Let $\varphi:\mathcal{F}\longrightarrow \mathcal{G}$ be a morphism of sheaves. Then the presheaf :$U\mapsto ker(\varphi(U))$ is a sheaf.
By the definition of sheaf, it's sufficiently to verify that for all $x\in ker(\varphi(U))$ satisfies the following conditions
If $U$ is an open set and $\{V_{i}\}$ is an open cover of $U$:
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If $x\in ker(\varphi(U))$, and $x|_{V_{i}}=0$, then $x=0$
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If $x_{i} \in ker(\varphi(V_i))$ for each $i$, with the property that for each $i,j$, $x_{i}|_{V_{i}\cap V_{j}}=x_{j}|_{V_{i}\cap V_{j}}$, then there is a element $x\in ker(\varphi(U))$ such that $x|_{V_{i}}=x_{i}$ for each i.
Could any one help me? Thank you advance!
Best Answer
What have you tried with this problem? I will help you out with (1) but you must show effort with (2). Here's how (1) goes. For every $x \in \ker \varphi(U)$ we can consider it as an element of $\mathcal{F}(U)$ since $\ker \varphi(U)$ is a subgroup of $\mathcal{F}(U)$. Since $\mathcal{F}$ is a sheaf the fact that $x|_{V_i} = 0$ implies immediately that $x = 0$. This proves (1).