The ring A is a commutative ring with identity.
I think ii) is true if they are not coprime.
because for every (x+a1,....,x+an) we can find a x such that f(x)= (x+a1,....,x+an).
Could you please show me a example that f(x) is not surjective？
2026-03-29 19:27:34.1774812454
is there a counterexample of this map isn't surjective?
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Consider $A=\mathbb{Z}$ and $a_1=a_2=2\mathbb{Z}$. Then the map is not surjective. For instance the element $(\overline 1, \overline 0)$ is not in the image.