TITLE: How to show that a presheaf kernel of a morphism of sheaves is a sheaf
QUESTION [7 upvotes]: 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!

REPLY [11 votes]: 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 $U$.