $\operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/}^{W}$.
$\square$