$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Corollary 8.3.5.8 (Functoriality of $\operatorname{Hom}$-Functors). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between $\infty $-categories. Choose natural transformations
\[ \alpha : \underline{\Delta ^0}_{\operatorname{Tw}(\operatorname{\mathcal{C}}) } \rightarrow \mathscr {H}_{\operatorname{\mathcal{C}}}|_{ \operatorname{Tw}(\operatorname{\mathcal{C}})} \quad \quad \beta : \underline{\Delta ^0}_{\operatorname{Tw}(\operatorname{\mathcal{D}}) } \rightarrow \mathscr {H}_{\operatorname{\mathcal{D}}}|_{ \operatorname{Tw}(\operatorname{\mathcal{D}})} \]
which exhibit $\mathscr {H}_{\operatorname{\mathcal{C}}}$ and $\mathscr {H}_{\operatorname{\mathcal{D}}}$ as $\operatorname{Hom}$-functors for $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$, respectively. Then there exists a natural transformation $\gamma : \mathscr {H}_{\operatorname{\mathcal{C}}}(-,-)\rightarrow \mathscr {H}_{\operatorname{\mathcal{D}}}( F(-), F(-) )$ for which the diagram
8.55
\begin{equation} \begin{gathered}\label{equation:Hom-functoriality} \xymatrix@R =50pt@C=50pt{ & \underline{ \Delta ^{0} }_{ \operatorname{Tw}(\operatorname{\mathcal{C}}) } \ar [dl]_-{ [\alpha ] } \ar [dr]^-{ [\beta ]} \\ \mathscr {H}_{\operatorname{\mathcal{C}}}|_{ \operatorname{Tw}(\operatorname{\mathcal{C}}) } \ar [rr]^{ [\gamma ] } & & \mathscr {H}_{\operatorname{\mathcal{D}}}|_{ \operatorname{Tw}(\operatorname{\mathcal{C}}) } } \end{gathered} \end{equation}
commutes (in the homotopy category $\mathrm{h} \mathit{ \operatorname{Fun}( \operatorname{Tw}(\operatorname{\mathcal{C}}), \operatorname{\mathcal{S}}) }$). Moreover, the natural transformation $\gamma $ is uniquely determined up to homotopy.
Proof.
This is a special case of Proposition 7.3.6.1, since $\alpha $ exhibits $\mathscr {H}_{\operatorname{\mathcal{C}}}$ as a left Kan extension of $\underline{ \Delta ^0}_{ \operatorname{Tw}(\operatorname{\mathcal{C}}) }$ along the left fibration $\operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}$ (Proposition 8.3.5.6).
$\square$