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Are the following two true?

(1) If a morphism of sheaves on a base(of a topological space) is surjective(I.e. $F(B_i)\to G(B_i)$ is surjective for all $B_i$) then the corresponding morphism of sheaves( not on the base but on the open sets) is epi.

(2) The converse of (1).

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    Thank you. There may be two versions of (2): (2a) fixed base or (2b) finer base can be chosen. Your explanation works for (2a). How about (2b)?2012-11-07

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Here is an example of a surjective morphism of sheaves $F\to G$ on $X$ such that for any open covering $\{ U_i\}_i$ of $X$, there is an index $i$ such that $F(U_i)\to G(U_i)$ is not surjective.

Let $X=\mathrm{Spec}(\mathbb C[t])$ be the affine line (say over $\mathbb C$) with origin $e$. Let $F=O_X$ be the structural sheaf of $X$. Let $R=O_{X,e}=\mathbb C[t]_{t\mathbb C[t]}$. Let $G$ be defined by $ G(U)= \left\lbrace\matrix{ 0 & \text{if } e\notin U \\ R & \text{if } e\in U.}\right.$ We can check that $G$ is actually a sheaf and $G_x=0$ if $x\ne e$ and $G_e=R$.

Consider the morphism $F\to G$ which on $U$ is the zero map if $e\notin U$ and is the localization map at $e$ otherwise. Then $F_x\to G_x$ is surjective for all $x\in X$. Hence $F\to G$ is surjective.

But for any open subset $U\ni e$, $F(U)\to G(U)=R$ is not surjective because if $U=D(f)$, then $F(U)=\mathbb C[t]_f$ is clearly strictly smaller than $R$.

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    @Tom: you are welcome !2012-11-16