Why $ (\operatorname{Im}\varphi)_x=\operatorname{Im}(\varphi_x) $?

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I don't understand the following fact (from Lemma 2.16 of Qing Liu ''Algebraic Geometry and Arithmetic Curves'').

Let $\mathcal{F},\mathcal{G}$ be sheaves (of Abelian groups) on $X$, $\varphi\colon\mathcal{F}\to\mathcal{G}$ a morphism. We have $ (\operatorname{Im}\varphi)_x=\operatorname{Im}(\varphi_x) $ (for all $x\in X$) and $\operatorname{Im}\varphi$ is a subsheaf of $\mathcal{G}$.

I already know $ (\operatorname{Im}^p\varphi)_x=\operatorname{Im}(\varphi_x) $ (for all $x\in X$) and $\operatorname{Im}^p\varphi$ is a subsheaf of $\mathcal{G}$ (where $\operatorname{Im}^p\varphi$ is a presheaf which sends open subset $U$ to $\operatorname{Im}\varphi(U)$), and let $F'$ be a sheafification of presheaf $F$ then $F_x\cong F'_x$ for all $x\in X$.

(And how about cokernel ?)