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Let $\mathcal F$ be a sheaf (of abelian groups) on a topological space $X$. We say that the support of $\mathcal F$ is

$$\text{Supp}(\mathcal F):=\{x\in X: \mathcal F_x\neq 0\}$$

Please help me to prove the following proposition:

If $\text{Supp}(\mathcal F)$ is a finite set (of closed points), then for every open set $U\subseteq X$ we have that $\mathcal F(U)=\prod_{x\in U}\mathcal F_x$.

Why that conditions on stalks influences the local sections?

The only reasonable map $\mathcal F(U)\rightarrow \prod_{x\in U}\mathcal F_x$ is $s\mapsto (s_x)_{x\in U}$ but I see only that it is injective. What about the surjectivity?

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1 Answer 1

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To show surjectivity, let $s_i \in \mathcal F _{x_i}$, where $i = 1, 2, \cdots n$ and $x_i \in U$. Then by definition of the stalk, there is an open set $V_i$ so that $x_i \in V_i$ and $s_i \in \mathcal F(V_i)$. As $x_j$ are closed points, one can assume that $V_i$ does not contain $x_j$ for $j\neq i$. By taking $V_i \cap U$, we may also assume $V_i \subset U$.

Now $s_i \in \mathcal F(V_i)$ satifies

$$s_i|_{V_{ij}} = s_{j}|_{V_{ij}}$$ trivially as $V_{ij}$ contains no $x_i$'s. Together with $$s_0 = 0 \in \mathcal F(U\setminus\{x_1, \cdots, x_n\})$$

The sheaf axiom implies the existence of a $s\in \mathcal F(U)$ so that $s|_{V_i} = s_i$. Thus the map is surjective.

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