Let $G$ be a torsion free, finitely generated abelian group. Suppose that $f : G \to G$ is an epimorphism (surjective homomorphism). Does it follow that $f$ is an isomorphism?
My thoughts: this is true if $G$ is a finite dimensional vector space. In the case $G$ is as above, $G \cong \mathbb{Z}^n$ for some $n \geq 0$. It resembles the case of a vector space...
On
Your thought is fine. For a given $f\colon \Bbb Z^n\to \Bbb Z^n$, which is uniquely determined by its values $f(1,0,\ldots), f(0,1,0,\ldots,0),\ldots, f(0,0,\ldots,1)$ on the standard base, we obtain a linear map $\tilde f\colon \Bbb Q^n\to \Bbb Q^n$ by sending the basis vectors to the same integer (hence rational) points as $f$. It follows that $\tilde f(x)=f(x)$ whenever $x\in\Bbb Z^n$. Conclude that $\tilde f$ is onto, hence is an isomorphimsm, hence is injective, hence to is $f$.
Tensor the exact sequence $0\to \ker f\to G\xrightarrow{f} G\to 0$ with $\mathbb{Q}$. Then you get $$ 0\to\ker f\otimes\mathbb{Q}\to G\otimes\mathbb{Q}\xrightarrow{f\otimes\mathbb{Q}} G\otimes\mathbb{Q}\to 0 $$ Then $f\otimes\mathbb{Q}$ is an isomorphism, because it is a surjective endomorphism of a finite-dimensional vector space. Therefore $\ker f\otimes\mathbb{Q}=0$, which means $\ker f$ is torsion. Since $G$ is torsionfree, we conclude $\ker f=0$.