I am trying to prove theat a surjective homomorphism between $\mathbb{Z}^3$ to itself is actually an isomorphism. The question was given after the material of free abelian groups was taught.
I actually do not have any idea of how to start reachinig this problem, and would very like your hints/help.
Thanks in advance!
Are you familiar with the rank of a free abelian group? It is defined as the size of a maximal linearly-independent set, and generalizes the notion of dimension of a vector space.
If $f : \mathbb{Z}^3 \to \mathbb{Z}^3$ is surjective, then $f(e_1)$, $f(e_2)$, $f(e_3)$ must form a spanning set of $\mathbb{Z}^3$ (here $\{e_i\}$ is the standard basis). But since 3 is the minimal size of a spanning set of $\mathbb{Z}^3$, this set must actually be a basis. Thus it is linearly independent, and $f$ is injective.
If you aren't familiar with this machinery, you can instead argue that $f(x) = e_1$, $f(y) = e_2$, and $f(z) = e_3$ for some $x,y,z \in \mathbb{Z}^3$, since $f$ is surjective. But this information is exactly what we need to define a homomorphism $g : \mathbb{Z}^3 \to \mathbb{Z}^3$ where $g(e_1) = x$, $g(e_2) = y$, $g(e_3) = z$. Can you see why this is an inverse to $f$, and thus $f$ was an isomorphism all along?
I hope this helps ^_^