The Stacks project

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

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)

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

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

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. $\square$

Proof. Represent $K_0$ by any complex $\mathcal{K}_0^\bullet $ of $\mathcal{O}_0$-modules. Apply Lemma 91.16.1 with $\mathcal{G}^ n = 0$ for all $n$. Denote $d : \mathcal{F}^ n \to \mathcal{F}^{n + 1}$ the maps produced by the lemma. Then we see that $d \circ d : \mathcal{F}^ n \to \mathcal{F}^{n + 2}$ is zero modulo $\mathcal{I}$. Since $\mathcal{F}^ n$ is flat, we see that $\mathcal{I}\mathcal{F}^ n = \mathcal{F}^ n \otimes _{\mathcal{O}} \mathcal{I} = \mathcal{F}^ n_0 \otimes _{\mathcal{O}_0} \mathcal{I}$. Hence we obtain a canonical map of complexes

Since $\mathcal{F}_0^\bullet $ is a bounded above complex of flat $\mathcal{O}_0$-modules, it is K-flat and may be used to compute derived tensor product. Moreover, the map of complexes $\mathcal{F}_0^\bullet \to \mathcal{K}_0^\bullet $ is a quasi-isomorphism by construction. Therefore the source and target of the map just constructed represent $K_0$ and $K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}[2]$ and we obtain our map $\omega (K_0)$.

Let us show that this procedure is compatible with maps of complexes. Namely, let $\mathcal{L}_0^\bullet $ represent another object of $D^-(\mathcal{O}_0)$ and suppose that

is a map of complexes. Apply Lemma 91.16.1 for the complex $\mathcal{L}_0^\bullet $, the flat modules $\mathcal{F}^ n$, the maps $\mathcal{F}^ n \to \mathcal{F}^{n + 1}$, and the compositions $\mathcal{F}^ n \to \mathcal{K}_0^ n \to \mathcal{L}_0^ n$ (we apologize for the reversal of letters used). We obtain flat modules $\mathcal{G}^ n$, maps $\mathcal{F}^ n \to \mathcal{G}^ n$, maps $\mathcal{G}^ n \to \mathcal{G}^{n + 1}$, and maps $\mathcal{G}^ n \to \mathcal{L}_0^ n$ with all properties as in the lemma. Then it is clear that

is a commutative diagram of complexes.

To see that $\omega (K_0)$ is well defined, suppose that we have two complexes $\mathcal{K}_0^\bullet $ and $(\mathcal{K}'_0)^\bullet $ of $\mathcal{O}_0$-modules representing $K_0$ and two systems $(\mathcal{F}^ n, d : \mathcal{F}^ n \to \mathcal{F}^{n + 1}, \mathcal{F}^ n \to \mathcal{K}_0^ n)$ and $((\mathcal{F}')^ n, d : (\mathcal{F}')^ n \to (\mathcal{F}')^{n + 1}, (\mathcal{F}')^ n \to \mathcal{K}_0^ n)$ as above. Then we can choose a complex $(\mathcal{K}''_0)^\bullet $ and quasi-isomorphisms $\mathcal{K}_0^\bullet \to (\mathcal{K}''_0)^\bullet $ and $(\mathcal{K}'_0)^\bullet \to (\mathcal{K}''_0)^\bullet $ realizing the fact that both complexes represent $K_0$ in the derived category. Next, we apply the result of the previous paragraph to

This produces a commutative diagram

Since the vertical arrows give quasi-isomorphisms on the summands we conclude the desired commutativity in $D(\mathcal{O}_0)$.

Having established well-definedness, the statement on compatibility with maps is a consequence of the result in the second paragraph. $\square$

Proof. Choose a distinguished triangle

In case (2) we see that $H^ i(C) = 0$ for $i \geq n$. Hence $H^ i(C \otimes _\mathcal {O}^\mathbf {L} \mathcal{F}) = 0$ for $i \geq n$ by (the dual of) Derived Categories, Lemma 13.16.1. This in turn shows that $H^ i(\alpha \otimes _\mathcal {O}^\mathbf {L} \text{id}_\mathcal {F})$ is an isomorphism for $i > n$ and surjective for $i = n$. In case (1) we moreover see that $H^{n - 1}(L) \to H^{n - 1}(C)$ is surjective. Considering the diagram

we conclude the lower horizontal arrow is surjective. Combined with what was said before this implies that $H^ n(\alpha \otimes _\mathcal {O}^\mathbf {L} \text{id}_\mathcal {F})$ is an isomorphism. $\square$

Proof. Let $K$ be as in (2). Then we can represent $K$ by a bounded above complex $\mathcal{F}^\bullet $ of flat $\mathcal{O}$-modules. Then $\mathcal{F}_0^\bullet = \mathcal{F}^\bullet \otimes _{\mathcal{O}} \mathcal{O}_0$ represents $K_0$ in $D(\mathcal{O}_0)$. Since $d_{\mathcal{F}^\bullet } \circ d_{\mathcal{F}^\bullet } = 0$ as $\mathcal{F}^\bullet $ is a complex, we see from the very construction of $\omega (K_0)$ that it is zero.

Assume (1). Let $\mathcal{F}^ n$, $d : \mathcal{F}^ n \to \mathcal{F}^{n + 1}$ be as in the construction of $\omega (K_0)$. The nullity of $\omega (K_0)$ implies that the map

is zero in $D(\mathcal{O}_0)$. By definition of the derived category as the localization of the homotopy category of complexes of $\mathcal{O}_0$-modules, there exists a quasi-isomorphism $\alpha : \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $ such that there exist $\mathcal{O}_0$-modules maps $h^ n : \mathcal{G}_0^ n \to \mathcal{F}_0^{n + 1} \otimes _\mathcal {O} \mathcal{I}$ with

We set

and we define

with obvious notation using that $\mathcal{F}_0^{n + 1} \otimes _{\mathcal{O}_0} \mathcal{I} = \mathcal{F}^{n + 1} \otimes _\mathcal {O} \mathcal{I} = \mathcal{I}\mathcal{F}^{n + 1} \subset \mathcal{F}^{n + 1}$. Then one checks $d' \circ d' = 0$ by our choice of $h^ n$ and definition of $\omega $. Hence $\mathcal{H}^\bullet $ defines an object in $D(\mathcal{O})$. On the other hand, there is a short exact sequence of complexes of $\mathcal{O}$-modules

We still have to show that $\mathcal{H}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0$ is isomorphic to $K_0$. Choose a quasi-isomorphism $\mathcal{E}^\bullet \to \mathcal{H}^\bullet $ where $\mathcal{E}^\bullet $ is a bounded above complex of flat $\mathcal{O}$-modules. We obtain a commutative diagram

We claim that $\delta $ is a quasi-isomorphism. Since $H^ i(\delta )$ is an isomorphism for $i \gg 0$, we can use descending induction on $n$ such that $H^ i(\delta )$ is an isomorphism for $i \geq n$. Observe that $\mathcal{E}^\bullet \otimes _\mathcal {O} \mathcal{I}$ represents $\mathcal{E}_0^\bullet \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}$, that $\mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}$ represents $\mathcal{G}_0^\bullet \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}$, and that $\beta = \delta \otimes _{\mathcal{O}_0}^\mathbf {L} \text{id}_\mathcal {I}$ as maps in $D(\mathcal{O}_0)$. This is true because $\beta = (\alpha \otimes \text{id}_\mathcal {I}) \circ (\delta \otimes \text{id}_\mathcal {I})$. Suppose that $H^ i(\delta )$ is an isomorphism in degrees $\geq n$. Then the same is true for $\beta $ by what we just said and Lemma 91.16.3. Then we can look at the diagram

Using Homology, Lemma 12.5.19 we see that $H^{n - 1}(\delta )$ is surjective. This in turn implies that $H^{n - 1}(\beta )$ is surjective by Lemma 91.16.3. Using Homology, Lemma 12.5.19 again we see that $H^{n - 1}(\delta )$ is an isomorphism. The claim holds by induction, so $\delta $ is a quasi-isomorphism which is what we wanted to show. $\square$

Proof. Set $\mathcal{F}_0^\bullet = \mathcal{F}^\bullet \otimes _\mathcal {O} \mathcal{O}_0$. By Derived Categories, Lemma 13.9.8 there exists a factorization

of the given map such that the first arrow has an inverse up to homotopy and the second arrow is termwise split surjective. Hence we may assume that $\mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet $ is termwise surjective. In that case we take

and everything is clear. $\square$

Proof. Finding $\alpha : K \to L$ lifing $\alpha _0$ is the same as finding $\alpha : K \to L$ such that the composition $K \xrightarrow {\alpha } L \to L_0$ is equal to the composition $K \to K_0 \xrightarrow {\alpha _0} L_0$. The short exact sequence $0 \to \mathcal{I} \to \mathcal{O} \to \mathcal{O}_0 \to 0$ gives rise to a canonical distinguished triangle

in $D(\mathcal{O})$. By Derived Categories, Lemma 13.4.2 the composition

is zero if and only if we can find $\alpha : K \to L$ lifting $\alpha _0$. The composition is an element in

by adjunction. $\square$

Proof. An automorphism of $(K, \alpha )$ is a map $\varphi : K \to K$ in $D(\mathcal{O})$ with $\varphi \otimes _\mathcal {O} \text{id}_{\mathcal{O}_0} = \text{id}$. This is the same thing as saying that

is zero. We conclude the group of automorphisms is the cokernel of a map

by the distinguished triangle

in $D(\mathcal{O})$ and Derived Categories, Lemma 13.4.2. To translate into the groups in the lemma use adjunction of the restriction functor $D(\mathcal{O}_0) \to D(\mathcal{O})$ and $- \otimes _\mathcal {O} \mathcal{O}_0 : D(\mathcal{O}) \to D(\mathcal{O}_0)$. This proves (1).

Proof of (2). Assume that $K_0 = K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0$ in $D(\mathcal{O})$. By Lemma 91.16.6 the map sending a lift $(K', \alpha _0)$ to the obstruction $o(\alpha _0)$ to lifting $\alpha _0$ defines a canonical injective map from the set of isomomorphism classes of pairs to $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_0}(K_0, K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})$. To finish the proof we show that it is surjective. Pick $\xi : K_0 \to (K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})[1]$ in the $\mathop{\mathrm{Ext}}\nolimits ^1$ of the lemma. Choose a bounded above complex $\mathcal{F}^\bullet $ of flat $\mathcal{O}$-modules representing $K$. The map $\xi $ can be represented as $t \circ s^{-1}$ where $s : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet $ is a quasi-isomorphism and $t : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1]$ is a map of complexes. By Lemma 91.16.5 we can assume there exists a quasi-isomorphism $\mathcal{G}^\bullet \to \mathcal{F}^\bullet $ of complexes of $\mathcal{O}$-modules such that $\mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $ factors through $s$ up to homotopy. We may and do replace $\mathcal{G}^\bullet $ by a bounded above complex of flat $\mathcal{O}$-modules (by picking a qis from such to $\mathcal{G}^\bullet $ and replacing). Then we see that $\xi $ is represented by a map of complexes $t : \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1]$ and the quasi-isomorphism $\mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $. Set

with differentials

This makes sense as $\mathcal{F}_0^{n + 1} \otimes _{\mathcal{O}_0} \mathcal{I} = \mathcal{F}^{n + 1} \otimes _\mathcal {O} \mathcal{I} = \mathcal{I}\mathcal{F}^{n + 1} \subset \mathcal{F}^{n + 1}$. We omit the computation that shows that $\mathcal{H}^\bullet $ is a complex of $\mathcal{O}$-modules. By construction there is a short exact sequence

of complexes of $\mathcal{O}$-modules. Exactly as in the proof of Lemma 91.16.4 one shows that this sequence induces an isomorphism $\alpha _0 : \mathcal{H}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \to \mathcal{G}_0^\bullet $ in $D(\mathcal{O}_0)$. In other words, we have produced a pair $(\mathcal{H}^\bullet , \alpha _0)$. We omit the verification that $o(\alpha _0) = \xi $; hint: $o(\alpha _0)$ can be computed explicitly in this case as we have maps $\mathcal{H}^ n \to \mathcal{F}^ n$ (not compatible with differentials) lifting the components of $\alpha _0$. This finishes the proof. $\square$