Lemma 36.13.9. Let $X$ be an affine scheme. Let $T \subset X$ be a closed subset such that $X \setminus T$ is quasi-compact. Let $U \subset X$ be a quasi-compact open. For every perfect object $F$ of $D(\mathcal{O}_ U)$ supported on $T \cap U$ the object $F \oplus F$ is the restriction of a perfect object $E$ of $D(\mathcal{O}_ X)$ supported in $T$.

Proof. Say $T = V(g_1, \ldots , g_ s)$. After replacing $g_ j$ by a power we may assume multiplication by $g_ j$ is zero on $F$, see Lemma 36.13.8. Choose $E$ as in Lemma 36.13.7. Note that $g_ j : E \to E$ restricts to zero on $U$. Choose a distinguished triangle

$E \xrightarrow {g_1} E \to C_1 \to E$

By Derived Categories, Lemma 13.4.11 the object $C_1$ restricts to $F \oplus F \oplus F \oplus F$ on $U$. Moreover, $g_1 : C_1 \to C_1$ has square zero by Derived Categories, Lemma 13.4.5. Namely, the diagram

$\xymatrix{ E \ar[r] \ar[d]_0 & C_1 \ar[d]_{g_1} \ar[r] & E \ar[d]_0 \\ E \ar[r] & C_1 \ar[r] & E }$

is commutative since the compositions $E \xrightarrow {g_1} E \to C_1$ and $C_1 \to E \xrightarrow {g_1} E$ are zero. Continuing, setting $C_{i + 1}$ equal to the cone of the map $g_ i : C_ i \to C_ i$ we obtain a perfect complex $C_ s$ on $X$ supported on $T$ whose restriction to $U$ gives

$F \oplus F^{\oplus s} \oplus F^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s]$

Choose morphisms of perfect complexes $\beta : C' \to C_ s$ and $\gamma : C' \to C_ s$ as in Lemma 36.13.6 such that $\beta |_ U$ is an isomorphism and such that $\gamma |_ U \circ \beta |_ U^{-1}$ is the morphism

$F \oplus F^{\oplus s} \oplus F^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s] \to F \oplus F^{\oplus s} \oplus F^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s]$

which is the identity on all summands except for $F$ where it is zero. By Lemma 36.13.6 we also have $C' = C_ s \otimes ^\mathbf {L} I$ for some perfect complex $I$ on $X$. Hence the nullity of $g_ j^2\text{id}_{C_ s}$ implies the same thing for $C'$. Thus $C'$ is supported on $T$ as well. Then $\text{Cone}(\gamma )$ is a solution. $\square$

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