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 $a \in \mathbf{Z}$.

  1. 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 $i < a$.

  2. 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 (2)

Comment #5949 by Dylan on

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


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