Lemma 9.2.6.12. Let $\kappa \leq \lambda $ be regular cardinals, let $\operatorname{\mathcal{C}}$ be an $\infty $-category which admits $\lambda $-small $\kappa $-filtered colimits, and let $\operatorname{\mathcal{C}}_0 \subseteq \operatorname{\mathcal{C}}$ be a full subcategory. Assume that $\operatorname{\mathcal{C}}_0$ generates $\operatorname{\mathcal{C}}$ under $\lambda $-small $\kappa $-filtered colimits (in the sense of Remark 9.2.6.11). Then every $(\kappa ,\lambda )$-compact object $C \in \operatorname{\mathcal{C}}$ is a retract of an object of $\operatorname{\mathcal{C}}_0$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Let $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ be the full subcategory spanned by those objects $D$ which satisfy the following condition:
- $(\ast )$
Every morphism $f: C \rightarrow D$ factors (up to homotopy) through an object of $\operatorname{\mathcal{C}}_0$.
Then $\operatorname{\mathcal{C}}'$ contains $\operatorname{\mathcal{C}}_0$, and our assumption that $C$ is $(\kappa ,\lambda )$-compact guarantees that $\operatorname{\mathcal{C}}'$ is closed under the formation of $\lambda $-small $\kappa $-filtered colimits. Since $\operatorname{\mathcal{C}}_0$ generates $\operatorname{\mathcal{C}}$ under $\lambda $-small $\kappa $-filtered colimits, we must have $\operatorname{\mathcal{C}}' = \operatorname{\mathcal{C}}$. In particular, $\operatorname{\mathcal{C}}'$ contains the object $C$, so the identity morphism $\operatorname{id}_{C}: C \rightarrow C$ factors through an object of $\operatorname{\mathcal{C}}_0$. $\square$