The Stacks project

73.6 Total direct image

The following lemma is the analogue of Cohomology of Spaces, Lemma 67.8.1.

Lemma 73.6.1. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-separated and quasi-compact morphism of algebraic spaces over $S$.

  1. The functor $Rf_*$ sends $D_\mathit{QCoh}(\mathcal{O}_ X)$ into $D_\mathit{QCoh}(\mathcal{O}_ Y)$.

  2. If $Y$ is quasi-compact, there exists an integer $N = N(X, Y, f)$ such that for an object $E$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ with $H^ m(E) = 0$ for $m > 0$ we have $H^ m(Rf_*E) = 0$ for $m \geq N$.

  3. In fact, if $Y$ is quasi-compact we can find $N = N(X, Y, f)$ such that for every morphism of algebraic spaces $Y' \to Y$ the same conclusion holds for the functor $R(f')_*$ where $f' : X' \to Y'$ is the base change of $f$.

Proof. Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. To prove (1) we have to show that $Rf_*E$ has quasi-coherent cohomology sheaves. This question is local on $Y$, hence we may assume $Y$ is quasi-compact. Pick $N = N(X, Y, f)$ as in Cohomology of Spaces, Lemma 67.8.1. Thus $R^ pf_*\mathcal{F} = 0$ for all quasi-coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ and all $p \geq N$. Moreover $R^ pf_*\mathcal{F}$ is quasi-coherent for all $p$ by Cohomology of Spaces, Lemma 67.3.1. These statements remain true after base change.

First, assume $E$ is bounded below. We will show (1) and (2) and (3) hold for such $E$ with our choice of $N$. In this case we can for example use the spectral sequence

\[ R^ pf_*H^ q(E) \Rightarrow R^{p + q}f_*E \]

(Derived Categories, Lemma 13.21.3), the quasi-coherence of $R^ pf_*H^ q(E)$, and the vanishing of $R^ pf_*H^ q(E)$ for $p \geq N$ to see that (1), (2), and (3) hold in this case.

Next we prove (2) and (3). Say $H^ m(E) = 0$ for $m > 0$. Let $V$ be an affine object of $Y_{\acute{e}tale}$. We have $H^ p(V \times _ Y X, \mathcal{F}) = 0$ for $p \geq N$, see Cohomology of Spaces, Lemma 67.3.2. Hence we may apply Lemma 73.5.8 to the functor $\Gamma (V \times _ Y X, -)$ to see that

\[ R\Gamma (V, Rf_*E) = R\Gamma (V \times _ Y X, E) \]

has vanishing cohomology in degrees $\geq N$. Since this holds for all $V$ affine in $Y_{\acute{e}tale}$ we conclude that $H^ m(Rf_*E) = 0$ for $m \geq N$.

Next, we prove (1) in the general case. Recall that there is a distinguished triangle

\[ \tau _{\leq -n - 1}E \to E \to \tau _{\geq -n}E \to (\tau _{\leq -n - 1}E)[1] \]

in $D(\mathcal{O}_ X)$, see Derived Categories, Remark 13.12.4. By (2) we see that $Rf_*\tau _{\leq -n - 1}E$ has vanishing cohomology sheaves in degrees $\geq -n + N$. Thus, given an integer $q$ we see that $R^ qf_*E$ is equal to $R^ qf_*\tau _{\geq -n}E$ for some $n$ and the result above applies. $\square$

Lemma 73.6.2. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-separated and quasi-compact morphism of algebraic spaces over $S$. Then $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ commutes with direct sums.

Proof. Let $E_ i$ be a family of objects of $D_\mathit{QCoh}(\mathcal{O}_ X)$ and set $E = \bigoplus E_ i$. We want to show that the map

\[ \bigoplus Rf_*E_ i \longrightarrow Rf_*E \]

is an isomorphism. We will show it induces an isomorphism on cohomology sheaves in degree $0$ which will imply the lemma. Choose an integer $N$ as in Lemma 73.6.1. Then $R^0f_*E = R^0f_*\tau _{\geq -N}E$ and $R^0f_*E_ i = R^0f_*\tau _{\geq -N}E_ i$ by the lemma cited. Observe that $\tau _{\geq -N}E = \bigoplus \tau _{\geq -N}E_ i$. Thus we may assume all of the $E_ i$ have vanishing cohomology sheaves in degrees $< -N$. Next we use the spectral sequences

\[ R^ pf_*H^ q(E) \Rightarrow R^{p + q}f_*E \quad \text{and}\quad R^ pf_*H^ q(E_ i) \Rightarrow R^{p + q}f_*E_ i \]

(Derived Categories, Lemma 13.21.3) to reduce to the case of a direct sum of quasi-coherent sheaves. This case is handled by Cohomology of Spaces, Lemma 67.5.2. $\square$

Remark 73.6.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of representable algebraic spaces $X$ and $Y$ over $S$. Let $f_0 : X_0 \to Y_0$ be a morphism of schemes representing $f$ (awkward but temporary notation). Then the diagram

\[ \xymatrix{ D_\mathit{QCoh}(\mathcal{O}_{X_0}) \ar@{=}[rrrrrr]_{\text{Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ X) \\ D_\mathit{QCoh}(\mathcal{O}_{Y_0}) \ar[u]^{Lf^*_0} \ar@{=}[rrrrrr]^{\text{Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ Y) \ar[u]_{Lf^*} } \]

(Lemma 73.5.5 and Derived Categories of Schemes, Lemma 36.3.8) is commutative. This follows as the equivalences $D_\mathit{QCoh}(\mathcal{O}_{X_0}) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ and $D_\mathit{QCoh}(\mathcal{O}_{Y_0}) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ of Lemma 73.4.2 come from pulling back by the (flat) morphisms of ringed sites $\epsilon : X_{\acute{e}tale}\to X_{0, Zar}$ and $\epsilon : Y_{\acute{e}tale}\to Y_{0, Zar}$ and the diagram of ringed sites

\[ \xymatrix{ X_{0, Zar} \ar[d]_{f_0} & X_{\acute{e}tale}\ar[l]^\epsilon \ar[d]^ f \\ Y_{0, Zar} & Y_{\acute{e}tale}\ar[l]_\epsilon } \]

is commutative (details omitted). If $f$ is quasi-compact and quasi-separated, equivalently if $f_0$ is quasi-compact and quasi-separated, then we claim

\[ \xymatrix{ D_\mathit{QCoh}(\mathcal{O}_{X_0}) \ar[d]_{Rf_{0, *}} \ar@{=}[rrrrrr]_{\text{Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[d]^{Rf_*} \\ D_\mathit{QCoh}(\mathcal{O}_{Y_0}) \ar@{=}[rrrrrr]^{\text{Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ Y) } \]

(Lemma 73.6.1 and Derived Categories of Schemes, Lemma 36.4.1) is commutative as well. This also follows from the commutative diagram of sites displayed above as the proof of Lemma 73.4.2 shows that the functor $R\epsilon _*$ gives the equivalences $D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_{X_0})$ and $D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_{Y_0})$.

Lemma 73.6.4. Let $S$ be a scheme. Let $f : X \to Y$ be an affine morphism of algebraic spaces over $S$. Then $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ reflects isomorphisms.

Proof. The statement means that a morphism $\alpha : E \to F$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$ is an isomorphism if $Rf_*\alpha $ is an isomorphism. We may check this on cohomology sheaves. In particular, the question is ├ętale local on $Y$. Hence we may assume $Y$ and therefore $X$ is affine. In this case the problem reduces to the case of schemes (Derived Categories of Schemes, Lemma 36.5.2) via Lemma 73.4.2 and Remark 73.6.3. $\square$

Lemma 73.6.5. Let $S$ be a scheme. Let $f : X \to Y$ be an affine morphism of algebraic spaces over $S$. For $E$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ we have $Rf_* Lf^* E = E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} f_*\mathcal{O}_ X$.

Proof. Since $f$ is affine the map $f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X$ is an isomorphism (Cohomology of Spaces, Lemma 67.8.2). There is a canonical map $E \otimes ^\mathbf {L} f_*\mathcal{O}_ X = E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X \to Rf_* Lf^* E$ adjoint to the map

\[ Lf^*(E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X) = Lf^*E \otimes ^\mathbf {L} Lf^*Rf_*\mathcal{O}_ X \longrightarrow Lf^* E \otimes ^\mathbf {L} \mathcal{O}_ X = Lf^* E \]

coming from $1 : Lf^*E \to Lf^*E$ and the canonical map $Lf^*Rf_*\mathcal{O}_ X \to \mathcal{O}_ X$. To check the map so constructed is an isomorphism we may work locally on $Y$. Hence we may assume $Y$ and therefore $X$ is affine. In this case the problem reduces to the case of schemes (Derived Categories of Schemes, Lemma 36.5.3) via Lemma 73.4.2 and Remark 73.6.3. $\square$


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 08F9. Beware of the difference between the letter 'O' and the digit '0'.