ScalingStacks

Proof. Write ${\mathcal{B}}\simeq\colim_{\alpha}{\mathcal{B}}_{\alpha}$ as a $\kappa$-filtered colimit of $\kappa$-compact stable $\infty$-categories ${\mathcal{B}}_{\alpha}$, and define ${\mathcal{A}}_{\alpha}={\mathcal{A}}\times_{\mathcal{B}}{\mathcal{B}}_{\alpha}$ to be the full subcategory of ${\mathcal{A}}$ consisting of those objects of ${\mathcal{A}}$ which lie in the image of ${\mathcal{B}}_{\alpha}$. Evidently, ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}/{\mathcal{A}}$ is the $\kappa$-filtered colimit of the exact sequences ${\mathcal{A}}_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}_{\alpha}/{\mathcal{A}}_{\alpha}$, and ${\mathcal{A}}_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}_{\alpha}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}_{\alpha}/{\mathcal{A}}_{\alpha}$ is strict-exact because if $Y\in{\mathcal{B}}_{\alpha}$ is a summand of $X\in{\mathcal{A}}_{\alpha}$ then $Y\in{\mathcal{A}}_{\alpha}$ because the image of $Y$ in ${\mathcal{B}}$ lies in ${\mathcal{A}}$. ∎

Proof. Write ${\mathcal{A}}\simeq\colim_{\alpha}{\mathcal{A}}_{\alpha}$ as a $\kappa$-filtered colimit of $\kappa$-compact small stable $\infty$-categories ${\mathcal{A}}_{\alpha}$. Then $\Idem({\mathcal{A}})\simeq\colim_{\alpha}\Idem({\mathcal{A}}_{\alpha})$, since $\Idem$ (viewed as an endofunctor of $\Cat_{\infty}^{\ex}$) commutes with $\kappa$-filtered colimits — this follows from the characterization of $\Idem$ in terms of a subcategory of the $\Ind$ category [52, 5.4.2.4] and the fact that filtered colimits in $\Cat_{\infty}^{\ex}$ can be computed in $\Cat_{\infty}$ [53, 1.1.4.6]. ∎