**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

Comment #6134 by Johan on

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Comment #9011 by Stacks project on