The Stacks project

Lemma 36.7.6. Let $f : X \to Y$ be a morphism of schemes. Assume $X$ and $Y$ are quasi-compact and have affine diagonal. Then, denoting

\[ \Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y)) \]

the right derived functor of $f_* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ the diagram

\[ \xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[d]_\Phi \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[d]^{Rf_*} \\ D(\mathit{QCoh}(\mathcal{O}_ Y)) \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ Y) } \]

is commutative.

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

\[ \Phi (j_*\mathcal{I}^\bullet ) = f_*j_*\mathcal{I}^\bullet = g_*\mathcal{I}^\bullet \]

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:

\[ 0 \to \mathop{\mathrm{Ker}}(c) \to \mathcal{F}^\bullet \to \mathop{\mathrm{Im}}(c) \to 0 \]

and

\[ 0 \to \mathop{\mathrm{Im}}(c) \to j_*j^*\mathcal{F}^\bullet \to \mathop{\mathrm{Coker}}(c) \to 0 \]

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. $\square$


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