The Stacks project

Lemma 21.43.8. In the situation above, there exists a cardinal $\kappa $ with the following properties:

  1. for every nonzero object $K$ of $\mathit{QC}(\mathcal{O})$ there exists a nonzero morphism $E \to K$ of $\mathit{QC}(\mathcal{O})$ such that $|E| \leq \kappa $,

  2. for every morphism $\alpha : E \to \bigoplus _ n K_ n$ of $\mathit{QC}(\mathcal{O})$ such that $|E| \leq \kappa $, there exist morphisms $E_ n \to K_ n$ in $\mathit{QC}(\mathcal{O})$ with $|E_ n| \leq \kappa $ such that $\alpha $ factors through $\bigoplus E_ n \to \bigoplus K_ n$.

Proof. Let $\kappa $ be an upper bound for the following set of cardinals:

  1. $|\coprod _ V j_{U!}\mathcal{O}_ U(V)|$ for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

  2. the cardinals $\kappa (\mathcal{O}(V) \to \mathcal{O}(U))$ found in More on Algebra, Lemma 15.102.5 for all morphisms $U \to V$ in $\mathcal{C}$,

  3. the cardinal found in Lemma 21.43.7.

We claim that for any complex $\mathcal{F}^\bullet $ representing an object of $\mathit{QC}(\mathcal{O})$ and any subcomplex $\mathcal{S}^\bullet \subset \mathcal{F}^\bullet $ with $|\mathcal{S}^\bullet | \leq \kappa $ there exists a subcomplex $\mathcal{H}^\bullet $ of $\mathcal{F}^\bullet $ containing $\mathcal{S}^\bullet $ such that $\mathcal{H}^\bullet $ represents an object of $\mathit{QC}(\mathcal{O})$ and such that $|\mathcal{H}^\bullet | \leq \kappa $. In the next two paragraphs we show that the claim implies the lemma.

As in (1) let $K$ be a nonzero object of $\mathit{QC}(\mathcal{O})$. Say $K$ is represented by the complex of $\mathcal{O}$-modules $\mathcal{F}^\bullet $. Then $H^ i(\mathcal{F}^\bullet )$ is nonzero for some $i$. Hence there exists an object $U$ of $\mathcal{C}$ and a section $s \in \mathcal{F}^ i(U)$ with $d(s) = 0$ which determines a nonzero section of $H^ i(\mathcal{F}^\bullet )$ over $U$. Then the image of $s : j_{U!}\mathcal{O}_ U[-i] \to \mathcal{F}^\bullet $ is a subcomplex $\mathcal{S}^\bullet \subset \mathcal{F}^\bullet $ with $|\mathcal{S}^\bullet | \leq \kappa $. Applying the claim we get $\mathcal{H}^\bullet \to \mathcal{F}^\bullet $ in $\mathit{QC}(\mathcal{O})$ nonzero with $|\mathcal{H}^\bullet | \leq \kappa $. Thus (1) holds.

Let $\alpha : E \to \bigoplus K_ n$ be as in (2). Choose any complexes $\mathcal{K}_ n^\bullet $ representing $K_ n$. Then $\bigoplus \mathcal{K}_ n^\bullet $ represents $\bigoplus K_ n$. By the construction of the derived category we can represent $E$ by a complex $\mathcal{E}^\bullet $ such that $\alpha $ is represented by a morphism $a : \mathcal{E}^\bullet \to \bigoplus \mathcal{K}_ n^\bullet $ of complexes. By Lemma 21.43.7 and our choice of $\kappa $ above we may assume $|\mathcal{E}^\bullet | \leq \kappa $. By the claim we get subcomplexes $\mathcal{E}_ n^\bullet \subset \mathcal{K}_ n^\bullet $ representing objects $E_ n$ of $\mathit{QC}(\mathcal{O})$ with $|E_ n| \leq \kappa $ containing the image of $a_ n : \mathcal{E}^\bullet \to \mathcal{K}_ n^\bullet $ as desired.

Proof of the claim. Let $\mathcal{F}^\bullet $ be a complex representing an object of $\mathit{QC}(\mathcal{O})$ and let $\mathcal{S}^\bullet \subset \mathcal{F}^\bullet $ be a subcomplex of size $\leq \kappa $. We are going to inductively construct subcomplexes

\[ \mathcal{S}^\bullet = \mathcal{S}_0^\bullet \subset \mathcal{S}_1^\bullet \subset \mathcal{S}_2^\bullet \subset \ldots \subset \mathcal{F}^\bullet \]

of size $\leq \kappa $ such that for every morphism $f : U \to V$ of $\mathcal{C}$ and every $i \in \mathbf{Z}$

  1. the kernel of the arrow $H^ i(\mathcal{S}_ n^\bullet (V) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U)) \to H^ i(\mathcal{S}_ n^\bullet (U))$ maps to zero in $H^ i(\mathcal{S}_{n + 1}^\bullet (V) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U))$,

  2. the image of the arrow $H^ i(\mathcal{S}_ n^\bullet (U)) \to H^ i(\mathcal{S}_{n + 1}^\bullet (U))$ is contained in the image of $H^ i(\mathcal{S}_{n + 1}^\bullet (V) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U)) \to H^ i(\mathcal{S}_{n + 1}^\bullet (U))$,

Once this is done we can set $\mathcal{H}^\bullet = \bigcup \mathcal{S}_ n^\bullet $. Namely, since derived tensor product and taking cohomology of complexes of modules over rings commute with filtered colimits, the conditions (1) and (2) together will guarantee that

\[ \mathcal{H}^\bullet (V) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U) \longrightarrow \mathcal{H}^\bullet (U) \]

is an isomorphism on cohomology in all degrees and hence an isomorphism in $D(\mathcal{O}(U))$ for all $f : U \to V$ in $\mathcal{C}$. Hence $\mathcal{H}^\bullet $ represents an object of $\mathit{QC}(\mathcal{O})$ as desired.

Construction of $\mathcal{S}_{n + 1}$ given $\mathcal{S}_ n$. For every morphism $f : U \to V$ of $\mathcal{C}$ we consider the commutative diagram

\[ \xymatrix{ \mathcal{S}_ n^\bullet (V) \ar[r] \ar[d] & \mathcal{S}_ n^\bullet (U) \ar[d] \\ \mathcal{F}^\bullet (V) \ar[r] & \mathcal{F}^\bullet (U) } \]

This is a diagram as in More on Algebra, Lemma 15.102.5 for the ring map $\mathcal{O}(V) \to \mathcal{O}(U)$, i.e., the bottom row induces an isomorphism

\[ \mathcal{F}^\bullet (V) \otimes _{\mathcal{O}(V)}^\mathbf {L} \mathcal{O}(U) \longrightarrow \mathcal{F}^\bullet (U) \]

in $D(\mathcal{O}(U))$. Thus we may choose subcomplexes

\[ \mathcal{S}_ n^\bullet (V) \subset M^\bullet _ f \subset \mathcal{F}^\bullet (V) \quad \text{and}\quad \mathcal{S}_ n^\bullet (U) \subset N^\bullet _ f \subset \mathcal{F}^\bullet (U) \]

as in More on Algebra, Lemma 15.102.5 and in particular we see that $|N^ i_ f|, |M^ i_ f| \leq \kappa $. Next, we apply Lemma 21.43.6 using the subsets

\[ \mathcal{S}_ n^ i(U) \amalg \coprod \nolimits _{f : U \to V} N^ i_ f \amalg \coprod \nolimits _{g : W \to U} M^ i_ g \subset \mathcal{F}^ i(U) \]

to find a subcomplex

\[ \mathcal{S}_ n^\bullet \subset \mathcal{S}_{n + 1}^\bullet \subset \mathcal{F}^\bullet \]

with containing those subsets and such that $|\mathcal{S}_{n + 1}^\bullet | \leq \kappa $. Conditions (1) and (2) hold because the corresponding statements hold for $\mathcal{S}_ n^\bullet (V) \subset M^\bullet _ f$ and $\mathcal{S}_ n^\bullet (U) \subset N^\bullet _ f$ by the construction in More on Algebra, Lemma 15.102.5. Thus the proof is complete. $\square$


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