Suppose $f:\Bbb Z \to \Bbb Z$ and $g:\Bbb Z^2 \to \Bbb Z$ are group homomorphisms, and $$0\to \Bbb Z \xrightarrow{(f,-f)} \Bbb Z^2 \xrightarrow{g} \Bbb Z \to 0$$ is exact. (Thus $f$ is at least injective, and $g$ is surjective.) Then is it true that $f$ is surjective, and hence an isomorphism?
I need this in my problem in topology, but I can't see whether this statement is true or not. Thanks in advance.
On
Every group homomorphism of the from $\mathbb{Z}^n \to \mathbb{Z}^m$ is given by a integer matrix, so the sequence is of the form $$ 0 \longrightarrow \mathbb{Z} \overset{ \begin{pmatrix} a \\ -a \end{pmatrix}}{\longrightarrow} \mathbb{Z}^2 \overset{ \begin{pmatrix} c & d \end{pmatrix}}{\longrightarrow} \mathbb{Z} \longrightarrow 0 $$ and exactness implies that $c=d=\pm 1$, so $\ker g = (1,-1)\mathbb{Z}$ and $a$ must be $\pm 1$, hence $f$ is an isomorphism.
We know that $f\colon\mathbb{Z}\to\mathbb{Z}$ is of the form $n\mapsto kn$ for some $k\in\mathbb{Z}$, as its image must be a subgroup of $\mathbb{Z}$. It will be an isomorphism if and only if $k=\pm1$, as then it will be bijective. Indeed, if $k=0$, it fails to be injective, contradicting the exactness of the sequence, and if $|k|>1$, then nothing will be sent to $1$. If $k=\pm1$, then we can choose $g\colon (m, n)\mapsto m+n$, as now $\ker(g)=\text{im}(f, -f)$, so you would need to show that such a $g$ won't exist if $|k|>1$.