The Stacks project

Proof. This is immediate from the definitions. $\square$

Proof. Omitted. $\square$

Proof. The pullback of a finite free module is finite free. The functor $f^*$ is additive functor hence preserves direct summands. The lemma follows. $\square$

Proof. We may assume $\mathcal{E} = \mathcal{O}_ X^{\oplus n}$ for some $n$. In this case finding the dotted arrow is equivalent to lifting the images of the basis elements in $\Gamma (X, \mathcal{F})$. This is locally possible by the characterization of surjective maps of sheaves (Sheaves, Section 6.16). $\square$

Proof. The first statement follows from the second, hence we only prove (2). We will prove this by induction on the length of the complex $\mathcal{E}^\bullet$. If $\mathcal{E}^\bullet \cong \mathcal{E}[-n]$ for some direct summand $\mathcal{E}$ of a finite free $\mathcal{O}_ X$-module and integer $n \geq a$, then the result follows from Lemma 20.46.5 and the fact that $\mathcal{F}^{n - 1} \to \mathop{\mathrm{Ker}}(\mathcal{F}^ n \to \mathcal{F}^{n + 1})$ is surjective by the assumed vanishing of $H^ n(\mathcal{F}^\bullet )$. If $\mathcal{E}^ i$ is zero except for $i \in [a, b]$, then we have a split exact sequence of complexes

which determines a distinguished triangle in $K(\mathcal{O}_ X)$. Hence an exact sequence

by the axioms of triangulated categories. The composition $\mathcal{E}^ b[-b] \to \mathcal{F}^\bullet$ is locally homotopic to zero, whence we may assume our map comes from an element in the left hand side of the displayed exact sequence above. This element is locally zero by induction hypothesis. $\square$

Proof. Our assumptions on $f$ imply the cone $C(f)^\bullet$ has vanishing cohomology sheaves in degrees $\geq a$. Hence Lemma 20.46.6 guarantees there is an open covering $X = \bigcup U_ i$ such that the composition $\mathcal{E}^\bullet \to \mathcal{F}^\bullet \to C(f)^\bullet$ is homotopic to zero over $U_ i$. Since

restricts to a distinguished triangle in $K(\mathcal{O}_{U_ i})$ we see that we can lift $\alpha |_{U_ i}$ up to homotopy to a map $\alpha _ i : \mathcal{E}^\bullet |_{U_ i} \to \mathcal{G}^\bullet |_{U_ i}$ as desired. $\square$

Proof. Proof of (1). By the construction of the derived category we can find a quasi-isomorphism $f : \mathcal{F}^\bullet \to \mathcal{G}^\bullet$ and a map of complexes $\beta : \mathcal{E}^\bullet \to \mathcal{G}^\bullet$ such that $\alpha = f^{-1}\beta$. Thus the result follows from Lemma 20.46.7. We omit the proof of (2). $\square$

Proof. Choose a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ into a K-injective complex. Let $(\mathcal{H}')^\bullet$ be the complex with terms

which represents $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ by the construction in Section 20.42. It suffices to show that the map

is a quasi-isomorphism. Given an open $U \subset X$ we have by inspection

By Lemma 20.46.8 the sheafification of $U \mapsto H^0(\mathcal{H}^\bullet (U))$ is equal to the sheafification of $U \mapsto H^0((\mathcal{H}')^\bullet (U))$. A similar argument can be given for the other cohomology sheaves. Thus $\mathcal{H}^\bullet$ is quasi-isomorphic to $(\mathcal{H}')^\bullet$ which proves the lemma. $\square$

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. $\square$

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. $\square$