The Stacks project

Proof. Choose a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ where $\mathcal{I}^\bullet$ is a bounded below complex of injectives. Note that $\mathcal{I}^\bullet$ is K-injective (Derived Categories, Lemma 13.31.4). Hence the construction in Section 20.42 shows that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ is represented by the complex $(\mathcal{H}')^\bullet$ with terms

(equality because there are only finitely many nonzero terms). Note that $\mathcal{H}^\bullet$ is the total complex associated to the double complex with terms $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-q}, \mathcal{F}^ p)$ and similarly for $(\mathcal{H}')^\bullet$. The natural map $(\mathcal{H}')^\bullet \to \mathcal{H}^\bullet$ comes from a map of double complexes. Thus to show this map is a quasi-isomorphism, we may use the spectral sequence of a double complex (Homology, Lemma 12.25.3)

converging to $H^{p + q}(\mathcal{H}^\bullet )$ and similarly for $(\mathcal{H}')^\bullet$. To finish the proof of the lemma it suffices to show that $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ induces an isomorphism

on cohomology sheaves whenever $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module. Since this is clear when $\mathcal{E}$ is finite free the result follows. $\square$