I am a mathematician working in analysis and my knowledge in algebra is rusty.
Is there a direct argument showing that if
$\phi:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$
is an epimorphism of abelian groups, then it is an isomorphism?
My argument goes as follows: Since $\phi$ is a surjection onto a free abelian group $\mathbb{Z}\oplus\mathbb{Z}$, the group $\mathbb{Z}\oplus\mathbb{Z}$ is isomorphisc to
$\mathbb{Z}\oplus\mathbb{Z}\oplus\ker\phi$. $\ker\phi$ is also a free abelian group (as a subgroup of free abelian) and hence has to be zero as otherwise bases of the two free abelian groups:
$\mathbb{Z}\oplus\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}\oplus\ker\phi$
would have different cardinality.
Adding proofs of all results that I am using here would make the proof somewhat lengthy. Is there a more direct argument?
Best Answer
Here's a different perspective, at least.
Let $M$ be the $2 \times 2$ matrix with $\mathbb Z$ coefficients such that $$\phi(v)=Mv $$ where $v \in \mathbb Z \oplus \mathbb Z$ is regarded as a column matrix.
If $\det(M)=0$, the map $\phi$ is not surjective.
If $|\det(M)| > 1$, the map $\phi$ is not surjective.
If $|\det(M)|=1$, the map $\phi$ is both injective and surjective.
All of these can be proved rather simply, using methods of solving 2 linear equations in 2 unknowns.
But frankly, your outline seems fine to me, and my outline may have as many details to fill in as yours, although not bringing in much in the way of "deep" mathematics. At least it's a different perspective