Proposition 5.15 (0NM2) · ScalingStacks

0NM2

Proposition 5.15. A sequence of $\kappa$ -cocomplete small stable $\infty$ -categories and $\kappa$ -small colimit preserving functors ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ is exact if and only if the associated sequence $\Ho({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{B}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{C}})$ of triangulated categories is exact, in the sense that the composite is trivial, $\Ho({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{B}})$ is fully faithful, and the map $\Ho({\mathcal{B}})/\Ho({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{C}})$ is an equivalence after idempotent completion.

0NM3

Proof. Suppose ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ is exact. Then the composite is trivial, ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}$ is fully faithful, and ${\mathcal{B}}/{\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{C}}$ is an equivalence up to idempotent completion, and so the same must be true on the level of triangulated homotopy categories. Thus it is enough to show that $\Ho({\mathcal{B}})/\Ho({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{C}})$ is an equivalence up to idempotent completion, which follows from proposition 5.14. Conversely, suppose that

\Ho({\mathcal{A}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{B}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Ho({\mathcal{C}})

is exact. Then ${\mathcal{A}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{B}}$ is fully faithful by proposition 5.10, and the equivalences $\Ho({\mathcal{B}}/{\mathcal{A}})\simeq\Ho({\mathcal{B}})/\Ho({\mathcal{A}})\simeq\Ho({\mathcal{C}})$ (the last up to idempotent completion) implies that ${\mathcal{B}}/{\mathcal{A}}\simeq{\mathcal{C}}$ by corollary 5.11. ∎