Lemma 48.4.1. In diagram (48.4.0.1) assume that $g$ is flat or more generally that $f$ and $g$ are Tor independent. Then $a \circ Rg_* \leftarrow Rg'_* \circ a'$ is an isomorphism.

Proof. In this case the base change map $Lg^* \circ Rf_* K \longrightarrow Rf'_* \circ L(g')^*K$ is an isomorphism for every $K$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.22.5. Thus the corresponding transformation between adjoint functors is an isomorphism as well. $\square$

## 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Your email address will not be published.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 0A9K. Beware of the difference between the letter 'O' and the digit '0'.