Kerodon

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Proposition 9.5.3.7. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a morphism of simplicial sets. Assume that, for each vertex $C \in \operatorname{\mathcal{C}}$, the fiber $\operatorname{\mathcal{E}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ is a presentable $\infty $-category. The following conditions are equivalent:

$(1)$

The morphism $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a presentable fibration (Definition 9.5.3.5). That is, it is a locally cocartesian fibration having the property that, for each edge $e: C \rightarrow C'$ of $\operatorname{\mathcal{C}}$, the covariant transport functor $e_{!}: \operatorname{\mathcal{E}}_{C} \rightarrow \operatorname{\mathcal{E}}_{C'}$ is cocontinuous.

$(2)$

The morphism $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a locally cartesian fibration. Moreover, for each edge $e: C \rightarrow C'$ of $\operatorname{\mathcal{C}}$, the contravariant transport functor $e^{\ast }: \operatorname{\mathcal{E}}_{C'} \rightarrow \operatorname{\mathcal{E}}_{C}$ is accessible and continuous.

Proof. Using Theorem 9.5.2.1, we obtain the following reformulations of conditions $(1)$ and $(2)$:

$(1')$

The morphism $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a locally cocartesian fibration. Moreover, for each edge $e: C \rightarrow C'$ of $\operatorname{\mathcal{C}}$, the covariant transport functor $e_{!}: \operatorname{\mathcal{E}}_{C} \rightarrow \operatorname{\mathcal{E}}_{C'}$ admits a right adjoint.

$(2')$

The morphism $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a locally cartesian fibration. Moreover, for each edge $e: C \rightarrow C'$ of $\operatorname{\mathcal{C}}$, the contravariant transport functor $e^{\ast }: \operatorname{\mathcal{E}}_{C'} \rightarrow \operatorname{\mathcal{E}}_{C}$ admits a left adjoint.

The equivalence of $(1')$ and $(2')$ is a special case of Corollary 6.2.5.5. $\square$