Lemma 13.13.8 (05S5)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 13: Derived Categories
Section 13.13: Filtered derived categories
Lemma 13.13.8 (cite)

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Lemma 13.13.8. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet \in K(\text{Fil}^ f(\mathcal{A}))$ .

If $H^ n(\text{gr}(K^\bullet )) = 0$ for all $n < a$ , then there exists a filtered quasi-isomorphism $K^\bullet \to L^\bullet$ with $L^ n = 0$ for all $n < a$ .
If $H^ n(\text{gr}(K^\bullet )) = 0$ for all $n > b$ , then there exists a filtered quasi-isomorphism $M^\bullet \to K^\bullet$ with $M^ n = 0$ for all $n > b$ .
If $H^ n(\text{gr}(K^\bullet )) = 0$ for all $|n| \gg 0$ , then there exists a commutative diagram of morphisms of complexes

$\xymatrix{ K^\bullet \ar[r] & L^\bullet \\ M^\bullet \ar[u] \ar[r] & N^\bullet \ar[u] }$

where all the arrows are filtered quasi-isomorphisms, $L^\bullet$ bounded below, $M^\bullet$ bounded above, and $N^\bullet$ a bounded complex.

Proof. Suppose that $H^ n(\text{gr}(K^\bullet )) = 0$ for all $n < a$ . By Homology, Lemma 12.19.15 the sequence

$K^{a - 1} \xrightarrow {d^{a - 2}} K^{a - 1} \xrightarrow {d^{a - 1}} K^ a$

is an exact sequence of objects of $\mathcal{A}$ and the morphisms $d^{a - 2}$ and $d^{a - 1}$ are strict. Hence $\mathop{\mathrm{Coim}}(d^{a - 1}) = \mathop{\mathrm{Im}}(d^{a - 1})$ in $\text{Fil}^ f(\mathcal{A})$ and the map $\text{gr}(\mathop{\mathrm{Im}}(d^{a - 1})) \to \text{gr}(K^ a)$ is injective with image equal to the image of $\text{gr}(K^{a - 1}) \to \text{gr}(K^ a)$ , see Homology, Lemma 12.19.13. This means that the map $K^\bullet \to \tau _{\geq a}K^\bullet$ into the truncation

$\tau _{\geq a}K^\bullet = (\ldots \to 0 \to K^ a/\mathop{\mathrm{Im}}(d^{a - 1}) \to K^{a + 1} \to \ldots )$

is a filtered quasi-isomorphism. This proves (1). The proof of (2) is dual to the proof of (1). Part (3) follows formally from (1) and (2). $\square$

Comments (1)

Comment #9777 by ZL on September 22, 2024 at 05:40

Typo: $K^{a - 2} \xrightarrow {d^{a - 2}} K^{a - 1} \xrightarrow {d^{a - 1}} K^ a$

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