The Stacks project

Lemma 91.16.1. Let $\mathcal{C}$ be a site. Let $\mathcal{O} \to \mathcal{O}_0$ be a surjection of sheaves of rings. Assume given the following data

  1. flat $\mathcal{O}$-modules $\mathcal{G}^ n$,

  2. maps of $\mathcal{O}$-modules $\mathcal{G}^ n \to \mathcal{G}^{n + 1}$,

  3. a complex $\mathcal{K}_0^\bullet $ of $\mathcal{O}_0$-modules,

  4. maps of $\mathcal{O}$-modules $\mathcal{G}^ n \to \mathcal{K}_0^ n$

such that

  1. $H^ n(\mathcal{K}_0^\bullet ) = 0$ for $n \gg 0$,

  2. $\mathcal{G}^ n = 0$ for $n \gg 0$,

  3. with $\mathcal{G}^ n_0 = \mathcal{G}^ n \otimes _\mathcal {O} \mathcal{O}_0$ the induced maps determine a complex $\mathcal{G}_0^\bullet $ and a map of complexes $\mathcal{G}_0^\bullet \to \mathcal{K}_0^\bullet $.

Then there exist

  1. flat $\mathcal{O}$-modules $\mathcal{F}^ n$,

  2. maps of $\mathcal{O}$-modules $\mathcal{F}^ n \to \mathcal{F}^{n + 1}$,

  3. maps of $\mathcal{O}$-modules $\mathcal{F}^ n \to \mathcal{K}_0^ n$,

  4. maps of $\mathcal{O}$-modules $\mathcal{G}^ n \to \mathcal{F}^ n$,

such that $\mathcal{F}^ n = 0$ for $n \gg 0$, such that the diagrams

\[ \xymatrix{ \mathcal{G}^ n \ar[r] \ar[d] & \mathcal{G}^{n + 1} \ar[d] \\ \mathcal{F}^ n \ar[r] & \mathcal{F}^{n + 1} } \]

commute for all $n$, such that the composition $\mathcal{G}^ n \to \mathcal{F}^ n \to \mathcal{K}_0^ n$ is the given map $\mathcal{G}^ n \to \mathcal{K}_0^ n$, and such that with $\mathcal{F}^ n_0 = \mathcal{F}^ n \otimes _\mathcal {O} \mathcal{O}_0$ we obtain a complex $\mathcal{F}_0^\bullet $ and map of complexes $\mathcal{F}_0^\bullet \to \mathcal{K}_0^\bullet $ which is a quasi-isomorphism.

Proof. We will prove by descending induction on $e$ that we can find $\mathcal{F}^ n$, $\mathcal{G}^ n \to \mathcal{F}^ n$, and $\mathcal{F}^ n \to \mathcal{F}^{n + 1}$ for $n \geq e$ fitting into a commutative diagram

\[ \xymatrix{ \ldots \ar[r] & \mathcal{G}^{e - 1} \ar[r] \ar@/_2pc/[dd] & \mathcal{G}^ e \ar[d] \ar[r] \ar@/_2pc/[dd] & \mathcal{G}^{e + 1} \ar[d] \ar[r] \ar@/_2pc/[dd]|\hole & \ldots \\ & & \mathcal{F}^ e \ar[d] \ar[r] & \mathcal{F}^{e + 1} \ar[d] \ar[r] & \ldots \\ \ldots \ar[r] & \mathcal{K}_0^{e - 1} \ar[r] & \mathcal{K}_0^ e \ar[r] & \mathcal{K}_0^{e + 1} \ar[r] & \ldots } \]

such that $\mathcal{F}_0^\bullet $ is a complex, the induced map $\mathcal{F}_0^\bullet \to \mathcal{K}_0^\bullet $ induces an isomorphism on $H^ n$ for $n > e$ and a surjection for $n = e$. For $e \gg 0$ this is true because we can take $\mathcal{F}^ n = 0$ for $n \geq e$ in that case by assumptions (a) and (b).

Induction step. We have to construct $\mathcal{F}^{e - 1}$ and the maps $\mathcal{G}^{e - 1} \to \mathcal{F}^{e - 1}$, $\mathcal{F}^{e - 1} \to \mathcal{F}^ e$, and $\mathcal{F}^{e - 1} \to \mathcal{K}_0^{e - 1}$. We will choose $\mathcal{F}^{e - 1} = A \oplus B \oplus C$ as a direct sum of three pieces.

For the first we take $A = \mathcal{G}^{e - 1}$ and we choose our map $\mathcal{G}^{e - 1} \to \mathcal{F}^{e - 1}$ to be the inclusion of the first summand. The maps $A \to \mathcal{K}^{e - 1}_0$ and $A \to \mathcal{F}^ e$ will be the obvious ones.

To choose $B$ we consider the surjection (by induction hypothesis)

\[ \gamma : \mathop{\mathrm{Ker}}(\mathcal{F}^ e_0 \to \mathcal{F}^{e + 1}_0) \longrightarrow \mathop{\mathrm{Ker}}(\mathcal{K}^ e_0 \to \mathcal{K}^{e + 1}_0)/ \mathop{\mathrm{Im}}(\mathcal{K}^{e - 1}_0 \to \mathcal{K}^ e_0) \]

We can choose a set $I$, for each $i \in I$ an object $U_ i$ of $\mathcal{C}$, and sections $s_ i \in \mathcal{F}^ e(U_ i)$, $t_ i \in \mathcal{K}^{e - 1}_0(U_ i)$ such that

  1. $s_ i$ maps to a section of $\mathop{\mathrm{Ker}}(\gamma ) \subset \mathop{\mathrm{Ker}}(\mathcal{F}^ e_0 \to \mathcal{F}^{e + 1}_0)$,

  2. $s_ i$ and $t_ i$ map to the same section of $\mathcal{K}^ e_0$,

  3. the sections $s_ i$ generate $\mathop{\mathrm{Ker}}(\gamma )$ as an $\mathcal{O}_0$-module.

We omit giving the full justification for this; one uses that $\mathcal{F}^ e \to \mathcal{F}^ e_0$ is a surjective maps of sheaves of sets. Then we set to put

\[ B = \bigoplus \nolimits _{i \in I} j_{U_ i!}\mathcal{O}_{U_ i} \]

and define the maps $B \to \mathcal{F}^ e$ and $B \to \mathcal{K}_0^{e - 1}$ by using $s_ i$ and $t_ i$ to determine where to send the summand $j_{U_ i!}\mathcal{O}_{U_ i}$.

With $\mathcal{F}^{e - 1} = A \oplus B$ and maps as above, this produces a diagram as above for $e - 1$ such that $\mathcal{F}_0^\bullet \to \mathcal{K}_0^\bullet $ induces an isomorphism on $H^ n$ for $n \geq e$. To get the map to be surjective on $H^{e - 1}$ we choose the summand $C$ as follows. Choose a set $J$, for each $j \in J$ an object $U_ j$ of $\mathcal{C}$ and a section $t_ j$ of $\mathop{\mathrm{Ker}}(\mathcal{K}^{e - 1}_0 \to \mathcal{K}^ e_0)$ over $U_ j$ such that these sections generate this kernel over $\mathcal{O}_0$. Then we put

\[ C = \bigoplus \nolimits _{j \in J} j_{U_ j!}\mathcal{O}_{U_ j} \]

and the zero map $C \to \mathcal{F}^ e$ and the map $C \to \mathcal{K}_0^{e - 1}$ by using $s_ j$ to determine where to the summand $j_{U_ j!}\mathcal{O}_{U_ j}$. This finishes the induction step by taking $\mathcal{F}^{e - 1} = A \oplus B \oplus C$ and maps as indicated. $\square$


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