Corollary 8.5.1.10. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ and $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be functors of $\infty $-categories. Suppose that $F$ is $U$-left Kan extended from a full subcategory $\operatorname{\mathcal{C}}^0 \subseteq \operatorname{\mathcal{C}}$. Then any functor $G: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ which is a retract of $F$ (in the $\infty $-category $\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$) is also $U$-left Kan extended from $\operatorname{\mathcal{C}}^0$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Let $\operatorname{ev}: \operatorname{\mathcal{C}}\times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{\mathcal{D}}$ denote the evaluation functor. By virtue of Remark 7.3.3.3, the functor $F$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ at an object $C \in \operatorname{\mathcal{C}}$ if and only if the functor $\operatorname{ev}$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0} \times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ at the object $(C,F)$. If this condition is satisfied, then Corollary 8.5.1.9 guarantees that $\operatorname{ev}$ is also $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0} \times \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ at the object $(C,G)$, so that $G$ is $U$-left Kan extended from $\operatorname{\mathcal{C}}^{0}$ at $C$. The desired result now follows by allowing the object $C \in \operatorname{\mathcal{C}}$ to vary. $\square$