3
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.