# Kerodon

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Example 8.6.5.5. Let $U: \operatorname{\mathcal{E}}\rightarrow \Delta ^1$ be a cocartesian fibration of $\infty$-categories. By virtue of Remark 5.2.4.3, the cocartesian fibration $U$ can be recovered (up to equivalence) from its homotopy transport representation, which we can identify with the functor $F: \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{E}}_1$ given by covariant transport along the nondegenerate edge of $\Delta ^1$. The fibration $U$ then a cocartesian dual $U': \operatorname{\mathcal{E}}' \rightarrow \Delta ^1$, whose covariant transport functor can be identified with the composition

$\operatorname{\mathcal{E}}'_{0} \simeq \operatorname{\mathcal{E}}_{0}^{\operatorname{op}} \xrightarrow { F^{\operatorname{op}} } \operatorname{\mathcal{E}}_{1}^{\operatorname{op}} \simeq \operatorname{\mathcal{E}}'_{1}$

(Corollary 8.6.4.10). Applying Proposition 6.2.3.5, we deduce the following:

$(a)$

The cocartesian fibration $U$ is a cartesian fibration if and only if the functor $F: \operatorname{\mathcal{E}}_0 \rightarrow \operatorname{\mathcal{E}}_1$ admits a right adjoint.

$(b)$

The cocartesian fibration $U'$ is a cartesian fibration if and only if the functor $F^{\operatorname{op}}: \operatorname{\mathcal{E}}_0^{\operatorname{op}} \rightarrow \operatorname{\mathcal{E}}_{1}^{\operatorname{op}}$ admits a right adjoint: that is, if and only if the functor $F$ admits a left adjoint.

Note that conditions $(a)$ and $(b)$ are not equivalent. If $(a)$ is satisfied and $(b)$ is not, then $U$ admits a cartesian dual $U'': \operatorname{\mathcal{E}}'' \rightarrow \Delta ^1$ which cannot be equivalent to $U'$ (since $U''$ is a cartesian fibration and $U'$ is not).