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

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

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

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