Epi, mono but not an isomorphism

Question

In category theorie there exists the notion of mono, epi and iso. In the category Ab of abelian groups we have 'mono $\Leftrightarrow$ injective' and 'epi $\Leftrightarrow$ surjective'.

We have the full subcategory of Ab of divisible group ie. an abelian group $A$ such that for any $a\in A$ and $n\in\mathbb{Z}_{\ge 0}$ there is a $b\in A$ with $nb=a.$

Of course $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are divisible. So we have the morphism (canonical projection) $$p:\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}$$ in the category of divisible groups.

The claim is that this is mono, epi but not an isomorphism (in the category of divisible groups).

1. mono. Let $D$ be a divisible group and $g,g':D\rightarrow\mathbb{Q}$ morphisms such that $p\circ g=p\circ g'.$ We need to get $g=g'$. We have $0=p(g(d))-p(g'(d))=p(g(d)-g'(d))\Rightarrow g(d)-g'(d)\in\mathbb{Z}$, so $g(d)-g'(d)=n$ for some $n$. How can we conclude $g(d)-g'(d)=0$? For $|n|\ge 0$ I find an $d'\in D$ with $|n|d'=d$. That is $|n|g(d')-|n|g'(d')=n.$
2. epi. Again $D$ a divisible group and $g,g':\mathbb{Q}/\mathbb{Z}\rightarrow D$ morphisms such that $g\circ p=g'\circ p.$ Let $p(q)=[q]\in\mathbb{Q}/\mathbb{Z}$. We have $g([q])=g(p(q))=g'(p(q))=g'([q])$, so $g=g'$.
3. not isom. Assume there is an inverse morphism $f:\mathbb{Q}/\mathbb{Z}\rightarrow\mathbb{Q}$ of $p$. What is then the problem?

Answer 1

So here's my proof for $p$ being a mono:

So assume you have $f, g$ such that $p\circ f=p\circ g$ where $D$ is their domain, a divisible group. So for every $x\in D$ you have $f(x)-g(x)\in \mathbb{Z}$. Now take $x\in D$ and suppose $n=f(x)-g(x)\neq 0$. Now take $p$ a prime which doesn't divide $n$ and $y\in D$ such that $py=x$. You get: $$n=p(f(y)-g(y))$$ since $f(y)-g(y)\in \mathbb Z$ it would imply that $p$ divides $n$ which is a contradiction.