Lemma 21.43.5. In the situation above, suppose that $K$ is an object of $\mathit{QC}(\mathcal{O})$ and $M$ arbitrary in $D(\mathcal{O})$. For every object $U$ of $\mathcal{C}$ we have

**Proof.**
We may replace $\mathcal{C}$ by $\mathcal{C}/U$. Thus we may assume $U = X$ is a final object of $\mathcal{C}$. By Lemma 21.43.4 we see that $K = Lf^*P$ where $P = R\Gamma (U, K) = R\Gamma (X, K) = Rf_*K$. Thus the result because $Lf^*$ is the left adjoint to $Rf_*(-) = R\Gamma (U, -)$.
$\square$

## 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Your email address will not be published.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.

## Comments (0)