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?

Thanks in advance.


No. In particular, a local cut cannot always be extended to a global cut.

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).


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