Given a morphism of sheaves $\alpha:F\rightarrow G$ of abelian groups over a topological space, I want to show that for any $x\in X$ holds $\ker(\alpha_x)=(\ker(\alpha))_x$.
Consider for any $x\in U\subseteq X$ open $$\ker(\alpha(U))\stackrel{\iota(U)}{\rightarrow}F(U)\stackrel{\alpha(U)}{\rightarrow}G(U)$$ inducing $$\ker(\alpha)_x\stackrel{\iota_x}{\rightarrow}F_x\stackrel{\alpha_x}{\rightarrow}G_x$$ since evidently $\alpha_x\iota_x=0$, $\ker(\alpha)_x$ is contained inside $\ker(\alpha_x)$ (or better there's an arrow by the universal property blabla).
Viceversa, take $s_x\in\ker(\alpha_x)$: then $s_x=(s)_x$ for some $s\in U$ open neighbourhood of $x$, and $\alpha_x(s_x)=(\alpha(U)(s))_x=0$: this means that there is an open neighbourhood $V\subseteq U$ of $x$ such that $0=(\alpha(U)(s))_{|V}=\alpha(V)(s_{|V})$: but this means exactly that $s_{|V}\in\ker(\alpha(V))$ and thus $s_x=(s_{|V})_x\in\ker(\alpha)_x$.