Example of non-balanced category. I have a claim that in the category of torsion-subgroup-free abelian groups any nonzero homomorphism from $\mathbb{Z}$ to $\mathbb{Z}$ is mono and epi. I am struggling to prove that it is indeed the case. I started with the mono case. So assume $G$ is a torsion-subgroup-free group and $g,h:G\rightarrow \mathbb{Z}$ s.t. $fg=fh$. Now I tried to go by contradiction and show that $g(x)\neq h(x)$ for some $x\in G$ and then maybe start looking at the subgroup generated by $g(x)-h(x)\in \mathbb{Z}$? But this leads nowhere. Also it is not obvious, where the assumption of torsion-freeness should be used. I would appreciate any hints.
Source of this claim: http://ncatlab.org/nlab/show/balanced+category
 A: In any concrete category $(\mathcal{C}, U)$ (here $U$ is a faithful forgetful functor $U : \mathcal{C} \to \mathsf{Set}$), we have a useful theorem:

If $Uf$ is an injection (resp. surjection), then $f$ must be mono (resp. epi).

This is a cute exercise (it follows immediately from faithfulness of $U$), or you can find a proof here. That link is also nice because it shows that whenever $U$ has a left (resp. right) adjoint, the converse also holds: if $f$ is mono (resp. epi), then $Uf$ is injective (resp. surjective), though I mention this more for cultural growth than anything.
Now any nonzero $f : \mathbb{Z} \to \mathbb{Z}$ is injective, therefore must be mono. If you want, we can make the abstract nonsense explicit and say that $fg = fh$. Then $Uf \circ Ug = Uf \circ Uh$, but $f$ is injective so $Uf$ is mono and $Ug = Uh$. Then since $U$ is faithful, we must have $g=h$.
To show that $f$ is also epi, we can't use this machinery (since it's not surjective), so we'll have to do it by hand. Since this category is enriched in $\mathsf{Ab}$ (that is, $\text{Hom}(G,H)$ is an abelian group, and this group structure respects composition, etc), we know that $gf = hf$ if and only if $gf - hf = 0$ if and only if $(g-h) \circ f = 0$. So it suffices to show that $kf = 0$ implies $k=0$.
Now say $f(x) = nx$ for $n \neq 0$. Then the image of $k$ is a subgroup of some torsion free abelian group $G$, but since $k(nx) = kf = 0$ we see that the image of $k$ has $n$-torsion. So the image of $k$ must actually be $0$, and $kf = 0$ implies $k=0$, as desired.

I hope this helps ^_^
