Cech Cohomology and module structure

Question

I just came across a question I had never thought about and that could be simple, but I can't answer it. If we consider a scheme $X$ and we pick a sheaf of $\mathcal{O}_{X}$-modules $\mathcal{F}$, under some assumptions we can compute its cohomology either via an injective resolution or with Cech cohomology. Both the answers carry an $\mathcal{O}_X$-module structure, but the theorem that states that the answers we get with the two approaches is the same is always stated saying that the abelian groups are the same. Therefore my question, is the isomorphism between Cech cohomology and cohomology as a derived functor an isomorphism of $\mathcal{O}_{X}$-modules?

Answer 1

New answer:

I'm going to explain in what sense the answer is yes, and why the "counterexample" I gave in my last answer was contrived.

First, recall that the cohomology $H^i(X, F) = R^i\Gamma(X, F)$ is only defined up to natural isomorphism, and depends on the choice of injective resolution of $F$. So, let's denote by $H^i(F,I)$ the cohomology with respect to an injective resolution $I$ of $F$.

If we pick a different resolution $J$ there is a natural map $H^i(F,I) \cong H^i(F,J)$ which doesn't depend on all the choices you make to construct it. So, if $I$ happens to be a resolution by $\mathcal O$-modules, this makes $H^i(F,I)$ an $\mathcal O$-module, and even if $J$ is not a resolution by $\mathcal O$-modules the natural isomorphism gives its cohomology an $\mathcal O$-module structure.

Since the map from the Cech sheaf to an injective resolution also gives natural maps $\check H^i(X,F) \to H^i(F,I)$ and $\check H^i(X,F) \to H^i(F,J)$, these will respect both the $\mathcal O$-action on $I$ and on $J$.

So in this sense the answer is yes, the map is always an isomorphism of $\mathcal O$-modules.

The example I gave originally is actually pretty contrived. It arose from this observation: If you take a complex of $\mathcal O$-modules $J$ which is an injective resolution of $F$ as abelian groups, but for which the augmentation $\varepsilon : F \to J$ is not $\mathcal O$-linear, then the map $\check H^i(X,F) \to H^i(F,J)$ will not be $\mathcal O$-linear, nor will the natural isomorphism $H^i(F,I) \cong H^i(F,J)$ with respect to be $\mathcal O$-actions.

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Original answer:

How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.

Basically you can build a resolution of $\mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $\mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $\mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?

The answer is no. Consider the case $X$ is a point. Let $\mathcal F = \mathcal O_X = \underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $\mathcal F$ with an injective resolution. One injective resolution is $\phi: \underline{k} \to \underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $\mathcal O$-linear by design because it coincides with $\phi$ itself.

So, if you want an $\mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $\Gamma(\mathcal O,X)$-linear.