Consider the sheaf $\mathcal{F}$ of polynomial functions on $\mathbb{R}^2$ endowed with the usual topology.
A sheaf is said to be "flasque" (or "flabby") if, given $V \subset U$ both open sets, the restriction map $\mathcal{F}(U) \to \mathcal{F}(V)$ is surjective.
I want to claim $\mathcal{F}$ is flasque, since any polynomial on $U$ restricts to $V$ and is entirely determined by its restriction to $V.$
The same argument would work for polynomial functions (or indeed analytic functions) on $\mathbb{R}^n$ (or $\mathbb{C}^n$) for any $n$.
Could anyone confirm that I am not writing nonsense? I am confused because this seemingly natural example is not given in my textbook, which suggests perhaps it is wrong...
For analytic functions, the restriction map is rarely surjective, a function analytic on $V$ in general does not have an analytic continuation to $U$ (but consider Hartogs figures to see that sometimes it is surjective for $V\subsetneqq U$).
For polynomials or entire analytic functions, the issue is with sheaves vs. presheaves. If $\mathcal{F}(U)$ is the set of restrictions of polynomials (or entire analytic functions) to $U$, then the restriction $\mathcal{F}(U) \to \mathcal{F}(V)$ is surjective for all open $V\subset U$, but $\mathcal{F}$ is not a sheaf, only a presheaf, since if $U$ is disconnected, and $U_1,U_2$ are two disjoint nonempty open sets with $U = U_1 \cup U_2$, then there is no $s\in \mathcal{F}(U)$ such that $s\lvert_{U_1} \equiv 1$ and $s\lvert_{U_2} \equiv 2$, although $1\lvert_{U_1\cap U_2} = 2\lvert_{U_1\cap U_2}$.
If $\mathcal{F}$ is the sheaf constructed from that presheaf, the restrictions are in general not surjective, consider a disconnected $V$ to see that.
So in neither case is $\mathcal{F}$ a flabby (or flasque) sheaf.