Let $0\to\ker f\to A \xrightarrow fB\to C \to 0$ be an exact sequence of abelian groups. Can we say that $C\cong\operatorname{coker}f$?

Question

Let $f:A\to B$ be a group homomorphism between abelian groups. Let $$0\to\ker f\to A\xrightarrow fB\to C\to0$$ be an exact sequence of abelian groups. Can we say that $C\cong \operatorname{coker}f$ ?

I think this does not hold in general, but I cannot find examples.

Answer 1

The answer is yes. Let $g$ denote the surjection from $B$ onto $C,$ then $$C\cong B/\ker g=B/\operatorname{im}f.$$

Answer 2

Let $g\colon B\to C$ be the given map. We know that $g$ is surjective, hence $B/\ker g\cong C$. But $\ker g=\operatorname{im} f$.