Category of torsion-free abelian groups not abelian In this article it says that the category of torsion-free abelian groups is not abelian since the map $\mu: \mathbb Z \to \mathbb Z: k \mapsto 2k$ is not a kernel. I have trouble showing this: 
If $\mu$ would be a kernel of some $f: \mathbb Z \to A$, I showed that $f = 0$. But another  kernel of $f$ would then be $id: \mathbb Z \to \mathbb Z$. Is this already a contradiction ?
 A: This isn't quite a contradiction yet, but it almost is: assuming that $\mu = \ker f$, it must be the case that $\mathrm{id}_{\mathbb{Z}}$ factors through $\mu$; that is, $\mathrm{id}_{\mathbb{Z}} = \mu \circ u$ for some (unique) $u : \mathbb{Z} \to \mathbb{Z}$. But this would imply that $u(1) = \frac{1}{2}$, which is nonsense.
A: If $f : A \to B$ is the kernel of $g : B \to C$ in $\mathsf{TorsFreeAb}$, then we may use the universal property for the object $\mathbb{Z}$ to see that $f : A \to B$ is also the kernel of $g : B \to C$ in $\mathsf{Ab}$ (after applying the forgetful functor). Besides, if $0 \neq n \in \mathbb{Z}$ and $b \in B$ are elements such that $n \cdot b$ lies in the image of $f$, then $b$ also lies in the image of $f$. This is because $0=g(n \cdot b)=n \cdot g(b)$ implies $g(b)=0$. In other words, the cokernel $B/\mathrm{im}(f)$ (taken in $\mathsf{Ab}$) is torsionfree. Conversely, if $f : A \to B$ is an injective homomorphism such that $B/\mathrm{im}(f)$ is torsionfree, then $f$ is the kernel of $B \twoheadrightarrow B/\mathrm{im}(f)$ in $\mathsf{TorsFreeAb}$.
For the specific example: $\mathbb{Z}/2\mathbb{Z}$ is not torsionfree.
