Theorem 9.2.5.15. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories. For every regular cardinal $\kappa $, the following conditions are equivalent:
The functor $F$ is $\kappa $-left exact.
For every object $D \in \operatorname{\mathcal{D}}$, the $\infty $-category $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{D/}$ is $\kappa $-cofiltered.
Proof. For each object $D \in \operatorname{\mathcal{D}}$, the $\infty $-category $\operatorname{\mathcal{D}}_{D/}$ has an initial object (Proposition 4.6.7.22), and is therefore $\kappa $-cofiltered (Example 9.1.1.6 ). Consequently, if $F$ is $\kappa $-left exact, then the $\infty $-category $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{D/}$ is also $\kappa $-cofiltered. We now prove the converse. Assume that, for each object $D \in \operatorname{\mathcal{D}}$, the $\infty $-category $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{D/}$ is $\kappa $-cofiltered. Let $U: \widetilde{\operatorname{\mathcal{D}}} \rightarrow \operatorname{\mathcal{D}}$ be a left fibration, where $\widetilde{\operatorname{\mathcal{D}}}$ is $\kappa $-cofiltered; we wish to show that the fiber product $\widetilde{\operatorname{\mathcal{C}}} = \operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \widetilde{\operatorname{\mathcal{D}}}$ is also $\kappa $-cofiltered. Applying Example 7.4.5.6, we deduce that $\widetilde{\operatorname{\mathcal{C}}}$ can be realized as the colimit of a diagram
For each object $X \in \widetilde{\operatorname{\mathcal{D}}}$, the $\infty $-category $\mathscr {F}(X)$ is equivalent to $\operatorname{\mathcal{C}}\times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{D}}_{ U(X)/ }$ (Corollary 4.6.4.20), and is therefore $\kappa $-cofiltered. Since $\widetilde{\operatorname{\mathcal{D}}}^{\operatorname{op}}$ is $\kappa $-filtered, Proposition 9.1.5.14 implies that $\widetilde{\operatorname{\mathcal{C}}}$ is also $\kappa $-cofiltered. $\square$