Let $\mathfrak{F}$ be a sheaf of abelian groups on a smooth scheme $X$. Suppose for some $i>1$, there exists a surjective morphism $H^i(\mathfrak{F}^{\otimes i1}) \to H^i(\mathfrak{F}^{\otimes i})$. Is it true that this means $h^i(\mathfrak{F}^{\otimes i1})=h^i(\mathfrak{F}^{\otimes i})=0$?
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closed as unclear what you're asking by Fernando Muro, Daniel Moskovich, David Roberts, Carlo Beenakker, Andres Caicedo Jun 30 '13 at 18:17
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No. Take $X$ to be an Abelian variety of dimension $g$, and $\mathfrak{F} = \mathcal{O}_X$. Then $h^i(\mathcal{O}_X) = \binom{g}{i}$. 

