Let $X$ be a topological space and let $\mathcal F$ be a sheaf of abelian groups on $X$. Suppose that $\mathcal G$ is a subsheaf of abelian groups of $\mathcal F$, then one can construct the quotient presheaf $\mathcal F/\mathcal G$ in the following way:

$$(\mathcal F/\mathcal G)(U):=\mathcal F(U)/\mathcal G(U)$$

Is this presheaf always separated? In other words I'm asking if the group homomorphism: $$(\mathcal F/\mathcal G)(U)\to\prod_{x\in U}(\mathcal F/\mathcal G)_x$$ is always injective.

My solution: I think that the answer is YES. We know that $(\mathcal F/\mathcal G)_x=\mathcal F_x/\mathcal G_x$, so consider the morphism: $$(\mathcal F/\mathcal G)(U)\to\prod_{x\in U}\mathcal F_x/\mathcal G_x$$ $$\mathcal G(U)+s\mapsto \mathcal G_x+s_x$$

and suppose that $\mathcal G(U)+s\in(\mathcal F/\mathcal G)(U)$ is an element such that $s_x\in\mathcal G_x$ for every $x\in U$. By the definition of stalks and by using the fact that $\mathcal G$ is a sheaf we conclude that $s\in\mathcal G(U)$.

Is this correct?


Yeah I think you are correct. Here are more details (I think one needs to be careful about the last step - boldfaced below). So you have a section $\mathscr{G}(U)+s$ of the quotient presheaf $\mathscr{F}(U)/\mathscr{G}(U)$ where $s\in \mathscr{F}(U)$. You have shown that if $\mathscr{G}(U)+s$ gets sent to $0$ under this map $$ (\mathcal F/\mathcal G)(U)\to\prod_{x\in U}\mathcal F_x/\mathcal G_x $$ then $s_x \in \mathcal{G}_x$ for all $x$, which means that for each $x\in X$, the germ $s_x$ has a representative $(U_x, t(x))$ with $t(x)\in\mathscr{G}(U)$. After shrinking $U_x$ if necessary, we may assume that $t(x) = s|_{U_x}$. Now, $\mathscr{G}$ is a sheaf, and $X=\bigcup_{x\in X} U_{x}$ and by construction $t(x) = t(y)$ on $U_x\cap U_y$ simply because they are both equal to $s|_{U_x\cap U_y}$. So the local sections glue to the section $s$ in $\mathcal{G}(U)$. Thus, $s\in\mathcal{G}(U)$ which proves the desired injectivity.


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