The Stacks project

Proof. We first use the induction principle to prove $i_ X$ is fully faithful. More precisely, we will use Lemma 75.9.6. Let $(U \subset W, V \to W)$ be an elementary distinguished square with $V$ affine and $U, W$ quasi-compact open in $X$. Assume that $i_ U$ is fully faithful. We have to show that $i_ W$ is fully faithful. We may replace $X$ by $W$, i.e., we may assume $W = X$ (we do this just to simplify the notation – observe that the condition in the statement of the lemma is preserved under this operation).

Suppose that $A, B$ are objects of $D(\mathit{QCoh}(\mathcal{O}_ X))$. We want to show that

is bijective. Let $T = |X| \setminus |U|$.

Assume first $i_ X(B)$ is supported on $T$. In this case the map

is a quasi-isomorphism (Lemma 75.10.7). By assumption we have an isomorphism $i_ X(\Phi (B|_ V)) \to Rj_{V, *}(i_ V(B|_ V))$ in $D(\mathcal{O}_ X)$. Moreover, $\Phi $ and ${-}|_ V$ are adjoint functors on the derived categories of quasi-coherent modules (by Derived Categories, Lemma 13.30.3). The adjunction map $B \to \Phi (B|_ V)$ becomes an isomorphism after applying $i_ X$, whence is an isomorphism in $D(\mathit{QCoh}(\mathcal{O}_ X))$. Hence

as desired. Here we have used that $i_ V$ is fully faithful (Lemma 75.11.3).

In general, choose any complex $\mathcal{B}^\bullet $ of quasi-coherent $\mathcal{O}_ X$-modules representing $B$. Next, choose any quasi-isomorphism $s : \mathcal{B}^\bullet |_ U \to \mathcal{C}^\bullet $ of complexes of quasi-coherent modules on $U$. As $j_ U : U \to X$ is quasi-compact and quasi-separated the functor $j_{U, *}$ transforms quasi-coherent modules into quasi-coherent modules (Morphisms of Spaces, Lemma 67.11.2). Thus there is a canonical map $\mathcal{B}^\bullet \to j_{U, *}(\mathcal{B}^\bullet |_ U) \to j_{U, *}\mathcal{C}^\bullet $ of complexes of quasi-coherent modules on $X$. Set $B'' = j_{U, *}\mathcal{C}^\bullet $ in $D(\mathit{QCoh}(\mathcal{O}_ X))$ and choose a distinguished triangle

in $D(\mathit{QCoh}(\mathcal{O}_ X))$. Since the first arrow of the triangle restricts to an isomorphism over $U$ we see that $B'$ is supported on $T$. Hence in the diagram

we have exact columns and the top and bottom horizontal arrows are bijective. Finally, choose a complex $\mathcal{A}^\bullet $ of quasi-coherent modules representing $A$.

Let $\alpha : i_ X(A) \to i_ X(B)$ be a morphism between in $D(\mathcal{O}_ X)$. The restriction $\alpha |_ U$ comes from a morphism in $D(\mathit{QCoh}(\mathcal{O}_ U))$ as $i_ U$ is fully faithful. Hence there exists a choice of $s : \mathcal{B}^\bullet |_ U \to \mathcal{C}^\bullet $ as above such that $\alpha |_ U$ is represented by an actual map of complexes $\mathcal{A}^\bullet |_ U \to \mathcal{C}^\bullet $. This corresponds to a map of complexes $\mathcal{A} \to j_{U, *}\mathcal{C}^\bullet $. In other words, the image of $\alpha $ in $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(i_ X(A), i_ X(B''))$ comes from an element of $\mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ X))}(A, B'')$. A diagram chase then shows that $\alpha $ comes from a morphism $A \to B$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$. Finally, suppose that $a : A \to B$ is a morphism of $D(\mathit{QCoh}(\mathcal{O}_ X))$ which becomes zero in $D(\mathcal{O}_ X)$. After choosing $\mathcal{B}^\bullet $ suitably, we may assume $a$ is represented by a morphism of complexes $a^\bullet : \mathcal{A}^\bullet \to \mathcal{B}^\bullet $. Since $i_ U$ is fully faithul the restriction $a^\bullet |_ U$ is zero in $D(\mathit{QCoh}(\mathcal{O}_ U))$. Thus we can choose $s$ such that $s \circ a^\bullet |_ U : \mathcal{A}^\bullet |_ U \to \mathcal{C}^\bullet $ is homotopic to zero. Applying the functor $j_{U, *}$ we conclude that $\mathcal{A}^\bullet \to j_{U, *}\mathcal{C}^\bullet $ is homotopic to zero. Thus $a$ maps to zero in $\mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ X))}(A, B'')$. Thus we may assume that $a$ is the image of an element of $b \in \mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ X))}(A, B'[-1])$. The image of $b$ in $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(i_ X(A), i_ X(B')[-1])$ comes from a $\gamma \in \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(A, B''[-1])$ (as $a$ maps to zero in the group on the right). Since we've seen above the horizontal arrows are surjective, we see that $\gamma $ comes from a $c$ in $\mathop{\mathrm{Hom}}\nolimits _{D(\mathit{QCoh}(\mathcal{O}_ X))}(A, B''[-1])$ which implies $a = 0$ as desired.

At this point we know that $i_ X$ is fully faithful for our original $X$. Since $RQ_ X$ is its right adjoint, we see that $RQ_ X \circ i_ X = \text{id}$ (Categories, Lemma 4.24.4). To finish the proof we show that for any $E$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ the map $i_ X(RQ_ X(E)) \to E$ is an isomorphism. Choose a distinguished triangle

in $D_\mathit{QCoh}(\mathcal{O}_ X)$. A formal argument using the above shows that $i_ X(RQ_ X(E')) = 0$. Thus it suffices to prove that for $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the condition $i_ X(RQ_ X(E)) = 0$ implies that $E = 0$. Consider an étale morphism $j : V \to X$ with $V$ affine. By Lemmas 75.11.3 and 75.11.2 and our assumption we have

Choose a distinguished triangle

Apply $RQ_ X$ to get a distinguished triangle

in other words the map in the middle is an isomorphism. Combined with the string of equalities above we find that our first distinguished triangle becomes a distinguished triangle

where the middle morphism is the adjunction map. However, the composition $E \to E'$ is zero, hence $E \to i_ X(RQ_ X(E'))$ is zero by adjunction! Since this morphism is isomorphic to the morphism $E \to Rj_*(E|_ V)$ adjoint to $\text{id} : E|_ V \to E|_ V$ we conclude that $E|_ V$ is zero. Since this holds for all affine $V$ étale over $X$ we conclude $E$ is zero as desired. $\square$