The Stacks project

Proof. Let $A^\bullet $ be a bounded complex of right $F$-acyclic objects. We claim that $RF$ is defined at $A^\bullet $ and that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^+(\mathcal{B})$. Namely, it holds for complexes with at most one nonzero right $F$-acyclic object for example by Lemma 13.16.4. Next, suppose that $A^ n = 0$ for $n \not\in [a, b]$. Using the “stupid” truncations we obtain a termwise split short exact sequence of complexes

see Homology, Section 12.15. Thus a distinguished triangle $(\sigma _{\geq a + 1} A^\bullet , A^\bullet , \sigma _{\leq a} A^\bullet )$. By induction hypothesis $RF$ is defined for the two outer complexes and these complexes compute $RF$. Then the same is true for the middle one by Lemma 13.14.12.

Suppose that $A^\bullet $ is a bounded below complex of acyclic objects such that $RF$ is defined at $A^\bullet $. To show that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^ i(F(A^\bullet )) \to H^ i(RF(A^\bullet ))$ is an isomorphism for all $i$. Pick $i$. Consider the termwise split short exact sequence of complexes

Note that this induces a termwise split short exact sequence

Hence we get distinguished triangles

Since $RF$ is defined at $A^\bullet $ (by assumption) and at $\sigma _{\leq i + 1}A^\bullet $ (by the first paragraph) we see that $RF$ is defined at $\sigma _{\geq i + 1}A^\bullet $ and we get a distinghuished triangle

See Lemma 13.14.6. Using these distinguished triangles we obtain a map of exact sequences

By the results of the first paragraph the map $\beta $ is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof it suffices to show that $H^ i(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$ and $H^{i + 1}(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$. This follows immediately from Lemma 13.16.1. $\square$