Lemma 52.6.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{I} \subset \mathcal{O}$ be a sheaf of ideals. If $K \in D(\mathcal{O})$ and $L \in D_{comp}(\mathcal{O})$, then $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, L) \in D_{comp}(\mathcal{O})$.

**Proof.**
Let $U$ be an object of $\mathcal{C}$ and let $f \in \mathcal{I}(U)$. Recall that

by Cohomology on Sites, Lemma 21.35.2. The right hand side is zero by Lemma 52.6.2 and the relationship between internal hom and actual hom, see Cohomology on Sites, Lemma 21.35.1. The same vanishing holds for all $U'/U$. Thus the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f}, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, L)|_ U)$ of $D(\mathcal{O}_ U)$ has vanishing $0$th cohomology sheaf (by locus citatus). Similarly for the other cohomology sheaves, i.e., $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f}, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, L)|_ U)$ is zero in $D(\mathcal{O}_ U)$. By Lemma 52.6.2 we conclude. $\square$

## 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Your email address will not be published.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)

There are also: