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suppose X is a smooth variety and F is a locally free sheaf on X. Let U be an open subset of X and i denote the inclusion map. Is i_*i^*F equal to F ?


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No. For example, let $X = P^1$, $U = A^1$ and $F = O_X$. Then $i^*F = O_U$ and the global sections of $i^*F$ is the algebra of polynomials $k[t]$. Therefore $\Gamma(X,i_*i^*F) = \Gamma(U,i^*F) = k[t]$, while $\Gamma(X,F) = k$.

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This is true if and only if the complement of $U$ has codimension at least $2$. To see that this condition is sufficient see this MO answer. To see that it is necessary, see Sasha's example, or take any $X$ and any Cartier divisor $D$ on $X$ and note that for $U=X\setminus \mathrm{Supp}D$, $i^*\mathscr O_X(mD)\simeq \mathscr O_U\simeq i^*\mathscr O_X(nD)$ for any $m,n\in \mathbb Z$, so $i_*i^*F$ can't be $F$ for both choices.

Remark for the codimension $2$ condition, you don't actually need smoothness. See the liked answer for more.

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In general, we have the socalled projection formula: if $f:X\to Y$ is a morphism of ringed spaces, $\mathcal F$ an $\mathcal O_X$-module, and $E$ be a locally-free $\mathcal O_Y$ module of finite rank, then $f_* (\mathcal F \otimes f^{*} E) \simeq f_{*}\mathcal F \otimes E$.

Edit (following Will's remark): The projection formula yields in the case of an open immersion $i:U \subset X$ the following identity : $i_* i^* F \simeq i_*\mathcal O_U \otimes F$. Therefore, if $U$ has codimension at least 2 in $X$, then $i_* i^* F\simeq F$ by normality of $X$.

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You need the assumption $f_* \mathcal O_U=\mathcal O_X$, which is not satisfied for all open immersions. – Will Sawin Nov 21 '12 at 8:39
Thanks, you are perfectly right! (I had in mind a morphism with connected fibers) So here here one can just say that $i_*i^*F= i_* \mathcal O_X \otimes F$. – Henri Nov 21 '12 at 10:06
In particular, as $X$ is smooth hence normal, the desired property holds as soon as $U$ has codimension at least $2$. – Henri Nov 21 '12 at 10:11

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