A surjective endomorphism (of a Noetherian ring) is injective.
Solution 1:
Let $S$ be a non-zero ring and take infinity many direct sum product $R = S^{\mathbb{N}}$. Then the homomorphism defined by
$$ f \colon R \to R,\ (r_1, r_2, r_3, \dotsc) \mapsto (r_2, r_3, r_4, \dotsc) $$
is clearly surjective but its kernel is $\ker f = S \times 0 \times 0 \times \dotsb \neq 0$.
So naive inference "$f \colon R \to R,\ \operatorname{im}f = R \implies \ker f = 0$" is wrong.