Proving that the kernel of a presheaf morphism is a subpresheaf

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Let $F\to G$ a morphism between presheaves in $X$.

I'm trying to prove that $\ker_\varphi$ is a subpresheaf of $F$. The easy part is prove that $\ker_\varphi (U)\subset F(U)$ for every open sets of $X$ (I proved very easily).

However, I couldn't prove that the following diagram commutes:

$$\require{AMScd} \begin{CD} \ker_\varphi(U) @>\text{inclusion}\;>> F(U) \\ @VVV @VVV\\ \ker_\varphi(V) @>>\text{inclusion}\;> F(V) \\ \end{CD}$$

I almost sure that I'm forgetting something really silly. If someone could help me, I would be very grateful.

Thanks a lot.