Lemma 48.30.4. Let $j : U \to X$ be an open immersion of Noetherian schemes. Let

\[ K \to L \to M \to K[1] \]

be a distinguished triangle of $D^ b_{\textit{Coh}}(\mathcal{O}_ U)$. Then there exists an inverse system of distinguished triangles

\[ K_ n \to L_ n \to M_ n \to K_ n[1] \]

in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ such that $(K_ n)$, $(L_ n)$, $(M_ n)$ are Deligne systems and such that the restriction of these distinguished triangles to $U$ is isomorphic to the distinguished triangle we started out with.

**Proof.**
Let $(K_ n)$ be as in Lemma 48.30.3 part (2). Choose an object $L'$ of $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ whose restriction to $U$ is $L$ (we can do this as the lemma shows). By Lemma 48.30.1 we can find an $n$ and a morphism $K_ n \to L'$ on $X$ whose restriction to $U$ is the given arrow $K \to L$. We conclude there is a morphism $K' \to L'$ of $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ whose restriction to $U$ is the given arrow $K \to L$.

By Derived Categories of Schemes, Proposition 36.11.2 we can find a morphism $\alpha ^\bullet : \mathcal{F}^\bullet \to \mathcal{G}^\bullet $ of bounded complexes of coherent $\mathcal{O}_ X$-modules representing $K' \to L'$. Choose a quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ whose vanishing locus is $X \setminus U$. Then we let $K_ n = \mathcal{I}^ n\mathcal{F}^\bullet $ and $L_ n = \mathcal{I}^ n\mathcal{G}^\bullet $. Observe that $\alpha ^\bullet $ induces a morphism of complexes $\alpha _ n^\bullet : \mathcal{I}^ n\mathcal{F}^\bullet \to \mathcal{I}^ n\mathcal{G}^\bullet $. From the construction of cones in Derived Categories, Section 13.9 it is clear that

\[ C(\alpha _ n)^\bullet = \mathcal{I}^ nC(\alpha ^\bullet ) \]

and hence we can set $M_ n = C(\alpha _ n)^\bullet $. Namely, we have a compatible system of distinguished triangles (see discussion in Derived Categories, Section 13.12)

\[ K_ n \to L_ n \to M_ n \to K_ n[1] \]

whose restriction to $U$ is isomorphic to the distinguished triangle we started out with by axiom TR3 and Derived Categories, Lemma 13.4.3.
$\square$

## Comments (0)