Lemma 36.40.2. In Situation 36.9.1 denote $j : U \to X$ the open immersion and let $K$ be the perfect object of $D(\mathcal{O}_ X)$ corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $a \in \mathbf{Z}$. Consider the following conditions
The canonical map $\tau _{\leq a}E \to \tau _{\leq a} Rj_*(E|_ U)$ is an isomorphism, and
$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], E) = 0$ for all $n \leq a$.
Then (2) implies (1) and (1) implies (2) with $a$ replaced by $a - 1$.
Proof.
Choose a distinguished triangle $E \to Rj_*(E|_ U) \to N \to E[1]$. Then (1) implies $\tau _{\leq a - 1}N = 0$ and (1) is implied by $\tau _{\leq a}N = 0$. Observe that
\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], Rj_*(E|_ U)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(K|_ U[-n], E) = 0 \]
for all $n$ as $K|_ U = 0$. Thus (2) is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \leq a$. Observe that there are distinguished triangles
\[ K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i + e''_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e''_ i}, \ldots , f_ r^{e_ r}) \to \ldots \]
of Koszul complexes, see More on Algebra, Lemma 15.28.11. Hence $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[-n], N) = 0$ for all $n \leq a$ is equivalent to $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K_ e[-n], N) = 0$ for all $n \leq a$ and all $e \geq 1$ with $K_ e$ as in Lemma 36.9.6. Since $N|_ U = 0$, that lemma implies that this in turn is equivalent to $H^ n(X, N) = 0$ for $n \leq a$. We conclude that (2) is equivalent to $\tau _{\leq a}N = 0$ since $N$ is determined by the complex of $A$-modules $R\Gamma (X, N)$, see Lemma 36.3.5. Thus we find that our lemma is true.
$\square$
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