Let $C$ be a site and $F$ an abelian presheaf on $C$. Suppose that for each object $U$ in $C$ there is a covering $\{ U\_i\to U \}$ such that $F(U\_i)=0$. Is it true that $F^{sh}=0$? 
This should be true, and is trivially if the presheaf is separated, but I don't see how to prove it in general.


edit: slight clarification



If presheaf is zero on a covering is the sheaf zero?
Indeed in the situation above 


$$F^\nmid=\varinjlim\_{\lbrace U\_i\to U\rbrace} \ker \left(\prod\_i F(U\_i)\rightrightarrows \prod\_{i,i'} F(U\_i\times\_U U\_{i'})\right)=:H(\lbrace U\_i\to U\rbrace, F)$$


is already zero. Fix a covering $\lbrace U\_i\to U \rbrace$. By assumption we find $\lbrace V\_{ij}\to U\_i\rbrace$ such that $F(V\_{ij})=0$ for all $i,j\in I\times J$. Also by the definition of a site $\lbrace V\_{ij}\to U\rbrace$ is a covering.
The canonical refinement $\lbrace V\_{ij}\to U\rbrace\to \lbrace U\_i\to U\rbrace$ induces


$$
H(\lbrace U\_i\to U\rbrace, F)\to H(\lbrace V\_{ij}\to U\rbrace, F)=0
$$


where the right side is zero because the product of $F(V\_{ij})$ is zero and so the kernel can't be anything else but zero.


Hence we have a system in which every object has a map to zero IN the system, and the colimit of such a system is zero as well.


