On the inclusion homomorphism $\mathbb Z\to\mathbb Q$ This question is about the group homomorphism $i: \mathbb Z\to\mathbb Q$ given by $m\mapsto m$.
1) Is the following proof of the fact that the arrow $i$ in $\mathbf {Ab}$ is  epic correct? Let $h,h':\mathbb Q\to G$ be group homomorphisms. Suppose $h\circ i=h'\circ i$. Then for all $n$, $h(i(n))=h'(i(n))$, i.e., $h(n)=h'(n)$ since $i(n)=n$. Thus $i$ is epic.
2) The group homomorphism $i: \mathbb Z\to\mathbb Q$ given by $m\mapsto m$ is known to be a monic that is not split in $\mathbf {Ab}$. Why is that? Assume that it is split monic. Then there is a group homo $l:\mathbb Q\to \mathbb Z$ such that $l\circ i=1_\mathbb Z$. What does this contradict to? For any $m\in \mathbb Z$ this says that $l(i(m))=m$, i.e., $l(m)=m$. So if $l$ exists, then it $l\restriction_\mathbb Z:\mathbb Z\to \mathbb Z$ must be given by $m\mapsto m$. So I guess the statement that needs to be proved is that there does not exist a group homomorphism $l:\mathbb Q\to\mathbb Z$ such that $l(m)=m$ for all $m\in\mathbb Z$. How to show that?

The inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$ is not epic in $\mathbf{Ab}$, but it is epic in $\mathbf{Ring}$ as well as in the category of torsion-free abelian groups.
Also, there is no nonzero homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$, because the image of any rational number would have to be an integer that is divisible by every nonzero integer, and no nonzero integer is divisible by any integer with a larger absolute value.
