Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Corollary 7.1.3.13. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories. The following conditions are equivalent:

$(1)$

The functor $F$ admits a right adjoint $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$.

$(2)$

For every object $Y \in \operatorname{\mathcal{D}}$, the right fibration $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ is representable.

$(3)$

For every representable right fibration $\widetilde{\operatorname{\mathcal{D}}} \rightarrow \operatorname{\mathcal{D}}$, the right fibration $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \widetilde{\operatorname{\mathcal{D}}} \rightarrow \operatorname{\mathcal{C}}$ is representable.

If these conditions are satisfied, then the functor $G$ carries each object $Y \in \operatorname{\mathcal{D}}$ to an object of $\operatorname{\mathcal{C}}$ which represents the right fibration $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$.

Proof. By virtue of Corollary 6.2.4.7, the functor $F$ admits a right adjoint if and only if, for every object $Y \in \operatorname{\mathcal{D}}$, the $\mathrm{h} \mathit{\operatorname{Kan}}$-enriched functor

$\theta : \mathrm{h} \mathit{\operatorname{\mathcal{C}}}^{\operatorname{op}} \rightarrow \mathrm{h} \mathit{\operatorname{Kan}} \quad \quad \theta (X) = \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(X), Y)$

is representable. Using Example 5.2.7.13 and Remark 5.2.7.5, we can identify $\theta$ with the homotopy transport representation for the right fibration $\operatorname{\mathcal{C}}\operatorname{\vec{\times }}_{\operatorname{\mathcal{D}}} \{ Y \} \rightarrow \operatorname{\mathcal{C}}$. The equivalence of $(1)$ and $(2)$ now follows from Proposition 7.1.3.12. The implication $(3) \Rightarrow (2)$ is immediate, and the reverse implication follows from Proposition 7.1.3.6. $\square$