It is well known that,in the category of $R$ modules if we have a short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$ the two following are equivalent(let denote for convenience $f:A\rightarrow B$ and $h:B\rightarrow C$):
$1)$There exists a morphism $g:B\rightarrow A$ such that $g \circ f=Id_{A}$
$2)$There exists an isomorphism between $B$ and the direct sum of $A$ and $C$
I was wondering if we could prove this result for arbitrary abelian categories.I tried to ''reproduce the proof'' in a categorical manner but i am stuck.Any help?
First, as paul blart math cop noted in the comments, 2) should be that there exists an isomorphism between the sequences $$0 \to A \xrightarrow{f} B \xrightarrow{h} C \to 0$$ and $$0 \to A \to A \oplus C \to C \to 0.$$
It is clear that 2) implies 1), because finite products and finite sums coincide in abelian categories.
So assume that there is a morphism $g: B \to A$ with $g \circ f = \operatorname{id}_A$. This allows us to define a morphism $g \oplus h: B \to A \oplus C$. If we use the identity maps on $A$ and $C$, this defines in fact a morphism of the two short exact sequences above, using the assumption $gf = \operatorname{id}_A$. By the 5-lemma, $g \oplus h$ is an isomorphism.