Lemma 13.30.2. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor of abelian categories. If

1. every object of $\mathcal{A}$ is a subobject of an object which is right acyclic for $F$,

2. there exists an integer $n$ such that $R^ nF = 0$,

then $RF : D(\mathcal{A}) \to D(\mathcal{B})$ exists. Any complex consisting of right acyclic objects for $F$ computes $RF$ and any complex is the source of a quasi-isomorphism into such a complex.

Proof. Note that the first condition implies that $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ exists, see Proposition 13.17.8. 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 13.30.1 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 13.30.1 is our assumption (1). The second assumption of Lemma 13.30.1 follows from the fact that $RF(A \oplus B) = RF(A) \oplus RF(B)$. The third assumption of Lemma 13.30.1 follows from the long exact cohomology sequence. Hence for every complex $K^\bullet$ there exists a quasi-isomorphism $K^\bullet \to L^\bullet$ with $L^ n$ right acyclic for $F$. 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 are done by Lemma 13.15.15. To prove the claim pick an integer $i \in \mathbf{Z}$. Consider the distinguished triangle

$\sigma _{\geq i - n - 1}L^\bullet \to \sigma _{\geq i - n - 1}M^\bullet \to Q^\bullet ,$

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 $R^ jF(Q^\bullet )$ 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 13.17.8 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. $\square$

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