The Stacks project

Proof. The functor $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ is exact, see Cohomology of Schemes, Lemma 30.2.3. 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. The diagram commutes by Lemma 36.5.1. $\square$

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 Schemes, Lemma 26.24.1. Recall that $\mathit{QCoh}(\mathcal{O}_ X)$ is a Grothendieck abelian category (Properties, Proposition 28.23.4). 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, Lemma 20.28.1. 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. The functor $Q_ X$ is the functor

by Schemes, Lemma 26.7.1. This immediately implies (1) and (2). The third assertion follows from (the proof of) Lemma 36.3.5. $\square$

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

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

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

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

and

Finally, the maps

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

Proof. Let $U \subset X$ be an affine open. Then the morphism $U \to X$ is affine by Morphisms, Lemma 29.11.11. Thus the assumption of Lemma 36.7.4 holds by Lemma 36.7.1 and we win. $\square$

Proof. Observe that the horizontal arrows in the diagram are equivalences of categories by Proposition 36.7.5. Hence we can identify these categories (and similarly for other quasi-compact schemes with affine diagonal). 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 $U \subset X$ be an affine open. Since the diagonal of $X$ is affine, the inclusion morphism $j : U \to X$ is affine (Morphisms, Lemma 29.11.11). Similarly, the composition $g = f \circ j : U \to Y$ is affine. Let $\mathcal{I}^\bullet $ be a K-injective complex in $\mathit{QCoh}(\mathcal{O}_ U)$. 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 $\mathit{QCoh}(\mathcal{O}_ X)$, see Derived Categories, Lemma 13.31.9. It follows that

By Lemma 36.7.1 we see that $j_*\mathcal{I}^\bullet $ represents $Rj_*\mathcal{I}^\bullet $ and $g_*\mathcal{I}^\bullet $ represents $Rg_*\mathcal{I}^\bullet $. On the other hand, we have $Rf_* \circ Rj_* = Rg_*$. Hence $f_*j_*\mathcal{I}^\bullet $ represents $Rf_*(j_*\mathcal{I}^\bullet )$. We conclude that the lemma is true for any complex of the form $j_*\mathcal{G}^\bullet $ with $\mathcal{G}^\bullet $ a complex of quasi-coherent modules on $U$. (Note that if $\mathcal{G}^\bullet \to \mathcal{I}^\bullet $ is a quasi-isomorphism, then $j_*\mathcal{G}^\bullet \to j_*\mathcal{I}^\bullet $ is a quasi-isomorphism as well since $j_*$ is an exact functor on quasi-coherent modules.)

Let $\mathcal{F}^\bullet $ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. Let $T \subset X$ be a closed subset such that the support of $\mathcal{F}^ p$ is contained in $T$ for all $p$. We will use induction on the minimal number $n$ of affine opens $U_1, \ldots , U_ n$ such that $T \subset U_1 \cup \ldots \cup U_ n$. The base case $n = 0$ is trivial. If $n \geq 1$, then set $U = U_1$ and denote $j : U \to X$ the open immersion as above. We consider the map of complexes $c : \mathcal{F}^\bullet \to j_*j^*\mathcal{F}^\bullet $. We obtain two short exact sequences of complexes:

and

The complexes $\mathop{\mathrm{Ker}}(c)$ and $\mathop{\mathrm{Coker}}(c)$ are supported on $T \setminus U \subset U_2 \cup \ldots \cup U_ n$ and the result holds for them by induction. The result holds for $j_*j^*\mathcal{F}^\bullet $ by the discussion in the preceding paragraph. We conclude by looking at the distinguished triangles associated to the short exact sequences and using the initial remark of the proof. $\square$