Let $X$ be a topological space and $F,G$ be presheaves of abelian groups on $X$. Then given two presheaf morphisms $\phi,\psi:F\to G$, we'd like to define the sum of the two morphisms, by taking (for $U$ open in $X$),
$$\newcommand{\res}{\operatorname{res}} (\phi+\psi)_U=\phi_U+\psi_U$$
where $(\phi_U+\psi_U)(f)=\phi_U(f)+\psi_U(f)$ for $f\in F(U)$.
To show this is a well-defined presheaf morphism, it looks like I need to use the property that for $V\subseteq U$ and $f,g\in G(U)$, $$\res_{U,V}(f+g)=\res_{U,V}(f)+\res_{U,V}(g)$$
However, this was never taken as an axiom for presheaves of abelian groups, and not proved (I'm using Vakil's notes). Does anybody know how to show this is true? Surely it isn't too complicated, but I'm not seeing it.
A (pre)sheaf of groups is a (pre)sheaf of sets which on each open set has a group as value and whose restriction maps are morphisms of groups. If the last condition as not used when defining presheaves of abelian groups, then the definition is simply wrong (in particular, the condition that restriction maps are morphisms of groups does not follow from the other axioms).
A simple example of this is take the a point space X, which has two open sets, $\emptyset$ and $X$ itself. There is a presheaf of sets $F$ on $X$ such that $F(\emptyset)=A$ and $F(X)=B$, where $A$ and $B$ are any abelian groups whatsoever, and the restriction map $F(X)\to F(\emptyset)$ is any function whatsoever.