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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.

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 $U=X\setminus \operatorname{Supp}D$, $i^*\mathscr O_X(mD)\simeq \mathscr O_U\simeq i^*\mathscr O_X(nD)$ 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 linked answer for more.

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.

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.

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 $D\subset X$ and note that for $U=X\setminus D$, $i^*\mathscr O_X(mD)=i^*\mathscr O_X(nD)$ 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.

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.