Lemma 13.11.5. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex.

1. If $H^ n(K^\bullet ) = 0$ for all $n \ll 0$, then there exists a quasi-isomorphism $K^\bullet \to L^\bullet$ with $L^\bullet$ bounded below.

2. If $H^ n(K^\bullet ) = 0$ for all $n \gg 0$, then there exists a quasi-isomorphism $M^\bullet \to K^\bullet$ with $M^\bullet$ bounded above.

3. If $H^ n(K^\bullet ) = 0$ 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 quasi-isomorphisms, $L^\bullet$ bounded below, $M^\bullet$ bounded above, and $N^\bullet$ a bounded complex.

Proof. Pick $a \ll 0 \ll b$ and set $M^\bullet = \tau _{\leq b}K^\bullet$, $L^\bullet = \tau _{\geq a}K^\bullet$, and $N^\bullet = \tau _{\leq b}L^\bullet = \tau _{\geq a}M^\bullet$. See Homology, Section 12.15 for the truncation functors. $\square$

Comment #4314 by Linyuan Liu on

I think we should pick $b\ll 0\ll a$ instead of $a\ll 0\ll b$. Besides, I don't think $N^{\bullet}=L^{\bullet}/M^{\bullet}$ makes sense. We could take $N^{\bullet}=M^{\bullet}/\tau_{\leq b} M^{\bullet}$.

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