Variant 9.1.4.7. Let $\kappa $, $\lambda $, and $\mu $ be infinite cardinals. Assume that $\kappa $ is uncountable and that $\mu $ has cofinality $\geq \kappa $ and exponential cofinality $\geq \lambda $. Let $\operatorname{\mathcal{K}}$ be a $\kappa $-small $\infty $-category and let $L$ be a $\lambda $-small simplicial set. If $\operatorname{\mathcal{K}}$-indexed colimits commute with $L$-indexed limits in the $\infty $-category $\operatorname{\mathcal{S}}^{< \mu }$, then the diagonal map $\operatorname{\mathcal{K}}\rightarrow \operatorname{Fun}( L^{\operatorname{op}}, \operatorname{\mathcal{K}})$ is right cofinal.
Proof. By virtue of Corollary 7.2.3.9, it will suffice to show that for every diagram $f: L^{\operatorname{op}} \rightarrow \operatorname{\mathcal{K}}$, the coslice $\infty $-category $\operatorname{\mathcal{K}}_{f/}$ is weakly contractible. Let $h^{\bullet }: \operatorname{\mathcal{K}}^{\operatorname{op}} \rightarrow \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{S}}^{< \mu } )$ be a contravariant Yoneda embedding for $\operatorname{\mathcal{K}}$. Since the $\infty $-category $\operatorname{\mathcal{S}}^{< \mu }$ admits $L$-indexed limits, the composite diagram
admits a limit $\mathscr {F} \in \operatorname{Fun}( \operatorname{\mathcal{K}}, \operatorname{\mathcal{S}}^{< \mu } )$. It follows from Corollary 8.4.2.8 that $\mathscr {F}$ is a covariant transport representation for the left fibration $\operatorname{\mathcal{K}}_{f/} \rightarrow \operatorname{\mathcal{K}}$. Consequently, the $\infty $-category $\operatorname{\mathcal{K}}_{f/}$ is weakly contractible if and only if the colimit $\varinjlim (\mathscr {F} )$ is contractible (Proposition 7.4.3.1). Invoking our assumption that $\operatorname{\mathcal{K}}$-indexed colimits commute with $L$-indexed limits, we are reduced to proving that for each vertex $\ell \in L$, the corepresentable functor $\operatorname{Hom}_{\operatorname{\mathcal{K}}}( f(\ell ), \bullet )$ has contractible colimit, which follows from Example 7.4.3.7. $\square$