Lemma 13.11.6. Let $\mathcal{A}$ be an abelian category. The subcategories $\text{Ac}^{+}(\mathcal{A})$, $\text{Ac}^{-}(\mathcal{A})$, resp. $\text{Ac}^ b(\mathcal{A})$ are strictly full saturated triangulated subcategories of $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, resp. $K^ b(\mathcal{A})$. The corresponding saturated multiplicative systems (see Lemma 13.6.10) are the sets $\text{Qis}^{+}(\mathcal{A})$, $\text{Qis}^{-}(\mathcal{A})$, resp. $\text{Qis}^ b(\mathcal{A})$.
The kernel of the functor $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is $\text{Ac}^{+}(\mathcal{A})$ and this induces an equivalence of triangulated categories
\[ K^{+}(\mathcal{A})/\text{Ac}^{+}(\mathcal{A}) = \text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{A}) \]
The kernel of the functor $K^{-}(\mathcal{A}) \to D^{-}(\mathcal{A})$ is $\text{Ac}^{-}(\mathcal{A})$ and this induces an equivalence of triangulated categories
\[ K^{-}(\mathcal{A})/\text{Ac}^{-}(\mathcal{A}) = \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{A}) \]
The kernel of the functor $K^ b(\mathcal{A}) \to D^ b(\mathcal{A})$ is $\text{Ac}^ b(\mathcal{A})$ and this induces an equivalence of triangulated categories
\[ K^ b(\mathcal{A})/\text{Ac}^ b(\mathcal{A}) = \text{Qis}^ b(\mathcal{A})^{-1}K^ b(\mathcal{A}) \longrightarrow D^ b(\mathcal{A}) \]
Proof.
The initial statements follow from Lemma 13.6.11 by considering the restriction of the homological functor $H^0$. The statement on kernels in (1), (2), (3) is a consequence of the definitions in each case. Each of the functors is essentially surjective by Lemma 13.11.5. To finish the proof we have to show the functors are fully faithful. We first do this for the bounded below version.
Suppose that $K^\bullet , L^\bullet $ are bounded above complexes. A morphism between these in $D(\mathcal{A})$ is of the form $s^{-1}f$ for a pair $f : K^\bullet \to (L')^\bullet $, $s : L^\bullet \to (L')^\bullet $ where $s$ is a quasi-isomorphism. This implies that $(L')^\bullet $ has cohomology bounded below. Hence by Lemma 13.11.5 we can choose a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet $ with $(L'')^\bullet $ bounded below. Then the pair $(s' \circ f, s' \circ s)$ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. Hence the functor is “full”. Finally, suppose that the pair $f : K^\bullet \to (L')^\bullet $, $s : L^\bullet \to (L')^\bullet $ defines a morphism in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$ which is zero in $D(\mathcal{A})$. This means that there exists a quasi-isomorphism $s' : (L')^\bullet \to (L'')^\bullet $ such that $s' \circ f = 0$. Using Lemma 13.11.5 once more we obtain a quasi-isomorphism $s'' : (L'')^\bullet \to (L''')^\bullet $ with $(L''')^\bullet $ bounded below. Thus we see that $s'' \circ s' \circ f = 0$ which implies that $s^{-1}f$ is zero in $\text{Qis}^{+}(\mathcal{A})^{-1}K^{+}(\mathcal{A})$. This finishes the proof that the functor in (1) is an equivalence.
The proof of (2) is dual to the proof of (1). To prove (3) we may use the result of (2). Hence it suffices to prove that the functor $\text{Qis}^ b(\mathcal{A})^{-1}K^ b(\mathcal{A}) \to \text{Qis}^{-}(\mathcal{A})^{-1}K^{-}(\mathcal{A})$ is fully faithful. The argument given in the previous paragraph applies directly to show this where we consistently work with complexes which are already bounded above.
$\square$
Comments (2)
Comment #8372 by Elías Guisado on
Comment #8978 by Stacks project on