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

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.

• have you proven that subgroups of free abelian groups are free and that free abelian groups have well defined ranks? Commented Aug 8, 2020 at 22:41
• I haven't. We did some stuff with finitely generated abelian groups, but nothing with free abelian groups. Commented Aug 8, 2020 at 22:44
• If this makes sense to you: a short exact sequence $0\to A \to B \to C \to 0$ splits if $C$ is free, in that case $B\cong A\oplus C$ (I think direct sum is more appropriate here, though the difference doesn't matter for these groups).
– user17892
Commented Aug 8, 2020 at 22:54

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 $$x = 0$$.