Remark 11.5.0.113. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. Then $U$ is a cartesian fibration if and only if the opposite functor $U^{\operatorname{op}}: \operatorname{\mathcal{E}}^{\operatorname{op}} \rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}}$ is a cocartesian fibration.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$