Presheaf quotient that is not a sheaf Let $ \mathcal F $ be a sheaf (of, say, abelian groups), and $ \mathcal G $ be a subsheaf of $ \mathcal F$. It is not true that, in general, the quotient presheaf $ U \mapsto \mathcal F(U) / \mathcal G(U) $ is a sheaf. Could someone show me a counterexample?
Thanks
 A: Let $X$ be $S^1$ and let $\mathcal{C}$ be the sheaf of continuous real-valued functions; inside of it, let $\mathcal{Z}$ be the sheaf of locally constant functions taking values in the group $2\pi\mathbb{Z}$. Now cover the circle with four big arcs $U_1$, $U_2$, $U_3$, $U_4$ where $U_i$ covers the part of the circle in the $i$th quadrant (and is $\epsilon$ bigger in angle, to be open -- by the way, three arcs would do for a counterexample but it's easier to describe it verbally with 4.)
Define continuous argument functions $\arg_i$ on $U_i$ as follows: in all cases, $\arg_i$ is the angle from the positive $x$-axis; in $U_1$, use a value between $(-\epsilon, \pi/2 + \epsilon)$, in $U_2$, between $(\pi/2 - \epsilon, \pi + \epsilon)$; in $U_3$, between $(\pi - \epsilon, 3\pi/2 + \epsilon)$, and in $U_4$, between $(3 \pi/2 + \epsilon, 2 \pi + \epsilon)$. 
Each function $\arg_i$ lives in $\mathcal{C}(U_i)$. Note that there is no function in $\mathcal{C}(X)$ restricting to $\arg_i$ on each $U_i$: there can't be, because $\arg_1$ and $\arg_4$ disagree on the overlap of $U_1$ and $U_4$.
Now consider the $\arg_i$ as members of $\mathcal{C}(U_i)/\mathcal{Z}(U_i)$. Here, they do agree on all the overlaps, because their difference is in all cases either $0$ or $2 \pi$. But there is no element of $\mathcal{C}(X)/\mathcal{Z}(X)$ restricting to this collection, so the second sheaf axiom is not satisfied.
