Show that the stalk of the kernel is the kernel of the stalks. This is Exercise 2.5.A. in Vakil's notes of Algebraic Geometry. This section is about sheaves of abelian groups on a topological space $X$.


  Show that the stalk of the kernel is the kernel of the stalks: for all $p\in X$, there is a natural isomorphism $$(\ker(\mathscr{F}\rightarrow\mathscr{G}))_p\cong \ker(\mathscr{F}_p\rightarrow \mathscr{G}_p).$$


My attempt:

We consider the following diagram:



where the downward arrow is the induced morphism on the stalks, as defined in Exercise 2.4.D. Since the three are all sheaves, the downward arrows are all injective. We want to show that $(\ker\phi)_p$ is the kernel of $\phi_p$ for all $p$. Since the second row is induced from the first row, we have $\phi_p\circ i_p=0$ for all $p$. Since all kernels are monomorphisms, by Exercise 2.4.N., the map $i_p$ is also a monomorphism. Hence it is a kernel of its cokernel.   

My question:
I am not sure how to show that $i_p$ is the kernel of $\phi_p$. I tried the universal property, but cannot connect it with the upper arrows, so cannot find a unique map. 

Thank you for your help!
If $p \in V \subset U$, then $\ker \varphi(U) \to \ker (\varphi_p)$ is the same as the composition $\ker \varphi(U) \to \ker\varphi(V) \to \ker (\varphi_p)$.
By the universal property of colimits, there exists a unique map $\alpha: (\ker \varphi)_p \to \ker (\varphi_p)$ where $[s,U] \mapsto [s,U]$.
It is straightforward to show that $\alpha$ is bijective.
If $p \in V \subset U$, then $\ker \varphi(U) \to \ker (\varphi_p)$ is the same as the composition $\ker \varphi(U) \to \ker\varphi(V) \to \ker (\varphi_p)$.
By the universal property of colimits, there exists a unique map $\alpha: (\ker \varphi)_p \to \ker (\varphi_p)$ where $[s,U] \mapsto [s,U]$.
It is straightforward to show that $\alpha$ is bijective.