ScalingStacks

Proof. First, by proposition 5.15, it suffices to check that

$$\Ho(\Ind({\mathcal{A}})^{\kappa})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho(\Ind({\mathcal{B}})^{\kappa})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho(\Ind({\mathcal{C}})^{\kappa})$$

is an exact sequence of triangulated categories. Again, we will deduce this from Neeman’s generalization of Thomason’s localization theorem (see [61, 4.4.9] or  [62]), as follows. First, observe that [53, 1.4.5.1] implies that there is an equivalence $\Ho(\Ind({\mathcal{A}})^{\kappa})\simeq\Ho(\Ind({\mathcal{A}}))^{\kappa}$ (and analogous equivalences for the other terms in the sequence). Next, since $\Ind({\mathcal{A}})$ and $\Ind({\mathcal{B}})$ are presentable, the criterion of [50] (characterizing well-generated triangulated categories) and [53, 1.4.5.2] imply that $\Ho(\Ind({\mathcal{B}}))$ is well-generated and (since the map $\Ind({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ind({\mathcal{B}})$ is fully-faithful) the image of $\Ho(\Ind({\mathcal{A}}))$ is a localizing subcategory generated by a small set of objects. Once again, the localization theorem [51, 7.2.1] implies that

$$\Ho(\Ind({\mathcal{B}}))^{\kappa}/\Ho(\Ind({\mathcal{A}}))^{\kappa}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho(\Ind({\mathcal{B}}/{\mathcal{A}}))^{\kappa}$$

is an equivalence up to idempotent completion. The hypothesis that ${\mathcal{B}}/{\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ is an equivalence up to idempotent completion now implies the result. ∎