Kernel and image of an endomorphism - when are they supplementary?

(1) If $f$ is invertible, then $\ker f=\{0\}$ and $\operatorname{Im} f=E$, so $\ker f+\operatorname{Im} f=E$ and $\ker f\cap\operatorname{Im} f=\{0\}$.

(2) On $\mathbb{R}^2$, let $p(x,y)=(x,0)$. Then $\ker p=\operatorname{span}((0,1))$, $\operatorname{Im} p=\operatorname{span}((1,0))$ and $\mathbb{R}^2=\ker p\oplus\operatorname{Im} p$, but $p$ is not invertible. (Any projection onto a proper subspace works.) So $E=\ker f\oplus\operatorname{Im} f$ does not imply that $f$ is invertible.

(3) (i)$\Rightarrow$(ii). Let $x\in\ker f^2$. Then $f(x)\in\ker f\cap\operatorname{Im} f=\{0\}$, so $x\in\ker f$. With $\ker f\subseteq\ker f^2$, we get $\ker f^2=\ker f$.

(ii)$\Rightarrow$(i). Let $w\in\ker f\cap\operatorname{Im} f$, say $w=f(y)$. Then $f^2(y)=f(w)=0$, so $y\in\ker f^2=\ker f$ and $w=f(y)=0$.

(i)$\Rightarrow$(iii). By the dimension formula for a sum, $\dim(\ker f+\operatorname{Im} f)=\dim\ker f+\dim\operatorname{Im} f-\dim(\ker f\cap\operatorname{Im} f)=\dim\ker f+\dim\operatorname{Im} f=\dim E$ (rank-nullity). So $\ker f+\operatorname{Im} f=E$, and the sum is direct since the intersection is $\{0\}$.

(iii)$\Rightarrow$(i). A direct sum has trivial intersection.

Conclusion. Invertibility implies $E=\ker f\oplus\operatorname{Im} f$, but not conversely. The decomposition holds exactly when $\ker f\cap\operatorname{Im} f=\{0\}$, equivalently $\ker f^2=\ker f$. For the nilpotent map of the previous exercise, $\ker f\cap\operatorname{Im} f\neq\{0\}$, so the decomposition fails there.