Let $f:\mathbb{Z}^2 \rightarrow \mathbb{Z}$ be a surjective homomorphism. Prove or disprove $\mathbb{Z}^2\cong\mathbb{Z}\times \ker(f)$

Question

I feel like Im overlooking some simple fact on this one, so any hints would be appreciated.

I saw this solution, but I'm wondering if there's a way to do it without short exact sequences.

Thanks in advance.

Answer 1

This is simple enough to be proved explicitly.

We pick any $v \in \Bbb Z^2$ such that $f(v) = 1$ and we define a homomorphism $h:\Bbb Z \times \ker(f) \rightarrow \Bbb Z^2$ by $h(a, x) = av + x$.

$h$ is surjective: if $y$ is any vector in $\Bbb Z^2$, then $h(f(y), y - f(y)v) = y$.

$h$ is injective: if $h(a, x) = 0$, then we have $av + x = 0$ and applying $f$ gives $a = 0$, which then implies $x = 0$.