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how can I see quickest that the following holds:

Let $X$ and $Y$ be sufficiently nice schemes (e.g. always noetherian or maybe varieties) and denote with $p$ the projection

$X\times Y \rightarrow X$.

Then for a (sufficiently nice) sheaf $F$ on $X\times Y$ the adjunction

$p^*p_*F\rightarrow F$

is an isomorphism.

Thanks!

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1  
Something is incorrect here. Is $F$ on $X$? If so, then $p^*p_*F$ doesn't make sense. –  Matt Aug 20 '11 at 15:03
    
Thank you, Matt, I already corrected it! –  Descartes Aug 20 '11 at 15:28
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In that case, I don't think the nicest situation will help. Suppose $X=\mathbb{P}^1$ and $Y=Spec(k)$ and $F=\mathcal{O}(1)$. –  Matt Aug 20 '11 at 15:37
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This can't be true: how could you reconstruct $F$ on $X\times Y$ with only a data on $X$? –  Henri Aug 20 '11 at 15:38
    
Ok, thanks for the counterexample. Actually this question arose when I considered the same situation, but the $p^*$ and $p_*$ are derived functors and you consider the derived categories of bounded coherent sheaves. I don't know if that changes something? I mean when you require only iso in the derived category. –  Descartes Aug 20 '11 at 16:08

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