Proof.
Assume K^\bullet satisfies the assumptions of (1). Let s : 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. From Categories, Lemma 4.22.11 we deduce that RF is the value of the essentially constant ind-object F(L^\bullet ) for these s. This means that \text{id} : RF(K^\bullet ) \to RF(K^\bullet ) factors through F(L^\bullet ) for some complex L^\bullet with L^ n = 0 for n < a. 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
Comment #8399 by Elías Guisado on
Comment #9011 by Stacks project on