# Is every restriction map in the sheaf theory surjective?

Let $$\mathcal{F}$$ be a sheaf on an $$n$$-dimensional manifold $$X$$. More precisely, $$\mathcal{F}$$ is a contravariant functor, defined on the category of open subsets of $$X$$ to the category of (finite dimensional) complex vector spaces. Then, for any two open subset $$U$$ and $$V$$ of $$X$$ such that $$U \subset V$$, there is a restriction map $$r : \mathcal{F}(V) \to \mathcal{F}(U)$$.

Could I say that the restriction map $$r : \mathcal{F}(V) \to \mathcal{F}(U)$$ is surjective for any $$U \subset V \subset X$$? If not, could you give me a counter example?

With $$X=\Bbb C$$, consider the sheaf of holomorphic functions. Then $$\mathcal F(\Bbb D)$$ contains $$f\colon z\mapsto \frac1{1-z}$$, but $$\mathcal F(\Bbb C)$$ (or $$\mathcal F(V)$$ for any $$V$$ containing also $$1$$) contains no holomorphic function that restricts to $$f$$.
However, you specifically look for sheaves where local cuts are finite dimensional vector spaces, so the holomorphic sheaf won't work, and in fact that severely restricts the possible topologies of $$X$$, or the interestingness of sheaves allowed (for exxmaple, you cannot have infinitely many disjoint open sets with non-trivial cuts).