Let $X$ be a smooth variety, $D^{b}(X)$ be the derived category of bounded coherent sheaves.Then there is a definition of $Hom(F^{\cdot},G^{\cdot})$ which is the derived functor of $Hom(F^{\cdot},-)$.

I think the way to compute $Hom(F^{\cdot},G^{\cdot})$ is: first one find a complex of injective objects $I^{\cdot}$ quasi-isomorphic to $G^{\cdot}$, and calculate $Hom(F^{\cdot},I^{\cdot})$ in the homotopy category $K^{b}(X)$ (Please correct me if anything I mentioned is wrong!).

My question is: if there is a complex of flasque sheaves $H^{\cdot}$ quasi-isomorphic to $G^{\cdot}$, can we still calculate $Hom(F^{\cdot},G^{\cdot})$ by using $Hom(F^{\cdot},H^{\cdot})$?

The motivation behind the problem is: in the computation of cohomology one can use flasque resolution instead of injective resolution.


I am a little bit confused by this question, but let me share my (limited) perspective. Almost all I know is included in Hartshorne's Algebraic Geometry. I will later write why this is a little bit limited.

The importance of flasque sheaves in computation of sheaf cohomology stems from the following fact that can be proved by hand

Let $0 \rightarrow \mathcal{A} \rightarrow \mathcal{B} \rightarrow \mathcal{C} \rightarrow 0$ be a short exact sequence of sheaves of abelian groups on some space X. If $\mathcal{A}$ is flasque, then the sequence $0 \rightarrow \Gamma(\mathcal{A}) \rightarrow \Gamma(\mathcal{B}) \rightarrow \Gamma(\mathcal{C}) \rightarrow 0$ is also exact.

This is a very special property of flasque sheaves and from this property if follows that every flasque sheaf $\mathcal{F}$ is acyclic, ie. $H^{p}(X, \mathcal{F}) = 0$ for all $p > 0$. This is important because this is a key step in establishing the isomorphism between Cech and derived functor cohomology, since injective sheaves are flasque, too.

Note that one proves - and this is crucial - that all flasque sheaves are acyclic, not necessarily quasi-coherent.

This is why I believe your question has a negative answer. Whether you mean the functor $Hom$ or $\mathcal{H}om$ (the sheaf version), it fails to be exact already at the module-level. I fail to see how flasqueness could salvage anything, but I'd love to be proven wrong! However, if your question is whether you can use flasque, quasi-coherent sheaves, than I am unable to come with any counterexamples - because my only known example of such a sheaf is $\widetilde{I}$ for $I$ an injective $A$ over a neotherian ring $A$.

If your question is about arbitrary flasque sheaves (because one can use arbitrary sheaves of abelian groups to compute cohomology), then there are trivial counterexamples. Let $k$ be an algebraically closed field of characteristic $2$ and consider the following flasque resolutions of $k$ on Spec(k):

$C_{1} ^{\bullet}: (0 \rightarrow k) \rightarrow k \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow _{*2} \mathbb{Z} \rightarrow \mathbb{Z}_{2} \rightarrow 0 \rightarrow \ldots$

$C_{2} ^{\bullet}: (0 \rightarrow k) \rightarrow k \rightarrow 0 \rightarrow \ldots$

Since $Hom(k, C_{1}^{\bullet})$ and $Hom(k, C_{2}^{\bullet})$ have different cohomology, they cannot conceivably both be equal to $RHom(k, k)$. I think it's not too hard to strenghten this example to one where all sheaves involved are sheaves of $\mathcal{O}_{X}$-modules ($\mathbb{Z}$ is not a $k$-module), but - as I already said - I don't know how to change it so that all sheaves involved are quasi-coherent.

  • $\begingroup$ Dear Piotr Pstragowski, thank you, I agree that it might not true that $Hom$ can be computed by flasque resolution because, basically, one cannot using flasque resolution to compute $Hom(\mathcal{F},\mathcal{G})$(this is not in the derived sense). $\endgroup$ – Li Yutong Mar 4 '13 at 20:37

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