Lemma 20.11.4. Let $X$ be a ringed space. Consider the functor $i : \textit{Mod}(\mathcal{O}_ X) \to \textit{PMod}(\mathcal{O}_ X)$. It is a left exact functor with right derived functors given by

see discussion in Section 20.7.

Lemma 20.11.4. Let $X$ be a ringed space. Consider the functor $i : \textit{Mod}(\mathcal{O}_ X) \to \textit{PMod}(\mathcal{O}_ X)$. It is a left exact functor with right derived functors given by

\[ R^ pi(\mathcal{F}) = \underline{H}^ p(\mathcal{F}) : U \longmapsto H^ p(U, \mathcal{F}) \]

see discussion in Section 20.7.

**Proof.**
It is clear that $i$ is left exact. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $. By definition $R^ pi$ is the $p$th cohomology *presheaf* of the complex $\mathcal{I}^\bullet $. In other words, the sections of $R^ pi(\mathcal{F})$ over an open $U$ are given by

\[ \frac{\mathop{\mathrm{Ker}}(\mathcal{I}^ p(U) \to \mathcal{I}^{p + 1}(U))}{\mathop{\mathrm{Im}}(\mathcal{I}^{p - 1}(U) \to \mathcal{I}^ p(U))}. \]

which is the definition of $H^ p(U, \mathcal{F})$. $\square$

Your email address will not be published.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 (2)

Comment #7790 by Jakob Werner on

Comment #8028 by Stacks Project on