Lemma 20.44.10. In the situation of Lemma 20.44.9 if $\mathcal{F}^\bullet$ is K-flat, then $\mathcal{H}^\bullet$ is K-flat.

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

$\gamma : \text{Tot}(\mathcal{K}^\bullet \otimes \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \mathcal{F}^\bullet )) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{E}^\bullet , \text{Tot}(\mathcal{K}^\bullet \otimes \mathcal{F}^\bullet ))$

of Lemma 20.39.3. Since $\mathcal{F}^\bullet$ is K-flat, the complex $\text{Tot}(\mathcal{K}^\bullet \otimes \mathcal{F}^\bullet )$ is acyclic, and hence by Lemma 20.44.8 (or Lemma 20.44.9 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

$0 \to \sigma _{\geq a + 1} \mathcal{E}^\bullet \to \mathcal{E}^\bullet \to \sigma _{\leq a} \mathcal{E}^\bullet \to 0$

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. $\square$

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