Applying an additive covariant functor to a sequence Let $F$ be an additive covariant functor and $0→A_1→A_1⊕A_2→A_2→0$ be the usual exact sequence in the category of $R$-modules with the first map injection on the first, and the second map the projection on the second component. Is it true that applying $F$ gives a split exact sequence? 
I need a detailed proof for that and appreciate any solver in advance.
 A: If your functor is exact, then the answer should be yes. Let's name the inclusion $\iota: A_1 \to A_1 \oplus A_2$ and the projection $p: A_1\oplus A_2 \to A_2$. 
Fact 1: Additive functors commute with finite coproducts:
This means that $F(A_1 \oplus A_2) \cong F(A_1) \oplus F(A_2)$.
Having applied the functor, we then have the exact sequence
$$ 0 \rightarrow F(A_1) \xrightarrow{F(\iota)} F(A_1) \oplus F(A_2) \xrightarrow{F(p)} F(A_2) \rightarrow 0.$$
Now if $s: A_2 \to A_1 \oplus A_2$ is the section of $p$ (so that $p\circ s = 1_{A_2}$), then $1_{F(A_2)} = F(1_{A_2}) = F(p\circ s) = F(p) \circ F(s)$ so that $F(s)$ is the splitting map.
If your functor maps to a category where right-split is not enough to split, do the same with the left-splitting map (which exists in $R$-mod).
Edit: Note that we did not need to know the maps $\iota, p$ explicitly (even though you gave them). The proof simply depends on additive functors commuting with finite coproducts, and that identity and compositions are preserved under functors. 
