Kerodon

Remark 7.7.3.16. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category which is locally cartesian closed. If $\operatorname{\mathcal{C}}$ has a final object $C$, then it is equivalent to the slice $\infty $-category $\operatorname{\mathcal{C}}_{ / C }$, and is therefore cartesian closed. Conversely, if $\operatorname{\mathcal{C}}$ is cartesian closed, then it has a final object (since it admits finite products). Beware that these conditions are not automatic: for example, the empty $\infty $-category $\operatorname{\mathcal{C}}= \emptyset $ is locally cartesian closed, but is not cartesian closed.