Is every abelian group a colimit of copies of Z? More precisely, is every abelian group a colimit $\text{colim}_{j \in J} F(j)$ over a diagram $F : J \to \text{Ab}$ where each $F(j)$ is isomorphic to $\mathbb{Z}$?
Note that this does not follow from the statement that every abelian group has a presentation, which is equivalent to the statement that every abelian group is a coequalizer of a pair of maps between free abelian groups, hence every abelian group is an iterated colimit of copies of $\mathbb{Z}$. A single colimit $A = \text{colim}_{j \in J} F(j)$ of copies of $\mathbb{Z}$ is in particular the coequalizer of a pair of maps between free abelian groups, but the maps have a very special form, which works out explicitly to imposing the following constraint:

$A$ must have a presentation by generators and relations in which the only relations say that some generator is a multiple of some other generator. 

Examples of abelian groups admitting a presentation of this form include cyclic groups and localizations of $\mathbb{Z}$, and the class of all such groups is closed under coproducts.
But I see no reason to believe that every abelian group admits a presentation of this form, and in particular I believe that the $p$-adic integers doesn't. Tyler Lawson sketched a proof of this in the homotopy theory chat but it had a gap; the subsequent discussion may have filled the gap but I didn't follow it, and in any case I'd like someone to write up the details. 
Mike Shulman wrote a lovely note about various different senses in which an object or objects of a category can generate it; in the terminology of that note, the question is whether $\mathbb{Z}$ is colimit-dense in $\text{Ab}$. Until a week or so ago, if you had asked me, I would have answered without hesitation that $R$ is colimit-dense in $\text{Mod}(R)$, and I doubt that I was alone in this...
 A: Here's my version of Tyler's argument that $\mathbb{Z}_p$ is a counterexample. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:
It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:
If $e_i$ is in $A$, I define $f(e_i) = e_i$
If $e_i$ is not in $A$, I define $f(e_i) = 0$
Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

*

*If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.

*If $n=0$, then the relation is respected.

*If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.
However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.
A: Abelian groups of this type are known under the name "simply presented abelian group" and there are various investigations on them. 
Among others, it is known that simply presented abelian $p$-groups that are reduced (that is the divisible supgroup is trvial) are characterized by their Ulm sequence. 
As a consequence, two distinct abelian $p$-groups that are reduced and have the same Ulm sequence cannot be both simply presented. 
An explcit example is the torsion subgroup of   $\prod_n\mathbb{Z}/p^n\mathbb{Z}$ that has the same Ulm sequence as $\oplus_n\mathbb{Z}/p^n\mathbb{Z}$, and the latter is simply presented. 
There is also a homological characterization of such groups; indeed, another name is totally projective $p$-groups. 
I believe some information on this is in Fuchs' classic books on abelian groups. A more recent source would be Loth "Classifications of Abelian Groups and Pontrjagin Duality" (1998)
A: Let $G$ be an abelian group with a presentation of abelian group given only by relators of the form $g^n=h$, $g,h$ generators and $n$ integer. Then $G$ is a direct sum of locally cyclic groups. We can assume we allow relators $g^n=1$ (because it can be encoded in adding a generator $k$ and adding the relators $h=k$ and $h^2=k$). Now given such a presentation (say "satisfying ($*$)") we can remove all generators representing the identity (replacing them with 1 in relators) and the resulting presentation still satisfies ($*$) and has the additional property that no generator maps to the identity. Now consider the equivalence relation on the set of generators  generated by $g\simeq h$ if the (nontrivial) generators $g,h$ occur in a single relator. If $C$ is an equivalence class, consider the group $G_C$ with the presentation with generators in $C$ and relators involving generators of $C$. Then $G$ is the coproduct of the $G_C$. Note that so far this does not make use of the torsion-freeness of $G$, and the argument works for modules over an arbitrary ring.
Now we are reduced to understand the case when the equivalence relation defined above has a single equivalence class (i.e., is the undiscrete one). Then assuming $G$ torsion-free results in the fact that $G$ is locally cyclic (i.e., for a torsion-free group, isomorphic to a subgroup of $\mathbf{Q}$). To see this, I use the fact that if $G$ is a torsion-free group generated by elements $g_i$ such that any two $g_i$ have a common power, then $G$ is locally cyclic. Unlike the original question, this reduces (if necessary) to finitely generated groups: if $G$ is a torsion-free abelian group generated by a finite subset $S$ such that any two elements of $S$ have a common power, then $G$ is cyclic. This in turn is proved by a simple argument showing that in a torsion-free abelian group, if two elements have a common power, then they are powers of a common element (indeed the subgroup they generate is torsion-free abelian, generated by 2 elements and not isomorphic to $\mathbf{Z}^2$, hence is cyclic).
Thus in general, assuming that the abelian group $G$ is torsion-free implies that $G$ is a direct sum of locally cyclic groups (and more generally, in the category of $A$-modules when $A$ is a domain, assuming that $M$ is torsion-free implies that $M$ is a direct sum of torsion-free modules of rank 1; when $A$ is a PID, a torsion-free module of rank 1 is the same as a torsion-free locally cyclic module).
Among torsion-free abelian groups, those that are direct sums of locally cyclic groups are pretty rare. For instance $\mathbf{Z}_p$ does not satisfy this property, as any torsion-free abelian group $A$ of $\mathbf{Q}$-rank $\ge 2$, not containing any copy of $\mathbf{Z}[1/p]$ and such that $A/pA$ is cyclic); many subgroups of $\mathbf{Z}[1/p]^2$ are ruled out by this criterion. Also any non-free subgroup of $ \mathbf{Z}^X$ fails to be such a direct sum; this includes $\mathbf{Z}^{X}$ itself when $X$ is infinite.
