The Stacks project

Proof. Let $L \to M \to N \to L[1]$ be a distinguished triangle in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$. Then we obtain a commutative diagram

whose rows are exact by Derived Categories, Lemma 13.4.2 and Algebra, Lemma 10.8.8. Hence if the statement of the lemma holds for $N[-1]$, $L$, $N$, and $L[1]$ then it holds for $M$ by the 5-lemma. Thus, using the distinguished triangles for the canonical truncations of $L$ (see Derived Categories, Remark 13.12.4) we reduce to the case that $L$ has only one nonzero cohomology sheaf.

Choose a bounded complex $\mathcal{F}^\bullet $ of coherent $\mathcal{O}_ X$-modules and a quasi-coherent ideal $\mathcal{I} \subset \mathcal{O}_ X$ cutting out $X \setminus U$ such that $K_ n$ is represented by $\mathcal{I}^ n\mathcal{F}^\bullet $. Using “stupid” truncations we obtain compatible termwise split short exact sequences of complexes

which in turn correspond to compatible systems of distinguished triangles in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$. Arguing as above we reduce to the case where $\mathcal{F}^\bullet $ has only one nonzero term. This reduces us to the case discussed in the next paragraph.

Given a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ and a coherent $\mathcal{O}_ X$-module $\mathcal{G}$ we have to show that the canonical map

is an isomorphism for all $i \geq 0$. For $i = 0$ this is Cohomology of Schemes, Lemma 30.10.5. Assume $i > 0$.

Injectivity. Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{I}^ n\mathcal{F}, \mathcal{G})$ be an element whose restriction to $U$ is zero. We have to show there exists an $m \geq n$ such that the restriction of $\xi $ to $\mathcal{I}^ m\mathcal{F} = \mathcal{I}^{m - n}\mathcal{I}^ n\mathcal{F}$ is zero. After replacing $\mathcal{F}$ by $\mathcal{I}^ n\mathcal{F}$ we may assume $n = 0$, i.e., we have $\xi \in \mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{F}, \mathcal{G})$ whose restriction to $U$ is zero. By Derived Categories of Schemes, Proposition 36.11.2 we have $D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D^ b(\textit{Coh}(\mathcal{O}_ X))$. Hence we can compute the $\mathop{\mathrm{Ext}}\nolimits $ group in the abelian category of coherent $\mathcal{O}_ X$-modules. This implies there exists an surjection $\alpha : \mathcal{F}'' \to \mathcal{F}$ such that $\xi \circ \alpha = 0$ (this is where we use that $i > 0$). Set $\mathcal{F}' = \mathop{\mathrm{Ker}}(\alpha )$ so that we have a short exact sequence

It follows that $\xi $ is the image of an element $\xi ' \in \mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ X(\mathcal{F}', \mathcal{G})$ whose restriction to $U$ is in the image of $\mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ U(\mathcal{F}''|_ U, \mathcal{G}|_ U) \to \mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ U(\mathcal{F}'|_ U, \mathcal{G}|_ U)$. By Artin-Rees the inverse systems $(\mathcal{I}^ n\mathcal{F}')$ and $(\mathcal{I}^ n \mathcal{F}'' \cap \mathcal{F}')$ are pro-isomorphic, see Cohomology of Schemes, Lemma 30.10.3. Since we have the compatible system of short exact sequences

we obtain a commutativew diagram

with exact rows. By induction on $i$ and the comment on inverse systems above we find that the left two vertical arrows are isomorphisms. Now $\xi $ gives an element in the top right group which is the image of $\xi '$ in the middle top group, which in turn maps to an element of the bottom middle group coming from some element in the left bottom group. We conclude that $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{I}^ n\mathcal{F}, \mathcal{G})$ for some $n$ as desired.

Surjectivity. Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^ i_ U(\mathcal{F}|_ U, \mathcal{G}|_ U)$. Arguing as above using that $i > 0$ we can find an surjection $\mathcal{H} \to \mathcal{F}|_ U$ of coherent $\mathcal{O}_ U$-modules such that $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^ i_ U(\mathcal{H}, \mathcal{G}|_ U)$. Then we can find a map $\varphi : \mathcal{F}'' \to \mathcal{F}$ of coherent $\mathcal{O}_ X$-modules whose restriction to $U$ is $\mathcal{H} \to \mathcal{F}|_ U$, see Properties, Lemma 28.22.4. Observe that the lemma doesn't guarantee $\varphi $ is surjective but this won't matter (it is possible to pick a surjective $\varphi $ with a little bit of additional work). Denote $\mathcal{F}' = \mathop{\mathrm{Ker}}(\varphi )$. The short exact sequence

shows that $\xi $ is the image of $\xi '$ in $\mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ U(\mathcal{F}'|_ U, \mathcal{G}|_ U)$. By induction on $i$ we can find an $n$ such that $\xi '$ is the image of some $\xi '_ n$ in $\mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ X(\mathcal{I}^ n\mathcal{F}', \mathcal{G})$. By Artin-Rees we can find an $m \geq n$ such that $\mathcal{F}' \cap \mathcal{I}^ m\mathcal{F}'' \subset \mathcal{I}^ n\mathcal{F}'$. Using the short exact sequence

the image of $\xi '_ n$ in $\mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ X(\mathcal{F}' \cap \mathcal{I}^ m\mathcal{F}'', \mathcal{G})$ maps by the boundary map to an element $\xi _ m$ of $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{I}^ m\mathop{\mathrm{Im}}(\varphi ), \mathcal{G})$ which maps to $\xi $. Since $\mathop{\mathrm{Im}}(\varphi )$ and $\mathcal{F}$ agree over $U$ we see that $\mathcal{F}/\mathcal{I}^ m\mathop{\mathrm{Im}}(\varphi )$ is supported on $X \setminus U$. Hence there exists an $l \geq m$ such that $\mathcal{I}^ l\mathcal{F} \subset \mathcal{I}^ m\mathop{\mathrm{Im}}(\varphi )$, see Cohomology of Schemes, Lemma 30.10.2. Taking the image of $\xi _ m$ in $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(\mathcal{I}^ l\mathcal{F}, \mathcal{G})$ we win. $\square$