# Functoriality of Ext-functor

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 $$X$$?

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 $$\neq 0$$.