I'm reading Vakil's the The Rising Sea note, and I'm stuck on a quite trivial question.
Let $\mathscr{F},\mathscr{G}$ be presheaves of abelian groups on a topological space $X$. Let $f:\mathscr{F}\rightarrow \mathscr{G}$ be a morphism between presheaves. The presheaf kernel $\ker_{\text{pre}}f$ is defined by $\ker_{\text{pre}}f(U)=\ker f(U)$ for each open subset $U\subseteq X$. Show that $\ker_{\text{pre}}f$ is a presheaf.
I'm guessing that the function in the hint is the induced map $\text{res}_{V, U}: \mathscr{G}(V)/f(V)(\mathscr{F}(V)) \rightarrow \mathscr{G}(U)/f(U)(\mathscr{F}(U))$ from the diagram chasing, but I'm not exactly sure what I should do with it. Should I try to compute $\ker_{\text{pre}}f(V) \rightarrow\ker_{\text{pre}}f(U)$ by writing down a composition through $\text{res}_{V, U}:\mathscr{G}(V)/f(V)(\mathscr{F}(V)) \rightarrow \mathscr{G}(U)/f(U)(\mathscr{F}(U))$?
Because restriction maps commute with our morphism arrows on sections, we have that if $$0\longrightarrow \ker_{\text{pre}} f(V) \xrightarrow{\phi}\mathscr{F}(V)\xrightarrow{f(V)}\mathscr{G}(V)$$ is the kernel of $f(V)$ (in the category theoretic sense, i.e. in whatever abelian category $\mathscr{F}(\cdot)$ belongs to), then $$f(U)\circ \text{res}_{V,U}\circ \phi = \text{res}_{V,U}\circ \underbrace{f(V)\circ \phi}_{0} = 0.$$ Thus by the universal property of $\ker_{\text{pre}} f(U)$, we get a unique arrow $\mu$ (up to unique isomorphism) $$\require{AMScd} \begin{CD} 0 @>>> \ker_{\text{pre}}f(V) @>\phi>> \mathscr{F}(V) @>f(V)>> \mathscr{G}(V) \\ \ @V\mu VV @V{\text{res}_{V,U}}VV @V{\text{res}_{V,U}}VV \\ 0 @>>> \ker_{\text{pre}}f(U) @>>> \mathscr{F}(U) @>f(U)>> \mathscr{G}(U) \end{CD}$$ We make this map $\mu$ our restriction map $\text{res}_{V,U} : \ker_{\text{pre}} f(V) \to \ker_{\text{pre}} f(U)$.
For proper composition, just chain the diagrams: $$\require{AMScd} \begin{CD} 0 @>>> \ker_{\text{pre}}f(V) @>>> \mathscr{F}(V) @>f(V)>> \mathscr{G}(V) \\ \ @VVV @V{\text{res}_{V,U}}VV @V{\text{res}_{V,U}}VV \\ 0 @>>> \ker_{\text{pre}}f(U) @>>> \mathscr{F}(U) @>f(U)>> \mathscr{G}(U) \\ \ @VVV @V{\text{res}_{U,W}}VV @V{\text{res}_{U,W}}VV \\ 0 @>>> \ker_{\text{pre}}f(W) @>>> \mathscr{F}(W) @>f(W)>> \mathscr{G}(W) \end{CD}$$ By uniqueness of the map induced by the universal property, we must have that $\text{res}_{U, W}\circ \text{res}_{V,U} = \text{res}_{V,W}$ since this diagram commutes.