$g \circ f = id_A$ implies $ker(g) \cong coker(f) $

Question

Let $R$ be a ring (for example $\mathbb{Z}$) and $A,B$ two $R$-modules.

Futhermore we have two morphisms $A \xrightarrow{f}B$ and $B \xrightarrow{g}A$ with property $g \circ f = id_A$.

My question is how to see that $ker(g) \cong coker(f) $?

My considerations:

We have $coker(f)= B/im(f)$ by definition and since $g \circ f = id_A$ $f$ is in injection so we can identify $A$ with $im(f)$.

So $coker(f) \cong B/A$. But I don't see how to settle $ker(g) \cong B/A$. Could anybody help?

Can the argument be generalized to any abelian category?

Answer 1

The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)\oplus {\rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.

If you need help with this equality, just let me know.

Answer 2

Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.