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Proposition 7.3.6.7. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be a functor of $\infty $-categories, and let $\operatorname{\mathcal{C}}^{0} \subseteq \operatorname{\mathcal{C}}$ be a full subcategory. Let $F,G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be functors having restrictions $F_0 = F|_{\operatorname{\mathcal{C}}^{0}}$ and $G_0 = G|_{\operatorname{\mathcal{C}}^{0}}$, so that we have a commutative diagram of Kan complexes

7.28
\begin{equation} \begin{gathered}\label{equation:relative-Kan-extension-mapping-property} \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})}( F, G ) \ar [r] \ar [d] & \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}}) }( F_0, G_0) \ar [d] \\ \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{E}}) }( U \circ F, U \circ G) \ar [r] & \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{E}}) }( U \circ F_0, U \circ G_0). } \end{gathered} \end{equation}

If $F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^0$ or $G$ is $U$-right Kan extended from $\operatorname{\mathcal{C}}^{0}$, then (7.28) is a homotopy pullback square.

Proof of Proposition 7.3.6.7. We will assume that the functor $F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ (the proof in the case where $G$ is $U$-right Kan extended from $\operatorname{\mathcal{C}}^{0}$ is similar). Using Corollary 4.5.2.23, we can factor the functor $U$ as a composition $\operatorname{\mathcal{D}}\xrightarrow {T} \operatorname{\mathcal{D}}' \xrightarrow {U'} \operatorname{\mathcal{E}}$, where $U'$ is an isofibration and $T$ is an equivalence of $\infty $-categories. Note that the functor $T \circ F$ is $U'$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ (Remark 7.3.3.15), and that the natural maps

\[ \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})}( F, G ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}') }( T \circ F, T \circ G) \]

\[ \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}})}( F_0, G_0 ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}}') }( T \circ F_0, T \circ G_0) \]

are homotopy equivalences. Consequently, we can replace $\operatorname{\mathcal{D}}$ by $\operatorname{\mathcal{D}}'$ and thereby reduce to proving Proposition 7.3.6.7 in the special case where the functor $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ is an isofibration of $\infty $-categories.

Let $V: \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}}) \times _{ \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{E}}) } \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}})$ be as in Remark 7.3.6.8; we wish to show that $F$ is a $V$-initial object of the $\infty $-category $\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$. Note that $V$ is also an isofibration (Proposition 4.4.5.1). By virtue of Corollary 7.1.5.17, it will suffice to show that every lifting problem

7.29
\begin{equation} \begin{gathered}\label{equation:relative-Kan-extension-mapping} \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^ n \ar [r]^-{ \sigma _0 } \ar [d] & \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \ar [d]^{V} \\ \Delta ^ n \ar@ {-->}[ur] \ar [r] & \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}}) \times _{ \operatorname{Fun}(\operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{E}}) } \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}}) } \end{gathered} \end{equation}

has a solution, provided that $n \geq 0$ and $\sigma _0(0) = F$. Unwinding the definitions, we can rewrite (7.29) as a lifting problem

7.30
\begin{equation} \begin{gathered}\label{equation:relative-Kan-extension-mapping2} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}^0 \ar [r]^-{ G_0 } \ar [d] & \operatorname{Fun}(\Delta ^ n, \operatorname{\mathcal{D}}) \ar [d]^{V'} \\ \operatorname{\mathcal{C}}\ar@ {-->}[ur]^{G} \ar [r] & \operatorname{Fun}(\operatorname{\partial \Delta }^ n, \operatorname{\mathcal{D}}) \times _{ \operatorname{Fun}(\operatorname{\partial \Delta }^ n, \operatorname{\mathcal{E}}) } \operatorname{Fun}( \Delta ^ n, \operatorname{\mathcal{E}}) } \end{gathered} \end{equation}

Note that $V'$ is also an isofibration of $\infty $-categories (Proposition 4.4.5.1).

We will complete the proof by showing that the lifting problem (7.30) admits a solution $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{Fun}(\Delta ^ n, \operatorname{\mathcal{D}})$ which is $V'$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$. By virtue of Proposition 7.3.5.5, it will suffice to show that for each object $C \in \operatorname{\mathcal{C}}$, the induced lifting problem

7.31
\begin{equation} \begin{gathered}\label{equation:relative-Kan-extension-mapping3} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}^0_{/C} \ar [r] \ar [d] & \operatorname{Fun}(\Delta ^ n, \operatorname{\mathcal{D}}) \ar [d]^{V'} \\ ( \operatorname{\mathcal{C}}^{0}_{/C})^{\triangleright } \ar@ {-->}[ur]^{Q} \ar [r] & \operatorname{Fun}(\operatorname{\partial \Delta }^ n, \operatorname{\mathcal{D}}) \times _{ \operatorname{Fun}(\operatorname{\partial \Delta }^ n, \operatorname{\mathcal{E}}) } \operatorname{Fun}( \Delta ^ n, \operatorname{\mathcal{E}}) } \end{gathered} \end{equation}

admits a solution $Q: (\operatorname{\mathcal{C}}^{0}_{/C})^{\triangleright } \rightarrow \operatorname{Fun}( \Delta ^ n, \operatorname{\mathcal{D}})$ which is a $V'$-colimit diagram. Our assumption that $\sigma _0(0) = F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ guarantees that the composite map

\[ ( \operatorname{\mathcal{C}}^{0}_{/C})^{\triangleright } \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^ n, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}( \{ 0\} , \operatorname{\mathcal{D}}) = \operatorname{\mathcal{D}} \]

is a $U$-colimit diagram. Applying Corollary 7.1.7.9, we conclude that the lifting problem (7.31) admits a solution $Q$, and Proposition 7.1.7.12 guarantees that $Q$ is automatically a $V'$-colimit diagram. $\square$