Lemma 36.7.4. Let $X$ be a quasi-compact and quasi-separated scheme. Suppose that for every affine open $U \subset X$ the right derived functor

$\Phi : D(\mathit{QCoh}(\mathcal{O}_ U)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$

of the left exact functor $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$ fits into a commutative diagram

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ U)) \ar[d]_\Phi \ar[r]_{i_ U} & D_\mathit{QCoh}(\mathcal{O}_ U) \ar[d]^{Rj_*} \\ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[r]^{i_ X} & D_\mathit{QCoh}(\mathcal{O}_ X) }$

Then the functor (36.3.0.1)

$D(\mathit{QCoh}(\mathcal{O}_ X)) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_ X)$

is an equivalence with quasi-inverse given by $RQ_ X$.

Proof. Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $A$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ X))$. We have to show that the adjunction maps

$RQ_ X(i_ X(A)) \to A \quad \text{and}\quad E \to i_ X(RQ_ X(E))$

are isomorphisms. Consider the hypothesis $H_ n$: the adjunction maps above are isomorphisms whenever $E$ and $i_ X(A)$ are supported (Definition 36.6.1) on a closed subset of $X$ which is contained in the union of $n$ affine opens of $X$. We will prove $H_ n$ by induction on $n$.

Base case: $n = 0$. In this case $E = 0$, hence the map $E \to i_ X(RQ_ X(E))$ is an isomorphism. Similarly $i_ X(A) = 0$. Thus the cohomology sheaves of $i_ X(A)$ are zero. Since the inclusion functor $\mathit{QCoh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$ is fully faithful and exact, we conclude that the cohomology objects of $A$ are zero, i.e., $A = 0$ and $RQ_ X(i_ X(A)) \to A$ is an isomorphism as well.

Induction step. Suppose that $E$ and $i_ X(A)$ are supported on a closed subset $T$ of $X$ contained in $U_1 \cup \ldots \cup U_ n$ with $U_ i \subset X$ affine open. Set $U = U_ n$. Consider the distinguished triangles

$A \to \Phi (A|_ U) \to A' \to A[1] \quad \text{and}\quad E \to Rj_*(E|_ U) \to E' \to E[1]$

where $\Phi$ is as in the statement of the lemma. Note that $E \to Rj_*(E|_ U)$ is a quasi-isomorphism over $U = U_ n$. Since $i_ X \circ \Phi = Rj_* \circ i_ U$ by assumption and since $i_ X(A)|_ U = i_ U(A|_ U)$ we see that $i_ X(A) \to i_ X(\Phi (A|_ U))$ is a quasi-isomorphism over $U$. Hence $i_ X(A')$ and $E'$ are supported on the closed subset $T \setminus U$ of $X$ which is contained in $U_1 \cup \ldots \cup U_{n - 1}$. By induction hypothesis the statement is true for $A'$ and $E'$. By Derived Categories, Lemma 13.4.3 it suffices to prove the maps

$RQ_ X(i_ X(\Phi (A|_ U))) \to \Phi (A|_ U) \quad \text{and}\quad Rj_*(E|_ U) \to i_ X(RQ_ X(Rj_*E|_ U))$

are isomorphisms. By assumption and by Lemma 36.7.2 (the inclusion morphism $j : U \to X$ is flat, quasi-compact, and quasi-separated) we have

$RQ_ X(i_ X(\Phi (A|_ U))) = RQ_ X(Rj_*(i_ U(A|_ U))) = \Phi (RQ_ U(i_ U(A|_ U)))$

and

$i_ X(RQ_ X(Rj_*(E|_ U))) = i_ X(\Phi (RQ_ U(E|_ U))) = Rj_*(i_ U(RQ_ U(E|_ U)))$

Finally, the maps

$RQ_ U(i_ U(A|_ U)) \to A|_ U \quad \text{and}\quad E|_ U \to i_ U(RQ_ U(E|_ U))$

are isomorphisms by Lemma 36.7.3. The result follows. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).