The Stacks project

Proof. Observe that $\mathcal{H}^\bullet $ is simply the hom complex $\mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ since the boundedness of the strictly prefect complex $\mathcal{E}^\bullet $ insures that the products in the definition of the hom complex turn into direct sums. Let $\mathcal{K}^\bullet $ be an acyclic complex of $\mathcal{O}_ X$-modules. Consider the map

of Lemma 20.41.3. Since $\mathcal{F}^\bullet $ is K-flat, the complex $\text{Tot}(\mathcal{K}^\bullet \otimes \mathcal{F}^\bullet )$ is acyclic, and hence by Lemma 20.46.8 (or Lemma 20.46.9 if you like) the target of $\gamma $ is acyclic too. Hence to prove the lemma it suffices to show that $\gamma $ is an isomorphism of complexes. To see this, we may argue by induction on the length of the complex $\mathcal{E}^\bullet $. If the length is $\leq 1$ then the $\mathcal{E}^\bullet $ is a direct summand of $\mathcal{O}_ X^{\oplus n}[k]$ for some $n \geq 0$ and $k \in \mathbf{Z}$ and in this case the result follows by inspection. If the length is $> 1$, then we reduce to smaller length by considering the termwise split short exact sequence of complexes

for a suitable $a \in \mathbf{Z}$, see Homology, Section 12.15. Then $\gamma $ fits into a morphism of termwise split short exact sequences of complexes. By induction $\gamma $ is an isomorphism for $\sigma _{\geq a + 1} \mathcal{E}^\bullet $ and $\sigma _{\leq a} \mathcal{E}^\bullet $ and hence the result for $\mathcal{E}^\bullet $ follows. Some details omitted. $\square$