For a surjective function $f: A \rightarrow A$, if there exist $n$ such that $\ker f^n = \ker f^{n+1} = \dots$ Is f injective?

Question

For a surjective function $f: A \rightarrow A$, where $A$ is a ring, if there exists $n$ such that $\ker f^n = \ker f^{n+1} = \dots$ Is f injective? (A is a ring)

Answer 1

Yes, even works for groups. Clearly the composition of surjective functions is surjective, so $f^{(n)}$ is surjective. Suppose $a\in \mathrm{ker}(f)$, and choose $b\in A$ such that $f^{(n)}(b)=a$. Then $f(a)=0$ and so $b=0$ by $\ker f^{(n)}=\ker f^{(n+1)}$, hence $a=0$.