The Stacks project

Proof. The functor $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ is exact, see Cohomology of Spaces, Lemma 69.8.2. Hence $f_*$ defines a derived functor $f_* : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ by simply applying $f_*$ to any representative complex, see Derived Categories, Lemma 13.16.9. For any complex of $\mathcal{O}_ X$-modules $\mathcal{F}^\bullet$ there is a canonical map $f_*\mathcal{F}^\bullet \to Rf_*\mathcal{F}^\bullet$. To finish the proof we show this is a quasi-isomorphism when $\mathcal{F}^\bullet$ is a complex with each $\mathcal{F}^ n$ quasi-coherent. The statement is étale local on $Y$ hence we may assume $Y$ affine. As an affine morphism is representable we reduce to the case of schemes by the compatibility of Remark 75.6.3. The case of schemes is Derived Categories of Schemes, Lemma 36.7.1. $\square$

Proof. We will prove this by showing that $RQ_ Y \circ Rf_*$ and $\Phi \circ RQ_ X$ are right adjoint to the same functor $D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathcal{O}_ X)$.

Since $f$ is quasi-compact and quasi-separated, we see that $f_*$ preserves quasi-coherence, see Morphisms of Spaces, Lemma 67.11.2. Recall that $\mathit{QCoh}(\mathcal{O}_ X)$ is a Grothendieck abelian category (Properties of Spaces, Proposition 66.32.2). Hence any $K$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$ can be represented by a K-injective complex $\mathcal{I}^\bullet$ of $\mathit{QCoh}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. Then we can define $\Phi (K) = f_*\mathcal{I}^\bullet$.

Since $f$ is flat, the functor $f^*$ is exact. Hence $f^*$ defines $f^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ and also $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$. The functor $f^* = Lf^* : D(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$ is left adjoint to $Rf_* : D(\mathcal{O}_ X) \to D(\mathcal{O}_ Y)$, see Cohomology on Sites, Lemma 21.19.1. Similarly, the functor $f^* : D(\mathit{QCoh}(\mathcal{O}_ Y)) \to D(\mathit{QCoh}(\mathcal{O}_ X))$ is left adjoint to $\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ by Derived Categories, Lemma 13.30.3.

Let $A$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ Y))$ and $E$ an object of $D(\mathcal{O}_ X)$. Then

This implies what we want. $\square$

Proof. Let $X_0 = \mathop{\mathrm{Spec}}(A)$ be the affine scheme representing $X$. Recall that there is a morphism of ringed sites $\epsilon : X_{\acute{e}tale}\to X_{0, Zar}$ which induces equivalences

see Lemma 75.4.2. Hence we see that $Q_ X = \epsilon ^* \circ Q_{X_0} \circ \epsilon _*$ by uniqueness of adjoint functors. Hence (1) follows from the description of $Q_{X_0}$ in Derived Categories of Schemes, Lemma 36.7.3 and the fact that $\Gamma (X_0, \epsilon _*\mathcal{F}) = \Gamma (X, \mathcal{F})$. Part (2) follows from (1) and the fact that the functor from $A$-modules to quasi-coherent $\mathcal{O}_ X$-modules is exact. The third assertion now follows from the result for schemes (Derived Categories of Schemes, Lemma 36.7.3) and Lemma 75.4.2. $\square$

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. $\square$

Proof. Let $V \to W$ be an étale morphism with $V$ affine and $W$ a quasi-compact open subspace of $X$. Then the morphism $V \to W$ is affine as $W$ has affine diagonal over $\mathbf{Z}$ and $V$ is affine (Morphisms of Spaces, Lemma 67.20.11). Lemma 75.11.1 then guarantees that the assumption of Lemma 75.11.4 holds. Hence we conclude. $\square$

Proof. Observe that the horizontal arrows in the diagram are equivalences of categories by Proposition 75.11.5. Hence we can identify these categories (and similarly for other quasi-compact algebraic spaces with affine diagonal) and then the statement of the lemma is that the canonical map $\Phi (K) \to Rf_*(K)$ is an isomorphism for all $K$ in $D(\mathit{QCoh}(\mathcal{O}_ X))$. Note that if $K_1 \to K_2 \to K_3 \to K_1[1]$ is a distinguished triangle in $D(\mathit{QCoh}(\mathcal{O}_ X))$ and the statement is true for two-out-of-three, then it is true for the third.

Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (X_{spaces, {\acute{e}tale}})$ be the set of objects which are quasi-compact and have affine diagonal. For $U \in \mathcal{B}$ and any morphism $g : U \to Z$ where $Z$ is a quasi-compact algebraic space over $S$ with affine diagonal, denote

the derived extension of $g_*$. Let $P(U) =$ “for any $K$ in $D(\mathit{QCoh}(\mathcal{O}_ U))$ and any $g : U \to Z$ as above the map $\Phi _ g(K) \to Rg_*K$ is an isomorphism”. By Remark 75.9.5 conditions (1), (2), and (3)(a) of Lemma 75.9.4 hold and we are left with proving (3)(b) and (4).

Checking condition (3)(b). Let $U$ be an affine scheme étale over $X$. Let $g : U \to Z$ be as above. Since the diagonal of $Z$ is affine the morphism $g : U \to Z$ is affine (Morphisms of Spaces, Lemma 67.20.11). Hence $P(U)$ holds by Lemma 75.11.1.

Checking condition (4). Let $(U \subset W, V \to W)$ be an elementary distinguished square in $X_{spaces, {\acute{e}tale}}$ with $U, W, V$ in $\mathcal{B}$ and $V$ affine. Assume that $P$ holds for $U$, $V$, and $U \times _ W V$. We have to show that $P$ holds for $W$. Let $g : W \to Z$ be a morphism to a quasi-compact algebraic space with affine diagonal. Let $K$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ W))$. Consider the distinguished triangle

in $D(\mathcal{O}_ W)$. By the two-out-of-three property mentioned above, it suffices to show that $\Phi _ g(Rj_{U, *}K|_ U) \to Rg_*(Rj_{U, *}K|_ U)$ is an isomorphism and similarly for $V$ and $U \times _ W V$. This is discussed in the next paragraph.

Let $j : U \to W$ be a morphism $X_{spaces, {\acute{e}tale}}$ with $U, W$ in $\mathcal{B}$ and $P$ holds for $U$. Let $g : W \to Z$ be a morphism to a quasi-compact algebraic space with affine diagonal. To finish the proof we have to show that $\Phi _ g(Rj_*K) \to Rg_*(Rj_*K)$ is an isomorphism for any $K$ in $D(\mathit{QCoh}(\mathcal{O}_ U))$. Let $\mathcal{I}^\bullet$ be a K-injective complex in $\mathit{QCoh}(\mathcal{O}_ U)$ representing $K$. From $P(U)$ applied to $j$ we see that $j_*\mathcal{I}^\bullet$ represents $Rj_*K$. Since $j_* : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_ X)$ has an exact left adjoint $j^* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ U)$ we see that $j_*\mathcal{I}^\bullet$ is a K-injective complex in $\mathit{QCoh}(\mathcal{O}_ W)$, see Derived Categories, Lemma 13.31.9. Hence $\Phi _ g(Rj_*K)$ is represented by $g_*j_*\mathcal{I}^\bullet = (g \circ j)_*\mathcal{I}^\bullet$. By $P(U)$ applied to $g \circ j$ we see that this represents $R_{g \circ j, *}(K) = Rg_*(Rj_*K)$. This finishes the proof. $\square$