The Stacks project

86.8 Right adjoint of pushforward for proper flat morphisms

For proper, flat, and finitely presented morphisms of quasi-compact and quasi-separated algebraic spaces the right adjoint of pushforward enjoys some remarkable properties.

Lemma 86.8.1. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 86.3.1. Then $a$ commutes with direct sums.

Proof. Let $P$ be a perfect object of $D(\mathcal{O}_ X)$. By Derived Categories of Spaces, Lemma 75.25.4 the complex $Rf_*P$ is perfect on $Y$. Let $K_ i$ be a family of objects of $D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, a(\bigoplus K_ i)) & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*P, \bigoplus K_ i) \\ & = \bigoplus \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rf_*P, K_ i) \\ & = \bigoplus \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P, a(K_ i)) \end{align*}

because a perfect object is compact (Derived Categories of Spaces, Proposition 75.16.1). Since $D_\mathit{QCoh}(\mathcal{O}_ X)$ has a perfect generator (Derived Categories of Spaces, Theorem 75.15.4) we conclude that the map $\bigoplus a(K_ i) \to a(\bigoplus K_ i)$ is an isomorphism, i.e., $a$ commutes with direct sums. $\square$

Lemma 86.8.2. Let $S$ be a scheme. Let $Y$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. The map (86.7.0.1) is an isomorphism for every object $K$ of $D_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. By Lemma 86.8.1 we know that $a$ commutes with direct sums. Hence the collection of objects of $D_\mathit{QCoh}(\mathcal{O}_ Y)$ for which (86.7.0.1) is an isomorphism is a strictly full, saturated, triangulated subcategory of $D_\mathit{QCoh}(\mathcal{O}_ Y)$ which is moreover preserved under taking direct sums. Since $D_\mathit{QCoh}(\mathcal{O}_ Y)$ is a module category (Derived Categories of Spaces, Theorem 75.17.3) generated by a single perfect object (Derived Categories of Spaces, Theorem 75.15.4) we can argue as in More on Algebra, Remark 15.59.11 to see that it suffices to prove (86.7.0.1) is an isomorphism for a single perfect object. However, the result holds for perfect objects, see Lemma 86.7.1. $\square$

Lemma 86.8.3. Let $Y$ be an affine scheme. Let $f : X \to Y$ be a morphism of algebraic spaces which is proper, flat, and of finite presentation. Let $a$ be the right adjoint for $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 86.3.1. Then

  1. $a(\mathcal{O}_ Y)$ is a $Y$-perfect object of $D(\mathcal{O}_ X)$,

  2. $Rf_*a(\mathcal{O}_ Y)$ has vanishing cohomology sheaves in positive degrees,

  3. $\mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ is an isomorphism.

Proof. For a perfect object $E$ of $D(\mathcal{O}_ X)$ we have

\begin{align*} Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}^\bullet ) & = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet ) \\ & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y) \\ & = (Rf_*E^\vee )^\vee \end{align*}

For the first equality, see Cohomology on Sites, Lemma 21.48.4. For the second equality, see Lemma 86.3.3, Remark 86.3.5, and Derived Categories of Spaces, Lemma 75.25.4. The third equality is the definition of the dual. In particular these references also show that the outcome is a perfect object of $D(\mathcal{O}_ Y)$. We conclude that $\omega _{X/Y}^\bullet $ is $Y$-perfect by More on Morphisms of Spaces, Lemma 76.52.14. This proves (1).

Let $M$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*a(\mathcal{O}_ Y)) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, a(\mathcal{O}_ Y)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*Lf^*M, \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*\mathcal{O}_ Y, \mathcal{O}_ Y) \end{align*}

The first equality holds by Cohomology on Sites, Lemma 21.19.1. The second equality by construction of $a$. The third equality by Derived Categories of Spaces, Lemma 75.20.1. Recall $Rf_*\mathcal{O}_ X$ is perfect of tor amplitude in $[0, N]$ for some $N$, see Derived Categories of Spaces, Lemma 75.25.4. Thus we can represent $Rf_*\mathcal{O}_ X$ by a complex of finite projective modules sitting in degrees $[0, N]$ (using More on Algebra, Lemma 15.74.2 and the fact that $Y$ is affine). Hence if $M = \mathcal{O}_ Y[-i]$ for some $i > 0$, then the last group is zero. Since $Y$ is affine we conclude that $H^ i(Rf_*a(\mathcal{O}_ Y)) = 0$ for $i > 0$. This proves (2).

Let $E$ be a perfect object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Then we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ X(E, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y)) & = \mathop{\mathrm{Hom}}\nolimits _ X(E \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y), a(\mathcal{O}_ Y)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} a(\mathcal{O}_ Y)), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , a(\mathcal{O}_ Y))), \mathcal{O}_ Y) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y), \mathcal{O}_ Y) \\ & = R\Gamma (Y, Rf_*E^\vee ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(E, \mathcal{O}_ X) \end{align*}

The first equality holds by Cohomology on Sites, Lemma 21.35.2. The second equality is the definition of $a$. The third equality comes from the construction of the dual perfect complex $E^\vee $, see Cohomology on Sites, Lemma 21.48.4. The fourth equality follows from the equality $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E^\vee , \omega _{X/Y}^\bullet ) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*E^\vee , \mathcal{O}_ Y)$ shown in the first paragraph of the proof. The fifth equality holds by double duality for perfect complexes (Cohomology on Sites, Lemma 21.48.4) and the fact that $Rf_*E$ is perfect by Derived Categories of Spaces, Lemma 75.25.4 The last equality is Leray for $f$. This string of equalities essentially shows (3) holds by the Yoneda lemma. Namely, the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Spaces, Lemma 75.13.10. Taking $E = \mathcal{O}_ X$ in the above we get a map $\alpha : \mathcal{O}_ X \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(a(\mathcal{O}_ Y), a(\mathcal{O}_ Y))$ corresponding to $\text{id}_{\mathcal{O}_ X} \in \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, \mathcal{O}_ X)$. Since all the isomorphisms above are functorial in $E$ we see that the cone on $\alpha $ is an object $C$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ such that $\mathop{\mathrm{Hom}}\nolimits (E, C) = 0$ for all perfect $E$. Since the perfect objects generate (Derived Categories of Spaces, Theorem 75.15.4) we conclude that $\alpha $ is an isomorphism. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E5S. Beware of the difference between the letter 'O' and the digit '0'.