Lemma 13.16.1 (05TC)—The Stacks project

Lemma 13.16.1. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $K^\bullet $ be a complex of $\mathcal{A}$ and $a \in \mathbf{Z}$.

If $H^ i(K^\bullet ) = 0$ for all $i < a$ and $RF$ is defined at $K^\bullet $, then $H^ i(RF(K^\bullet )) = 0$ for all $i < a$.
If $RF$ is defined at $K^\bullet $ and $\tau _{\leq a}K^\bullet $, then $H^ i(RF(\tau _{\leq a}K^\bullet )) = H^ i(RF(K^\bullet ))$ for all $i \leq a$.

Proof. Assume $K^\bullet $ satisfies the assumptions of (1). Let $K^\bullet \to L^\bullet $ be any quasi-isomorphism. Then it is also true that $K^\bullet \to \tau _{\geq a}L^\bullet $ is a quasi-isomorphism by our assumption on $K^\bullet $. Hence in the category $K^\bullet /\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : K^\bullet \to L^\bullet $ with $L^ n = 0$ for $n < a$ are cofinal. Thus $RF$ is the value of the essentially constant ind-object $F(L^\bullet )$ for these $s$ it follows that $H^ i(RF(K^\bullet )) = 0$ for $i < a$.

To prove (2) we use the distinguished triangle

\[ \tau _{\leq a}K^\bullet \to K^\bullet \to \tau _{\geq a + 1}K^\bullet \to (\tau _{\leq a}K^\bullet )[1] \]

of Remark 13.12.4 to conclude via Lemma 13.14.6 that $RF$ is defined at $\tau _{\geq a + 1}K^\bullet $ as well and that we have a distinguished triangle

\[ RF(\tau _{\leq a}K^\bullet ) \to RF(K^\bullet ) \to RF(\tau _{\geq a + 1}K^\bullet ) \to RF(\tau _{\leq a}K^\bullet )[1] \]

in $D(\mathcal{B})$. By part (1) we see that $RF(\tau _{\geq a + 1}K^\bullet )$ has vanishing cohomology in degrees $< a + 1$. The long exact cohomology sequence of this distinguished triangle then shows what we want. $\square$

Comments (4)

Comment #5949 by Dylan on February 26, 2021 at 13:21

I believe it should be "for i<a" in the last sentence of the first paragraph.

Comment #6134 by Johan on April 30, 2021 at 14:15

Thanks for pointing out this confusing typo. Fix is here.

Comment #8399 by Elías Guisado on May 16, 2023 at 11:06

Typos: instead of «Thus $RF$ is» I think it should be «Thus $RF(K^\bullet)$ is». After, instead of «for these $s$ it follows» it should be «for these $s$ . It follows» On the other hand, in «it follows that $H^i(RF(K^\bullet))=0$ for $i<a$ », maybe one could give more detail? I guess saying «there is a quasi-isomorphism $s:K^\bullet\to L^\bullet$ with $L^n=0$ for $n<a$ and a map $RF(K^\bullet)\to L^\bullet$ as in Categories, Definition 4.22.1, (1). In particular $RF(K^\bullet)\to L^\bullet$ is monic, so $L^\bullet\cong RF(K^\bullet)\oplus E^\bullet$ for some $E^\bullet$ by Lemma 13.4.12, and the claim follows.»

Comment #9011 by Stacks project on April 17, 2024 at 15:23

Very good. Yes, I think this was an incomplete arguement. For example, if you look at the essentially constant system $Z^2 \to Z^2 \to Z^2 \to \ldots$ given following Definition 4.22.2, then the system lives in a subcategory (namely free $Z$ -modules of even rank) but the value does not! Fixed here.

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