The Stacks project

Proof. Let us say a filtered complex $L^\bullet $ is basic if each $L^ n$, $F^ iL^ n$, $\text{gr}^ iL^ n$ is a free $R$-module and if all differentials are zero.

There exists a basic filtered complex $P_0^\bullet $ and a map $f_0 : P_0^\bullet \to K^\bullet $ of filtered complexes such that $f_0$ and $F^ if_0$, $i \in \mathbf{Z}$ are surjective on cohomology. To see this set

with zero differentials and the map $f_0$ defined by $f_0(\xi _ z) = z$. As for the filtration, we set

We leave it to the reader to check that this gives $f_0 : P_0^\bullet \to K^\bullet $ as claimed.

By induction on $m \geq 0$ we are going to construct embeddings

of filtered complexes and maps $f_ m : P_ m^\bullet \to K^\bullet $ with the following properties

1. the filtered complex $P_{m + 1}^\bullet / P_ m^\bullet $ is basic,

2. the kernel of $H^ n(f_ m)$ and $H^ n(F^ if_ m)$ maps to zero in $H^ n(P_{m + 1}^\bullet )$ and $H^ n(F^ iP_{m + 1}^\bullet )$,

3. the map $f_{m + 1} : P_{m + 1}^\bullet \to K^\bullet $ extends the map $f_ m$.

To to this, set

Note that $\Omega _{n, m}$ surjects onto the kernel of $H^ n(f_ m)$ and that $\Omega _{n, i, m}$ surjects onto the kernel of $H^ n(F^ if_ n)$. For each $z \in \Omega _{n, m}$, resp. $z \in \Omega _{n, i, m}$ we choose a $y_ z \in K^{n - 1}$, resp. $y_ z \in F^ iK^{n - 1}$ such that $d_ K(y_ z) = f_ m(z)$. Then we set

We set $d_{P_ m}(\eta _ z) = z$ and we set $f_{m + 1}(\eta _ z) = y_ z$. Finally, we set

We leave it to the reader to check that this gives $P_ m^\bullet \subset P_{m + 1}^\bullet $ and $f_{m + 1} : P_{m + 1}^\bullet \to K^\bullet $ as claimed.

At this point we simply take

as a filtered complex with map $f : P^\bullet \to K^\bullet $ given by $\bigcup f_ m$.

Part (1) of the statement of the lemma holds because $P^ n$ as a filtered module is isomorphic to the direct sum of $P_0^ n$ and $P_{m + 1}^ n/P_ m^ n$ for $m \geq 0$. Small detail omitted.

Part (3) of the statement. Observe that $H^ n(P^\bullet ) = \mathop{\mathrm{colim}}\nolimits H^ n(P_ m^\bullet )$. The map $f_0$ is surjective on cohomology; whence the same holds for each $f_ m$. For each $m$ by construction the embedding $P_ m^\bullet \subset P_{m + 1}^\bullet $ kills the kernel of $H^ n(f_ m)$. Combining these facts the reader easily concludes that $H^ n(f)$ is an isomorphism. Similarly for $H^ n(F^ if_ n)$. Then also $H^ n(\text{gr}^ if)$ must be an isomorphism because of the short exact sequence $0 \to F^{i + 1} \to F^ i \to \text{gr}^ i \to 0$ (of functors on the category of filtered complexes, say). Small detail omitted.

Part (2) of the statement. To see that $P^\bullet $ is K-flat, by Lemma 15.59.8, it suffices to show that $P_ m^\bullet $ is K-flat. By Lemma 15.59.6 and induction it suffices to note that a complex with zero differentials and free terms is K-flat. The same argument works to show that $F^ iP^\bullet $ is K-flat for all $i \in \mathbf{Z}$. Finally, we see that $\text{gr}^ iP^\bullet $ is K-flat by another application of Lemma 15.59.6. $\square$