Kerodon

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$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 6.2.3.6. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a cocartesian fibration of simplicial sets. The following conditions are equivalent:

$(1)$

The morphism $U$ is a cartesian fibration of simplicial sets.

$(2)$

For every edge $e: C \rightarrow D$ of the simplicial set $\operatorname{\mathcal{C}}$, the covariant transport functor $e_{!}: \operatorname{\mathcal{E}}_{C} \rightarrow \operatorname{\mathcal{E}}_{D}$ of Notation 5.2.2.9 admits a right adjoint.

Moreover, if these conditions are satisfied and $e: C \rightarrow D$ is an edge of $\operatorname{\mathcal{C}}$, then the contravariant transport functor $e^{\ast }: \operatorname{\mathcal{E}}_{D} \rightarrow \operatorname{\mathcal{E}}_{C}$ of Notation 5.2.2.18 is right adjoint to $e_{!}$.

Proof. Assume first that condition $(1)$ is satisfied and let $e: C \rightarrow D$ be an edge of the simplicial set $\operatorname{\mathcal{C}}$, which we identify with a morphism $\Delta ^1 \rightarrow \operatorname{\mathcal{C}}$. Applying Proposition 6.2.3.4 to the projection map $\Delta ^1 \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}\rightarrow \Delta ^1$, we deduce that the covariant transport functor $e_{!}: \operatorname{\mathcal{E}}_{C} \rightarrow \operatorname{\mathcal{E}}_{D}$ is right adjoint to the contravariant transport functor $e^{\ast } \operatorname{\mathcal{E}}_{D} \rightarrow \operatorname{\mathcal{E}}_{C}$, which proves $(2)$.

We now show that $(2)$ implies $(1)$. Let $e: C \rightarrow D$ be an edge of $\operatorname{\mathcal{C}}$, and let $Y$ be an object of the $\infty $-category $\operatorname{\mathcal{E}}_{D}$. If the covariant transport functor $e_{!}$ admits a right adjoint $e^{\ast }$, then Proposition 6.2.3.4 guarantees that the projection map $U_{e}: \Delta ^1 \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}\rightarrow \Delta ^1$ is a cartesian fibration. In particular, we can choose an object $X = e^{\ast }(Y)$ together with a locally $U$-cartesian edge $f: X \rightarrow Y$ satisfying $U(f) = e$. Applying Lemma 6.2.3.5, we conclude that $f$ is $U$-cartesian. $\square$