The Stacks project

Lemma 48.10.1. In the situation above, the map (48.10.0.1) is an isomorphism if and only if the base change map

\[ Lf^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ Z, K) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{O}_{Z'}, Lf^*K) \]

of Cohomology, Remark 20.42.13 is an isomorphism.

Proof. The statement makes sense because $\mathcal{O}_{Z'} = Lf^*\mathcal{O}_ Z$ by the assumed tor independence. Since $i'_*$ is exact and faithful we see that it suffices to show the map (48.10.0.1) is an isomorphism after applying $Ri'_*$. Since $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ by the assumed tor independence and Derived Categories of Schemes, Lemma 36.22.9 we obtain a map

\[ Lf^*Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow Ri'_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Z'}, Lf^*K) \]

whose source and target are as in the statement of the lemma by Lemma 48.9.3. We omit the verification that this is the same map as the one constructed in Cohomology, Remark 20.42.13. $\square$

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Your email address will not be published.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 0E2L. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 48.10.1 as pdf