**Proof.**
Assume (1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G[-i], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(G, E[i])$ we see that this is zero for $i \gg 0$ by Lemma 75.17.2. This proves that (1) implies (2).

Parts (2), (3), (4) are equivalent by the discussion in Cohomology on Sites, Section 21.36. Part (5) and (6) are equivalent as $H^ i(X, -) = H^ i(R\Gamma (X, -))$ by definition. The equivalent conditions (2), (3), (4) are equivalent to the equivalent conditions (5), (6) by Cohomology on Sites, Lemma 21.48.4 and the fact that $(G[-i])^\vee = G^\vee [i]$.

It is clear that (7) implies (2). Conversely, let us prove that the equivalent conditions (2) – (6) imply (7). Recall that $G$ is a classical generator for $D_{perf}(\mathcal{O}_ X)$ by Remark 75.16.2. For $P \in D_{perf}(\mathcal{O}_ X)$ let $T(P)$ be the assertion that $R\mathop{\mathrm{Hom}}\nolimits _ X(P, E)$ is in $D^-(\mathbf{Z})$. Clearly, $T$ is inherited by direct sums, satisfies the 2-out-of-three property for distinguished triangles, is inherited by direct summands, and is perserved by shifts. Hence by Derived Categories, Remark 13.36.7 we see that (4) implies $T$ holds on all of $D_{perf}(\mathcal{O}_ X)$. The same argument works for all other properties, except that for property (7)(b) and (7)(c) we also use that $P \mapsto P^\vee $ is a self equivalence of $D_{perf}(\mathcal{O}_ X)$. Small detail omitted.

We will prove the equivalent conditions (2) – (7) imply (1) using the induction principle of Lemma 75.9.3.

First, we prove (2) – (7) $\Rightarrow $ (1) if $X$ is affine. This follows from the case of schemes, see Derived Categories of Schemes, Proposition 36.40.5.

Now assume $(U \subset X, j : V \to X)$ is an elementary distinguished square of quasi-compact and quasi-separated algebraic spaces over $S$ and assume the implication (2) – (7) $\Rightarrow $ (1) is known for $U$, $V$, and $U \times _ X V$. To finish the proof we have to show the implication (2) – (7) $\Rightarrow $ (1) for $X$. Suppose $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ satisfies (2) – (7). By Lemma 75.15.3 and Theorem 75.15.4 there exists a perfect complex $Q$ on $X$ such that $Q|_ U$ generates $D_\mathit{QCoh}(\mathcal{O}_ U)$.

Say $V = \mathop{\mathrm{Spec}}(A)$. Let $Z \subset V$ be the reduced closed subscheme which is the inverse image of $X \setminus U$ and maps isomorphically to it (see Definition 75.9.1). This is a retrocompact closed subset of $V$. Choose $f_1, \ldots , f_ r \in A$ such that $Z = V(f_1, \ldots , f_ r)$. Let $K \in D(\mathcal{O}_ V)$ be the perfect object corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. Note that since $K$ is supported on $Z$, the pushforward $K' = Rj_*K$ is a perfect object of $D(\mathcal{O}_ X)$ whose restriction to $V$ is $K$ (see Lemmas 75.14.3 and 75.10.7). By assumption, we know $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(Q, E)$ and $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K', E)$ are bounded above.

By Lemma 75.10.7 we have $K' = j_!K$ and hence

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K'[-i], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ V)}(K[-i], E|_ V) = 0 \]

for $i \gg 0$. Therefore, we may apply Derived Categories of Schemes, Lemma 36.40.1 to $E|_ V$ to obtain an integer $a$ such that $\tau _{\geq a}(E|_ V) = \tau _{\geq a} R (U \times _ X V \to V)_* (E|_{U \times _ X V})$. Then $\tau _{\geq a} E = \tau _{\geq a} R (U \to X)_* (E |_ U)$ (check that the canonical map is an isomorphism after restricting to $U$ and to $V$). Hence using Lemma 75.29.1 twice we see that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(Q|_ U [-i], E|_ U) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Q[-i], R (U \to X)_* (E|_ U)) = 0 \]

for $i \gg 0$. Since the Proposition holds for $U$ and the generator $Q|_ U$, we have $E|_ U \in D^-_\mathit{QCoh}(\mathcal{O}_ U)$. But then since the functor $R (U \to X)_*$ preserves $D^-_\mathit{QCoh}$ (by Lemma 75.6.1), we get $\tau _{\geq a}E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$. Thus $E \in D^-_\mathit{QCoh}(\mathcal{O}_ X)$.
$\square$

## Comments (0)