The Stacks project

Lemma 36.21.2. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of schemes. If the right adjoints $DQ_ X$ and $DQ_ Y$ of the inclusion functors $D_\mathit{QCoh}\to D$ exist for $X$ and $Y$, then

\[ Rf_* \circ DQ_ X = DQ_ Y \circ Rf_* \]

Proof. The statement makes sense because $Rf_*$ sends $D_\mathit{QCoh}(\mathcal{O}_ X)$ into $D_\mathit{QCoh}(\mathcal{O}_ Y)$ by Lemma 36.4.1. The statement is true because $Lf^*$ similarly maps $D_\mathit{QCoh}(\mathcal{O}_ Y)$ into $D_\mathit{QCoh}(\mathcal{O}_ X)$ (Lemma 36.3.8) and hence both $Rf_* \circ DQ_ X$ and $DQ_ Y \circ Rf_*$ are right adjoint to $Lf^* : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D(\mathcal{O}_ X)$. $\square$

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Your email address will not be published.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 0CR1. Beware of the difference between the letter 'O' and the digit '0'.