Lemma 20.7.3 (01E4)—The Stacks project

Lemma 20.7.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a $\mathcal{O}_ X$ -module. The sheaves $R^ if_*\mathcal{F}$ are the sheaves associated to the presheaves

$V \longmapsto H^ i(f^{-1}(V), \mathcal{F})$

with restriction mappings as in Equation (20.7.1.1). There is a similar statement for $R^ if_*$ applied to a bounded below complex $\mathcal{F}^\bullet$ .

Proof. Let $\mathcal{F} \to \mathcal{I}^\bullet$ be an injective resolution. Then $R^ if_*\mathcal{F}$ is by definition the $i$ th cohomology sheaf of the complex

$f_*\mathcal{I}^0 \to f_*\mathcal{I}^1 \to f_*\mathcal{I}^2 \to \ldots$

By definition of the abelian category structure on $\mathcal{O}_ Y$ -modules this cohomology sheaf is the sheaf associated to the presheaf

$V \longmapsto \frac{\mathop{\mathrm{Ker}}(f_*\mathcal{I}^ i(V) \to f_*\mathcal{I}^{i + 1}(V))}{\mathop{\mathrm{Im}}(f_*\mathcal{I}^{i - 1}(V) \to f_*\mathcal{I}^ i(V))}$

and this is obviously equal to

$\frac{\mathop{\mathrm{Ker}}(\mathcal{I}^ i(f^{-1}(V)) \to \mathcal{I}^{i + 1}(f^{-1}(V)))}{\mathop{\mathrm{Im}}(\mathcal{I}^{i - 1}(f^{-1}(V)) \to \mathcal{I}^ i(f^{-1}(V)))}$

which is equal to $H^ i(f^{-1}(V), \mathcal{F})$ and we win. $\square$

Comments (4)

Comment #1811 by Keenan Kidwell on January 30, 2016 at 19:10

Is there something non-trivial being used here about sheafification and quotients? The second displayed formula is the presheaf quotient of the kernel of $f_*\mathcal{I}^i\to f_*\mathcal{I}^{i+1}$ by the subpresheaf given by the presheaf image of $f_*\mathcal{I}^{i-1}\to f_*\mathcal{I}^i$ . But by definition, the cohomology sheaf $R^if_*\mathcal{F}$ is the sheaf associated to the presheaf quotient of the aforementioned kernel by the sheaf-theoretic image of $f_*\mathcal{I}^{i-1}\to f_*\mathcal{I}^i$ . That these are the same follows by e.g. exactness of sheafification, right? Or is there something simpler that I'm missing?

Comment #1829 by Johan on February 04, 2016 at 20:12

Nope, you are not missing anything. The problem with statements such as these is really that once you understand everything well enough, it actually becomes confusing when you try to spell out what is going on. At least that is my take on it.

Comment #4643 by Andy on November 03, 2019 at 16:58

Adding to the discussion here, I think another point of view is that pullback of sheaves is forgetting it's a sheaf, pulling back, and sheafifying. And the second and third functor do not contribute in the derived setting, hence the statement. I'm not claiming this is any easier than the proof given of course.

Comment #4644 by Andy on November 03, 2019 at 17:00

Sorry I meant to say pushforward, but still should be exact in the presheaf setting.

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