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](https://stacks.math.columbia.edu/tag/087M). 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 $X$?



Functoriality of Ext-functor
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 $\neq 0$.


