Corollary 7.3.1.16 (02YN)

Corollary 7.3.1.16. Let $G: \operatorname{\mathcal{C}}^{0} \rightarrow \operatorname{\mathcal{C}}$, $F_0: \operatorname{\mathcal{C}}^{0} \rightarrow \operatorname{\mathcal{D}}$, and $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be functors of $\infty $-categories, where $G$ is fully faithful. Then:

If $\alpha : F \circ G \rightarrow F_0$ is a natural transformation which exhibits $F$ as a right Kan extension of $F_0$ along $G$, then $\alpha $ is an isomorphism in the $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}})$.
If $\beta : F_0 \rightarrow F \circ G$ is a natural transformation which exhibits $F$ as a left Kan extension of $F_0$ along $G$, then $\beta $ is an isomorphism in the $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}^{0}, \operatorname{\mathcal{D}})$.

Proof. Let $\alpha : F \circ G \rightarrow F_0$ be a natural transformation which exhibits $F$ as a right Kan extension of $F_0$ along $G$. Applying Proposition (in the special case where $K = \operatorname{\mathcal{C}}^0$), we deduce that $\alpha $ also exhibits $F \circ G$ as a right Kan extension of $F_0$ along the identity functor $\operatorname{id}_{ \operatorname{\mathcal{C}}^{0} }: \operatorname{\mathcal{C}}^{0} \rightarrow \operatorname{\mathcal{C}}^{0}$. Invoking Example 7.3.1.8, we see that $\alpha $ is an isomorphism. This proves the first assertion; the second follows by a similar argument. $\square$