Let $X$ be a topological space, $\mathcal F$ a sheaf of abelian groups on $X$ and $X=\bigcup_iU_i$ an open covering such that $\mathcal F(U_i)=0$ for all $i$. Does this imply $\mathcal F=0$?
I think it doesn't, but I haven't managed to come up with a counterexample.
No. Suppose $X = \Bbb P^1,$ and let $\mathcal{F} = \mathcal{O}(-1).$ $\mathcal{F}$ is nontrivial, as $\mathcal{F}(\Bbb A^1)\neq 0$ for any $\Bbb A^1\subseteq\Bbb P^1,$ but $\mathcal{F}(X) = \{0\}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.