I think the following is true. (Sheaves are of abelian groups) Let $\mathscr{F}$ be a sheaf and $\mathscr{F'}$ is a subsheaf of $\mathscr{F}$ such that $\mathscr{F'}$ is flasque. (The restriction maps of $\mathscr{F'}$ are surjective. ) Show that the presheaf $U \mapsto \mathscr{F}(U)/\mathscr{F'}(U)$ is a sheaf.
Please prove it directly without using the exactness properties.(I want to use this to prove one of them) I am having problems proving sheaf property (II).
What I have tried is this (for showing sheaf property II):
Let $V_i$, $i \in I$, be an open covering for an open subset $U$ of $X$. We have to show that if there are $x_i \in \mathscr{F}(V_i)$ for every $i \in I$ such that $x_i \mid_{V_i \cap V_j} - x_j\mid_{V_i \cap V_j} \in \mathscr{F'}({V_i \cap V_j})$ for every $i,j \in I$. Then we have to show that there exists an $s \in \mathscr{F}(U)$ such that $s\mid _{V_i} - x_i \in \mathscr{F}'(V_i)$ for every $i \in I$.
I will be done if I can prove that there exists $t_i \in \mathscr{F}'(V_i)$ such that $x_i \mid_{V_i \cap V_j} - x_j\mid_{V_i \cap V_j} = t_i \mid_{V_i \cap V_j} - t_j \mid_{V_i \cap V_j} $ for all $i,j$. I'm unable to prove this.
Let us write $\mathcal{G} = \mathcal{F}/\mathcal{F}^{'}$ an remember that it is a presheaf so far.
We want to show that it is a sheaf. Take two intersecting open sets $U_{1}, U_{2}$ and take sections $s_{i} \in \mathcal{G}(U_{i}), i =1,2$ which agree on $\mathcal{G}(U_{1} \cap U_{2})$. Now we can lift these sections to $\widetilde{s_{i}} \in \mathcal{F}(U_{i})$, but they will in general not agree on $U_{1}\cap U_{2}$. Their difference restricted to $U_{1}\cap U_{2}$, $\widetilde{s}_{12} =\widetilde{s_{1}} - \widetilde{s_{2}} |_{$U_{1}\cap U_{2}} \in \mathcal{F}^{'}(U_{1} \cap U_{2})$. Using flasque property you can find a section $t \in \mathcal{F}^{'}(U_{1} \cup U_{2})$ that restricts to $\widetilde{s}_{12}$. Now the sections $\widetilde{s_{i}} -t \in \mathcal{F}(U_{i})$ patch nicely to a section of $\mathcal{F}(U_{1} \cup U_{2})$ and its image in $\mathcal{G}(U_{1} \cup U_{2})$ is the section extending $s_{i} \in \mathcal{G}(U_{i})$. The uniqueness follows from the fact that everything is unique mod $\mathcal{F}^{'}(U_{1} \cup U_{2})$.