# Surjectivity of morphisms of sheaves on a base

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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What do you guess? Actually it seems that (2) is rather simple. And I guess that the answer to (1) follows simply applying the properties of sheaves to the definition of topological basis. – Giovanni De Gaetano Nov 7 '12 at 10:05
You mean both are true? – Tom Nov 7 '12 at 12:16
I also thought so, but Vakil's notes p85(3.7.E) disagree with it. – Tom Nov 7 '12 at 12:27
The converse of (1) is false: being an epimorphism of sheaves is a property that is in general detected at the level of stalks. But if it is surjective at the level of sections it will automatically be surjective on the level of stalks. – Zhen Lin Nov 7 '12 at 12:31
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)? – Tom Nov 7 '12 at 13:33

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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Thank you. But I do not understand the last line. What does $\mathbb C[X]_f$ mean? $\mathbb C[X]$ is the set of continuous functions on $X$? – Tom Nov 8 '12 at 1:47
@Tom: $\mathbb C[X]$ is the ring of polynomials in one variable $X$ and $\mathbb C[X]_f$ are the ring of rational fonctional with denominators a positive power of $f$. But I see that $X$ also denotes the algebraic variety $X$! I will change the notation. – user18119 Nov 8 '12 at 6:48
Thank you for your answer. Sorry for not responding soon. It is because I didn't have enough knowledge(structural sheaf) to understand yours. But now I understand, thank you. – Tom Nov 15 '12 at 4:59
@Tom: you are welcome ! – user18119 Nov 16 '12 at 0:32