Let $X$ be a normal, projective variety and $U$ be the regular locus of $X$. Let $\mathcal{F},\mathcal{G}$ be reflexive sheaves on $X$ and $f:\mathcal{F} \to \mathcal{G}$ be a morphism. Suppose that the restriction of $f$ to $U$ is surjective i.e., $f|\_U:\mathcal{F}|\_U \to \mathcal{G}|\_U$ is surjective. 


Is it true that $f:\mathcal{F} \to \mathcal{G}$ is surjective? The problem that I have is, $U$ need not be affine. 


The second question is: Is there any criterion when $U$ is going to be affine? More generally, does there exist an open subcheme $V$ contained in $U$ which is affine and satisfies $U\backslash V$ is of codimension at least $2$?



On a morphism between reflexive sheaves
No, that is not true. Let $X$ be a normal, quasi-projective variety over a field $k$. Let $U\subset X$ be the regular locus. Denote by $Z\subset X$ the closed complement of $U$ with its reduced, induced structure. Assume that $Z$ is not empty. Denote by $\mathcal{I}\subset \mathcal{O}\_X$ the ideal sheaf of $Z$. Since $X$ is quasi-projective, there exists an integer $d\geq 0$, and integer $N>0$, and a surjection $\widetilde{f}:\mathcal{O}\_X(-d)^{\oplus N} \to \mathcal{I}$. 


Denote $\mathcal{O}\_X(-d)^{\oplus N}$ by $\mathcal{F}$. Denote by $\mathcal{G}$ the structure sheaf $\mathcal{O}\_X$. Denote by $f$ the composition of $\widetilde{f}$ and the inclusion $\mathcal{I}\subset \mathcal{O}\_X$. Then $f$ is a morphism of reflexive sheaves that is surjective on $U$, yet it is not surjective on all of $X$.


