Godement Resolution don't see $\mathcal{O}_X$-module structure

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?

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.