How to show that a presheaf kernel of a morphism of sheaves is a sheaf 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 $U$:


*

*If $x\in ker(\varphi(U))$, and $x|_{V_{i}}=0$, then $x=0$

*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!
 A: 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).


Hint for (2): Do exactly the same, if you have local sections of $x_i \in \ker \varphi(V_i)$ that agree on the intersection, you can think of these local sections as elements of $\mathcal{F}(V_i)$ and so the $x_i$ glue to give a global section on $U$. Now show that this global section actually lies in $\ker\varphi(U)$.


A: To show (2) notice that the global section $x$ on $U$ that glues together the various $x_i$ have the property that $\phi(x_i)$ is zero for all $i$, so $\phi(x)$ restricted to $V_i$ is zero for all $i$ (use commutative diagram for $V_i$ in $U$, and $\phi$).  Therefore $\phi(x)$ is zero by the sheaf property of $\mathcal G$.  Therefore $x$ is in kernel $\phi$ on $U$.  
