Let $\mathcal{F}$ a sheaf on $X$. We can define the Godement sheaf $\mathcal{F^+}$ by $\Gamma(U, \mathcal{F^+}) := \prod _{a \in U} \mathcal{F_a}$ for every open subset $U$.

Then we have the canonical inclusion $\mathcal{F} \subset \mathcal{F^+}$ by $s_U \to (s_a)_{a \in U}$. Set $\mathcal{F^0} := \mathcal{F^+}$.

Recursively we can define $\mathcal{F^{r+1}}:= (\mathcal{F^r}/\mathcal{F^{r-1}})^+$. This provides the Godement resolution

$0 \to \mathcal{F} \to \mathcal{F^0} \to \mathcal{F^1} \to ...$

Their cohomology is defined by $H^r(X, \mathcal{F}):= \frac{Ker(\Gamma(X, \mathcal{F^r}) \to \Gamma(X, \mathcal{F^{r+1}})}{Im(\Gamma(X, \mathcal{F^{r-1}}) \to \Gamma(X, \mathcal{F^r}))}$.

My question is: If $\mathcal{F} $ has a $\mathcal{O}_X$-module structure, why $H^r(X, \mathcal{F})$ is only depending on $ \mathcal{F}$ as abelian sheaf, therefore $H^r(X, \mathcal{F})$ "forgets" the $\mathcal{O}_X$-module structure?

  • $\begingroup$ Where do you use the $\mathcal{O}_X$-module structure in the definition of $\mathcal{F}^+$, the Godement resolution and its cohomology ? $\endgroup$
    – Roland
    Dec 4, 2017 at 17:38
  • $\begingroup$ The $\Gamma(X, \mathcal{F}^r) $ inherit the $\mathcal{O}_X(X)$ so their kernels and images, too, don't they? $\endgroup$
    – KarlPeter
    Dec 11, 2017 at 17:44
  • 2
    $\begingroup$ Yes but this is not relevant to your question. In the definition of $H^r(X,\mathcal{F})$ you can remember that $\mathcal{F}$ has an $\mathcal{O}_X$-module structure, or forget it, it will not change the result, because nowhere in the definition you use it (even if it exists). $\endgroup$
    – Roland
    Dec 11, 2017 at 17:51
  • $\begingroup$ But what is the special feature of the Godement resolution concerning the forgeting the $\mathcal{O}_X$- module structure? I can maybe choose another injective resolution $0 \to \mathcal{F} \to \mathcal{I}^0 \to \mathcal{I}^1 \to ...$ of $\mathcal{O}_X$- modules and consider the cohomology $H^r(X, \mathcal{F}):= \frac{Ker(\Gamma(X, \mathcal{I^r}) \to \Gamma(X, \mathcal{I^{r+1}})}{Im(\Gamma(X, \mathcal{I^{r-1}}) \to \Gamma(X, \mathcal{I^r}))}$ where I can again consider $H^r(X, \mathcal{F})$ again as $\mathcal{O}_X$-module as well as abelian group by forgetting the module structure... $\endgroup$
    – KarlPeter
    Dec 11, 2017 at 21:33
  • $\begingroup$ Godement resolution has several nice features : - it is historically an important construction - it is functorial ! No need to make choices of injectives. - you can show that this gives the right cohomology groups. But since Godement resolution does not depend on the $\mathcal{O}_X$-module structure, this show that the cohomology of $\mathcal{F}$ does not depend on the $\mathcal{O}_X$-structure. This is not obvious at all ! In fact it fails for coherent sheaves on some weird schemes : the derived functor of $\Gamma:Coh(X)\rightarrow Ab$ is not the same as the usual $H^i$. $\endgroup$
    – Roland
    Dec 11, 2017 at 22:14

1 Answer 1


There is a very subtle point in homological algebra which can easily be overlooked. Let's say that $\mathcal{F}$ is a quasi-coherent sheaf over a scheme $X$. What does $H^i(X,\mathcal{F})$ means ?

Easy one might say : this is the derived functor of the global section functor $\Gamma(X,.)$. But what is the source category of $\Gamma(X,.)$ ? Is is $QCoh(X), \mathcal{O}_X$-mod or $\mathcal{Ab}_X$ ? Does that change something ?

From a more abstract view point, let $\mathcal{A}\subset \mathcal{B}$ be abelian categories such that the inclusion functor $i$ is exact. Let $F:\mathcal{B}\rightarrow\mathcal{C}$ be a left exact functor and $X\in\mathcal{A}$. What does $R^kF(X)$ means ? Is it $(R^kF)(i(X))$ or $R^k(F\circ i)(X)$. What makes things really confusing is the fact that in general we don't write $i$, but we should since these functors are not equals in general.

Before giving examples, let see where there might be a problem. By definition, $R^k(F\circ i)(X)$ are computed this way :

  • choose an injective resolution of $X$ in $\mathcal{A}$. Write it $X\rightarrow I^\bullet$.
  • apply $i$, in other words, see $I^\bullet$ as objects of $\mathcal{B}$.
  • apply $F$
  • take cohomology.

Since the inclusion functor is exact, $i(X)\rightarrow i(I^\bullet)$ is still a resolution in $\mathcal{B}$. So it looks like we can use it to compute $(R^kF)(i(X))$. But $i(I^\bullet)$ might not be injective in $\mathcal{B}$. Worse, it might not be acyclic for $F$, thus, it can't be used to compute $(R^kF)(i(X))$.

Here is an easy example where it fails, an example which is not far away from your situation ! Instead of sheaves, consider just $\mathbb{F}_2$-vector spaces and $\mathbb{Z}$-modules (in other words let $X=\operatorname{Spec}\mathbb{F}_2$). We have the inclusion functor $\mathbb{F}_2$-mod$\subset\mathbb{Z}$-mod which is exact. Now consider the functor $F:\mathbb{Z}$-mod$\rightarrow\mathbb{Z}$-mod such that $F(A)=\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},A)$.

The functor $F\circ i$ is exact (in fact it is canonically isomorphic to $i$). So its derived functor is trivial.

However $R^1F(\mathbb{F}_2):=(R^1F)(i(\mathbb{F}_2))=\operatorname{Ext}^1_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}$ as you can easily see using a projective resolution in the first argument.

As I said previously, an injective resolution of $\mathbb{F}_2$ in $\mathbb{F}_2$-mod is simply $\mathbb{F}_2$ (in fact every object is injective). But this object is not injective anymore in $\mathbb{Z}$-mod, worse, it is not acyclic.

What is going on for sheaves and for quasi-coherent modules on schemes ?

A bad news : on some weird schemes (I don't have counter-examples), the derived functor of $\Gamma(X,.):QCoh(X)\rightarrow\mathrm{Ab}$ does not give the right thing.

A good news : the derived functor of $\Gamma(X,.):\mathcal{O}_X\mathrm{-mod}\rightarrow\mathrm{Ab}$ does not depend on the $\mathcal{O}_X$-module structure, and give the same thing as the derived functor of $\mathcal{Ab}_X\rightarrow\mathrm{Ab}$. This is what is usually called $H^i(X,.)$.

How to see this : in $\mathcal{O}_X$-mod, injectives modules are flabby (flasque). A proof is given here : https://stacks.math.columbia.edu/tag/01EA. (You will see that it rely on sheaves $j_!\mathcal{O}_U$. But these are not quasi-coherent, this proof fails in $QCoh(X)$ : in $QCoh(X)$ injective are not necassarily flabby). Moreover, in $\mathcal{Ab}_X$, flasque sheaves are acyclics, so can be used to compute cohomology.

Or you can use Godement resolution : it computes the right thing (in fact this is a resolution by flabby sheaves), and does not depend on the $\mathcal{O}_X$-module structure.

  • $\begingroup$ Thank you for your detailed answer. Concretly, by independence of $\Gamma(X, .)$ from $\mathcal{O}_X$-strucure you mean that for all $\mathcal{O}_X$-modules $\mathcal{F}$ holds: $\Gamma(X, \mathcal{F}|_{Ab})=\Gamma(X, \mathcal{F})|_{Ab}$ where $|_{Ab}$ means applying the forgetful functor, right? So to prove this it would be enough to find a resolution which provides the cohomologies in both categories? $\endgroup$
    – KarlPeter
    Dec 13, 2017 at 0:07
  • $\begingroup$ If I understood your argument correctly we can wlog consider a flasque resolution which stays flasque under "forgetting" by $|_{Ab}$ and is therefore in BOTH categories acycyclical. So our strike is that we can take this (therefore the same) resolution for calculating cohomology in $(\mathcal{O}_X)$ as well as in $(Ab-Sh)$? Or did I misunderstood your argument? $\endgroup$
    – KarlPeter
    Dec 13, 2017 at 0:07
  • 1
    $\begingroup$ Yes that is correct. The thing is, you can take Godement resolution for the definition of sheaf cohomology, this has the advantage that you don't have to bother with this subtle point. $H^i(X,\mathcal{F})$ is independent of the $\mathcal{O}_X$-module structure because it is not used in the definition of the resolution, and that's it. Many books use this viewpoint. $\endgroup$
    – Roland
    Dec 13, 2017 at 6:21

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