Let $X$ be a normal, integral variety and $U \subset X$ an open subset such that the complement of $U$ is of codimension at least $2$. Let $F$ be a coherent sheaf on $X$ such that $\mathcal{E}xt^1_U(F|_U,\mathcal{O}_U)=0$, where $\mathcal{E}xt^1$ denotes sheaf Ext. Does this imply that $\mathcal{E}xt^1_X(F,\mathcal{O}_X)=0$? I know that (locally) $Ext$ commutes with flat base change. But, since the natural inclusion map from $U$ to $X$ is not affine, I do not know if I can use this statement to answer my question (if I could then using the fact the $i_*\mathcal{O}_U=\mathcal{O}_X$ and then the base change statement, would give a positive answer to my question). Is there an analogous sheaf theoretic functoriality statement of Ext? To summarize, can I say that: $$\mathcal{E}xt^1_X(F,i_*\mathcal{O}_U) \cong i_*\mathcal{E}xt^1_U(i^*F,\mathcal{O}_U),$$ where $i$ is the natural inclusion of $U$ in $X$?
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This is definitely false, even in very simple situations: take $X$ smooth, $Z$ a smooth subvariety of $X$ of codimension 2, and $\mathscr{F}=\mathscr{I}_Z$. Then obviously $\mathscr{E}xt^1_U(\mathscr{I}_{Z}{}_{|U}, \mathscr{O}_{U})=0$, but because of the exact sequence $0\rightarrow \mathscr{I}_Z\rightarrow \mathscr{O}_X\rightarrow \mathscr{O}_Z\rightarrow 0$, $\mathscr{E}xt^1_X(\mathscr{I}_Z,\mathscr{O}_X)$ is isomorphic to $\mathscr{E}xt^2_X(\mathscr{O}_Z,\mathscr{O}_X)$ which is $\neq 0$.
