The Stacks project

Lemma 13.13.9. Let $\mathcal{A}$ be an abelian category. The subcategories $\text{FAc}^{+}(\mathcal{A})$, $\text{FAc}^{-}(\mathcal{A})$, resp. $\text{FAc}^ b(\mathcal{A})$ are strictly full saturated triangulated subcategories of $K^{+}(\text{Fil}^ f\mathcal{A})$, $K^{-}(\text{Fil}^ f\mathcal{A})$, resp. $K^ b(\text{Fil}^ f\mathcal{A})$. The corresponding saturated multiplicative systems (see Lemma 13.6.10) are the sets $\text{FQis}^{+}(\mathcal{A})$, $\text{FQis}^{-}(\mathcal{A})$, resp. $\text{FQis}^ b(\mathcal{A})$.

  1. The kernel of the functor $K^{+}(\text{Fil}^ f\mathcal{A}) \to DF^{+}(\mathcal{A})$ is $\text{FAc}^{+}(\mathcal{A})$ and this induces an equivalence of triangulated categories

    \[ K^{+}(\text{Fil}^ f\mathcal{A})/\text{FAc}^{+}(\mathcal{A}) = \text{FQis}^{+}(\mathcal{A})^{-1}K^{+}(\text{Fil}^ f\mathcal{A}) \longrightarrow DF^{+}(\mathcal{A}) \]
  2. The kernel of the functor $K^{-}(\text{Fil}^ f\mathcal{A}) \to DF^{-}(\mathcal{A})$ is $\text{FAc}^{-}(\mathcal{A})$ and this induces an equivalence of triangulated categories

    \[ K^{-}(\text{Fil}^ f\mathcal{A})/\text{FAc}^{-}(\mathcal{A}) = \text{FQis}^{-}(\mathcal{A})^{-1}K^{-}(\text{Fil}^ f\mathcal{A}) \longrightarrow DF^{-}(\mathcal{A}) \]
  3. The kernel of the functor $K^ b(\text{Fil}^ f\mathcal{A}) \to DF^ b(\mathcal{A})$ is $\text{FAc}^ b(\mathcal{A})$ and this induces an equivalence of triangulated categories

    \[ K^ b(\text{Fil}^ f\mathcal{A})/\text{FAc}^ b(\mathcal{A}) = \text{FQis}^ b(\mathcal{A})^{-1}K^ b(\text{Fil}^ f\mathcal{A}) \longrightarrow DF^ b(\mathcal{A}) \]

Proof. This follows from the results above, in particular Lemma 13.13.8, by exactly the same arguments as used in the proof of Lemma 13.11.6. $\square$

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