The Stacks project

Lemma 86.10.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $S$. Assume $X$ and $Y$ are representable and let $f_0 : X_0 \to Y_0$ be a morphism of schemes representing $f$ (awkward but temporary notation). Let $a : D_\mathit{QCoh}(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ be the right adjoint of $Rf_*$ from Lemma 86.3.1. Let $a_0 : D_\mathit{QCoh}(\mathcal{O}_{Y_0}) \to D_\mathit{QCoh}(\mathcal{O}_{X_0})$ be the right adjoint of $Rf_*$ from Duality for Schemes, Lemma 48.3.1. Then

\[ \xymatrix{ D_\mathit{QCoh}(\mathcal{O}_{X_0}) \ar@{=}[rrrrrr]_{\text{Derived Categories of Spaces, Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ X) \\ D_\mathit{QCoh}(\mathcal{O}_{Y_0}) \ar[u]^{a_0} \ar@{=}[rrrrrr]^{\text{Derived Categories of Spaces, Lemma 071Q}} & & & & & & D_\mathit{QCoh}(\mathcal{O}_ Y) \ar[u]_ a } \]

is commutative.

Proof. Follows from uniqueness of adjoints and the compatibilities of Derived Categories of Spaces, Remark 75.6.3. $\square$


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Your email address will not be published.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 0E6E. Beware of the difference between the letter 'O' and the digit '0'.