Remark 4.6.2.20. Suppose we are given a commutative diagram of $\infty $-categories
satisfying the following conditions:
- $(a)$
The functor $q$ is an inner fibration and $q'$ is an isofibration.
- $(b)$
The functor $\overline{F}$ is essentially surjective.
- $(c)$
For each object $D \in \operatorname{\mathcal{D}}$, the induced functor $F_{D}: \operatorname{\mathcal{C}}_{D} \rightarrow \operatorname{\mathcal{C}}'_{ \overline{F}(D) }$ is essentially surjective.
Then the functor $F$ is essentially surjective. To prove this, consider an arbitrary object $Z \in \operatorname{\mathcal{C}}'$. Assumption $(b)$ guarantees that there exists an object $D \in \operatorname{\mathcal{D}}$ and an isomorphism $\overline{u}: \overline{F}(D) \rightarrow q'(Z)$ in the $\infty $-category $\operatorname{\mathcal{D}}'$. Assumption $(a)$ guarantees that we can lift $\overline{u}$ to an isomorphism $u: Y \rightarrow Z$ in the $\infty $-category $\operatorname{\mathcal{C}}'$, where $Y$ belongs to the fiber $\operatorname{\mathcal{C}}'_{ \overline{F}(D)}$. Applying $(c)$, we can choose an object $X \in \operatorname{\mathcal{C}}_{D}$ and an isomorphism $v: F(X) \rightarrow Y$ in the $\infty $-category $\operatorname{\mathcal{C}}'_{ \overline{F}(D) }$. It follows that $Z$ is isomorphic to $F(X)$ in the $\infty $-category $\operatorname{\mathcal{C}}'$.