$X=\mathbb{C}$ with Zariski topology and $\mathcal{F}=\underline{\mathbb{Z}_X}$ the locally constant sheaf. $\mathcal F$ is flasque

Question

Let $X=\mathbb{C}$ with Zariski topology and $\mathcal{F}=\underline{\mathbb{Z}_X}$ the locally constant sheaf.

I want to prove that $\mathcal F$ is flasque, i.e. the restrictions are surjective.

I take $U\subseteq V $ two open subsets.

$\mathcal{F}(U)$={$ f:U\to \mathbb{Z}| \quad \forall x\in U \quad \exists W\in \mathcal{V}_x \quad \exists g:W\to \mathbb{Z}$ constant $ s.t. \quad f|_W=g$}

I have to prove that for any $f\in \mathcal{F}(U)$ there exists $h\in \mathcal{F}(V)$ s.t. $h|_U=f$.

Now, $U=\mathbb{C}$ \ {$a_1,...,a_n$} and $V=\mathbb{C}$ \ {$ a_1,...,a_m$}, $n\geq m$ and $a_i\in\mathbb{C},\forall i$.

How to continue?

I need examples and counterexamples of flasque sheaves.

Answer 1

It is trivial to see:

A non-zero locally constant sheaf on a connected space is flasque if and only if the space is irreducible (i.e. if the locally constant sheaf is actually the constant presheaf)

So all you have to do is check that $X$ is irreducible and I think this is well known to you.