Remark 20.7.5. Here is a different approach to the proofs of Lemmas 20.7.2 and 20.7.3 above. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i_ X : \textit{Mod}(\mathcal{O}_ X) \to \textit{PMod}(\mathcal{O}_ X)$ be the inclusion functor and let $\#$ be the sheafification functor. Recall that $i_ X$ is left exact and $\#$ is exact.

1. First prove Lemma 20.11.4 below which says that the right derived functors of $i_ X$ are given by $R^ pi_ X\mathcal{F} = \underline{H}^ p(\mathcal{F})$. Here is another proof: The equality is clear for $p = 0$. Both $(R^ pi_ X)_{p \geq 0}$ and $(\underline{H}^ p)_{p \geq 0}$ are delta functors vanishing on injectives, hence both are universal, hence they are isomorphic. See Homology, Section 12.12.

2. A restatement of Lemma 20.7.2 is that $(\underline{H}^ p(\mathcal{F}))^\# = 0$, $p > 0$ for any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$. To see this is true, use that ${}^\#$ is exact so

$(\underline{H}^ p(\mathcal{F}))^\# = (R^ pi_ X\mathcal{F})^\# = R^ p(\# \circ i_ X)(\mathcal{F}) = 0$

because $\# \circ i_ X$ is the identity functor.

3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The presheaf $V \mapsto H^ p(f^{-1}V, \mathcal{F})$ is equal to $R^ p (i_ Y \circ f_*)\mathcal{F}$. You can prove this by noticing that both give universal delta functors as in the argument of (1) above. Hence Lemma 20.7.3 says that $R^ p f_* \mathcal{F}= (R^ p (i_ Y \circ f_*)\mathcal{F})^\#$. Again using that $\#$ is exact a that $\# \circ i_ Y$ is the identity functor we see that

$R^ p f_* \mathcal{F} = R^ p(\# \circ i_ Y \circ f_*)\mathcal{F} = (R^ p (i_ Y \circ f_*)\mathcal{F})^\#$

as desired.

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