Lemma 13.11.6 (05RW)—The Stacks project

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 below 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 May 04, 2023 at 04:11

Typo: the first sentence of the second paragraph should be "suppose $K^\bullet,L^\bullet$ are bounded below complexes". Also, in the first paragraph I would write "the statement on kernels in (1), (2), (3) is a consequence of the definitions in each case plus Homology, Lemma 12.8.3". Finally, in the second paragraph, I would invoke Lemma 4.27.6 to justify the sentence "this means that there exists a quasi-isomorphism $s':(L')^\bullet\to (L'')^\bullet$ such that $s'\circ f=0$ ". (An alternative to invocation of Lemma 4.27.6 would be creating a corollary of 4.27.6 in Section 12.8 that states "if $\mathcal{C}$ is a preadditive category and $S$ is a LMS in $\mathcal{C}$ , then for a morphism $f:X\to Y$ in $\mathcal{C}$ , the morphism $Q(f)\in S^{-1}\mathcal{C}$ is zero if and only if there is $s:Y\to Z$ in $S$ with $s\circ f=0$ " and then referring to it.)

Comment #8978 by Stacks project on April 13, 2024 at 15:29

Thanks for the typo which is fixed here. I think that looking at Lemma 13.6.11 whose statement refers to Lemma 13.6.10 we do see that the kernel of each localization functor is the set of acyclic complexes. I think the reader should at this point be sufficiently familiar with computations in localizations of categories to be able to see why such an $s'$ exists, but yes I agree that this can be done using Lemma 4.27.6.

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