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$\newcommand{\oh}{\mathcal{O}} \newcommand{\QCoh}{\mathsf{QCoh}} \newcommand{\ra}{\rightarrow} \newcommand{\F}{\mathcal{F}} \newcommand{\Mod}{\text{-}\mathsf{Mod}}$I'm working through the proof of Proposition 10 of Murfet's notes. The proposition is as follows:

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!

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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.

Q3: Yes, this is essentially correct.

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  • $\begingroup$ Great - everything makes sense now. Thank you for your answer! $\endgroup$
    – mathphys
    May 31, 2019 at 12:51

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