The Stacks project

Proof. Note that the first assumption implies that $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ exists, see Proposition 13.16.8. Let $A$ be an object of $\mathcal{A}$. Choose an injection $A \to A'$ with $A'$ acyclic. Then we see that $R^{n + 1}F(A) = R^ nF(A'/A) = 0$ by the long exact cohomology sequence. Hence we conclude that $R^{n + 1}F = 0$. Continuing like this using induction we find that $R^ mF = 0$ for all $m \geq n$.

We are going to use Lemma 13.32.1 with the function $d : \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}) \to \{ 0, 1, 2, \ldots \}$ given by $d(A) = \max \{ 0\} \cup \{ i \mid R^ iF(A) \not= 0\}$. The first assumption of Lemma 13.32.1 is our assumption (1). The second assumption of Lemma 13.32.1 follows from the fact that $RF(A \oplus B) = RF(A) \oplus RF(B)$. The third assumption of Lemma 13.32.1 follows from the long exact cohomology sequence. Hence for every complex $K^\bullet$ there exists a quasi-isomorphism $K^\bullet \to L^\bullet$ into a complex of objects right acyclic for $F$. This proves statement (3).

We claim that if $L^\bullet \to M^\bullet$ is a quasi-isomorphism of complexes of right acyclic objects for $F$, then $F(L^\bullet ) \to F(M^\bullet )$ is a quasi-isomorphism. If we prove this claim then we get statements (1) and (2) of the lemma by Lemma 13.14.15. To prove the claim pick an integer $i \in \mathbf{Z}$. Consider the distinguished triangle

i.e., let $Q^\bullet$ be the cone of the first map. Note that $Q^\bullet$ is bounded below and that $H^ j(Q^\bullet )$ is zero except possibly for $j = i - n - 1$ or $j = i - n - 2$. We may apply $RF$ to $Q^\bullet$. Using the second spectral sequence of Lemma 13.21.3 and the assumed vanishing of cohomology (2) we conclude that $H^ j(RF(Q^\bullet ))$ is zero except possibly for $j \in \{ i - n - 2, \ldots , i - 1\}$. Hence we see that $RF(\sigma _{\geq i - n - 1}L^\bullet ) \to RF(\sigma _{\geq i - n - 1}M^\bullet )$ induces an isomorphism of cohomology objects in degrees $\geq i$. By Proposition 13.16.8 we know that $RF(\sigma _{\geq i - n - 1}L^\bullet ) = \sigma _{\geq i - n - 1}F(L^\bullet )$ and $RF(\sigma _{\geq i - n - 1}M^\bullet ) = \sigma _{\geq i - n - 1}F(M^\bullet )$. We conclude that $F(L^\bullet ) \to F(M^\bullet )$ is an isomorphism in degree $i$ as desired.

Part (4)(a) follows from Lemma 13.16.1.

For part (4)(b) let $E$ be represented by the complex $L^\bullet$ of objects right acyclic for $F$. By part (2) $RF(E)$ is represented by the complex $F(L^\bullet )$ and $RF(\sigma _{\geq c}L^\bullet )$ is represented by $\sigma _{\geq c}F(L^\bullet )$. Consider the distinguished triangle

of Remark 13.12.4. The vanishing established above gives that $H^ i(RF(\tau _{\geq b - n}L^\bullet ))$ agrees with $H^ i(RF(\tau _{\geq b - n + 1}L^\bullet ))$ for $i \geq b$. Consider the short exact sequence of complexes

Using the distinguished triangle associated to this (see Section 13.12) and the vanishing as before we conclude that $H^ i(RF(\tau _{\geq b - n}L^\bullet ))$ agrees with $H^ i(RF(\sigma _{\geq b - n}L^\bullet ))$ for $i \geq b$. Since the map $RF(\sigma _{\geq b - n}L^\bullet ) \to RF(L^\bullet )$ is represented by $\sigma _{\geq b - n}F(L^\bullet ) \to F(L^\bullet )$ we conclude that this in turn agrees with $H^ i(RF(L^\bullet ))$ for $i \geq b$ as desired.

Proof of (4)(c). Under the assumption on $E$ we have $\tau _{\leq a - 1}E = 0$ and we get the vanishing of $H^ i(RF(E))$ for $i \leq a - 1$ from part (4)(a). Similarly, we have $\tau _{\geq b + 1}E = 0$ and hence we get the vanishing of $H^ i(RF(E))$ for $i \geq b + n$ from part (4)(b). $\square$