Cokernel of 0-morphism is isomorphism

I need to prove that in an abelian category, the co-kernel of a 0-morphism is an isomorphism.

Let $A\rightarrow B$ be a $0$-morphism and $f:B\rightarrow C$ its cokernel. Then by the universal property of the the cokernel, I can show that there is a unique morphism $g:C\rightarrow B$ such that $g\circ f =id_B$. But I don't know how to prove that $f\circ g=id_C$.

I appreciate any help.

What you proved is that $f$ is split monomorphism. Now it is enough to show that it is epimorphism, but every cokernel is epimorphism by uniqueness from the universal property. Thus, $f$ is isomorphism.
By universal property of the cokernel, since $f : B \to C$ composed with $A \to B$ is the zero map, there exists a unique morphism $h : C \to C$ such that $h \circ f = f$.
Clearly the identity map satisfies this, so by unicity $h = \operatorname{id}_C$. But you also have $f \circ g \circ f = f \circ \operatorname{id}_B = f$, therefore $f \circ g$ also satisfies the property, therefore by unicity $h = f \circ g$. It follows that $f \circ g = h = \operatorname{id}_C$.