Corollary 4.7.9.16 (Transitivity of Local Smallness). Let $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{D}}$ and $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be inner fibrations of simplicial sets, let $\kappa $ be an uncountable regular cardinal, and suppose that $U$ is locally $\kappa $-small. Then $V$ is locally $\kappa $-small if and only if $U \circ V$ is locally $\kappa $-small.
Proof. Suppose first that $V$ is locally $\kappa $-small. Choose an $n$-simplex $\sigma : \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$; we wish to show that the $\infty $-category $\Delta ^ n \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ is locally $\kappa $-small. This follows by applying Proposition 4.7.9.15 to the inner fibration of $\infty $-categories $(\operatorname{id}\times V): \Delta ^ n \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}\rightarrow \Delta ^ n \times _{ \operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$, which is locally $\kappa $-small by virtue of Remark 4.7.9.3.
Now suppose that $U \circ V$ is locally $\kappa $-small, and choose an $n$-simplex $\widetilde{\sigma }: \Delta ^ n \rightarrow \operatorname{\mathcal{D}}$; we wish to show that the fiber product $\Delta ^ n \times _{ \operatorname{\mathcal{D}}} \operatorname{\mathcal{E}}$ is locally $\kappa $-small. To prove this, we are free to replace $\operatorname{\mathcal{D}}$ and $\operatorname{\mathcal{E}}$ by $\Delta ^ n \times _{ \operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ and $\Delta ^ n \times _{ \operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$, respectively, and thereby reduce to the case where $\operatorname{\mathcal{C}}= \Delta ^ n$ is a locally $\kappa $-small $\infty $-category. In this case, our assumptions on $U$ and $U \circ V$ guarantee that the $\infty $-categories $\operatorname{\mathcal{D}}$ and $\operatorname{\mathcal{E}}$ are also locally $\kappa $-small (Proposition 4.7.9.15), so that $V$ is automatically locally $\kappa $-small (by Proposition 4.7.9.15 again). $\square$