# Showing that $R^pf_*\mathcal{F} \cong \widetilde{H^p(X, \mathcal{F})}$


Let $$f : X \ra Y$$ be a morphism of schemes, where $$X$$ is noetherian and $$Y = \text{Spec} A$$ is affine. Then for any $$\F \in \QCoh(X)$$ and $$i \geq 0$$ there exists a canonical isomorphism of sheaves of modules on $$Y$$ natural in $$\F$$ $$\beta : R^if_* \F \longrightarrow \widetilde{H^i(X, \F)}.$$

The proof goes as follows (I've denoted my questions with "Q:"):

$$R^if_* \F$$ has a canonical $$\oh_Y$$-module structure and $$H^i(X, \F)$$ has a canonical $$\Gamma(X, \F)$$-module structure, and since $$\Gamma(X, \F)$$ has the structure of an $$A$$-module, $$H^i(X, \F)$$ has an $$A$$-module structure. Also, $$X$$ noetherian implies that $$f_* \F \in \QCoh(Y)$$. Therefore, we have a canonical isomorphism of sheaves $$f_* \F \cong \widetilde{\Gamma(X, \F)}.$$ Q: I'm not sure why we have this. I know that if in general $$\F$$ is a quasi-coherent sheaf on $$X$$, then for $$U_i$$ opens of a particular cover of $$X$$, we have $$\F|_{U_i} \cong \widetilde{F(U_i)}$$. But I'm not sure as to how we derive the above? I've tried $$f_* \F|_{U_i} = \F(f^{-1}(-))|_{U_i}$$ but this doesn't really get me anywhere.

Proceeding with the proof, we have for $$i=0$$ a canonical isomorphism natural in $$\F$$ $$\mu^0 : R^0f_* \F \cong f_* \F \cong \widetilde{\Gamma(X, \F)} = \widetilde{H^0(X, \F)} \quad \checkmark$$ Now, since the tilde functor $$\widetilde{-}: A \Mod \ra \oh_Y \Mod$$ is exact, we have two cohomological $$\delta$$-functors $$\{ R^i f_*(-) \}_{i \geq 0}$$ and $$\{ \widetilde{H^i(X, -)} \}_{i \geq 0}$$ between $$\QCoh(X)$$ and $$\oh_Y \Mod$$.

Q: Why does this follow from exactness of the tilde functor? Apologies; this may be obvious (I'm not too brushed up on my $$\delta$$-functor knowledge).

Quasi-coherent sheaves may be embedded into flasque quasi-coherent sheaves. Hence, both functors are effaceable for $$i>0$$.

Q: Why does the $$\delta$$-functors being effaceable follow from this? Effaceable means (in this case) that for any object $$\F \in \QCoh(X)$$ there exists a monomorphism $$u : \F \ra \mathcal{G}$$ such that $$\{ R^i f_*(u) \}_{i \geq 0} = 0$$ and $$\{\widetilde{H^i(X, u)} \}_{i \geq 0}=0$$, some $$\mathcal{G}$$. I think that this is because we can say $$u$$ is the embedding into a flasque quasi-coherent sheaf, and since sheaf cohomology vanishes for flasque sheaves and higher direct images, we get the result. Is this correct modulo details?

Then by a theorem of Grothendieck, both of the $$\delta$$-functors are universal (i.e. a universal $$\delta$$-functor is characterised by the property that giving any morphism from it to any other $$\delta$$-functor is equivalent to giving just the $$0$$th degree). Therefore, $$\mu^0$$ gives rise to the canonical natural equivalence that we require.

Apologies for the lengthy post and thank you for any answers!

Q1: On an affine scheme $$Z$$, we know that for any quasicoherent sheaf $$\mathcal{A}$$ we have that $$\mathcal{A}\cong \widetilde{\mathcal{A}(Z)}$$. Applying that to the case at hand, we know that $$f_*\mathcal{F}$$ is a quasicoherent sheaf on $$Y$$, so $$f_*\mathcal{F}\cong \widetilde{f_*\mathcal{F}(Y)}$$. But by the definition of the pushforward, we have that $$f_*\mathcal{F}(Y) = \mathcal{F}(f^{-1}(Y)) = \mathcal{F}(X)$$, and so we may conclude that $$f_*\mathcal{F} \cong \widetilde{\mathcal{F}(X)}$$.
Q2: One functor described here is just $$R^\bullet f_*(-)$$, which is known to be a cohomological $$\delta$$-functor basically by definition (it's a right derived functor of a left exact functor). The other functor here is the functor $$\widetilde{H^\bullet(X,-)}$$, which can be written as the composite of the two functors $$\widetilde{-}$$ and $$H^\bullet(X,-)$$. The first functor, $$H^\bullet(X,-)$$ is already a $$\delta$$-functor, and $$\widetilde{-}$$ being exact implies that it preserves the exactness of all the sequences and diagrams required to verify $$H^\bullet(X,-)$$ as a $$\delta$$-functor, so the composite is a $$\delta$$-functor.