Does the category $\mathbf{FreeAb}$ of free abelian groups have sequential colimits?
A sequential colimit is a colimit of the countable shape
$\bullet \to \bullet \to \bullet \to \cdots,$
i.e. indexed by $(\mathbb{N},\leq)$. This is a special case of a filtered colimit. A few weeks ago Jeremy Rickard showed in MSE/5025660 that $\mathbf{FreeAb}$ does not have filtered colimits, and although the filtered colimit there (the finitely generated subgroups of $\mathbb{Z}^{\mathbb{N}}$) was countable, it is clearly not sequential. I am not sure if it can be made sequential somehow, but I doubt it, since I think we would need the cardinal $\aleph_1$ to index such a sequence, not $\aleph_0$, which I need.
Notice that, as in the linked question, I am not asking if $\mathbf{FreeAb}$ is closed under sequential colimits taken in $\mathbf{Ab}$. This is a stronger property, and it fails miserably: the colimit of $\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \cdots$ in $\mathbf{Ab}$ is not free, it is $\mathbb{Z}[1/2]$. But the colimit of that sequence in $\mathbf{FreeAb}$ does exist and is zero.
Of course, some other sequential colimits also exist and are even the same as in $\mathbf{Ab}$, such as $\mathbb{Z} \hookrightarrow \mathbb{Z}^2 \hookrightarrow \mathbb{Z}^3 \hookrightarrow \cdots$ with transition maps $x \mapsto (\dotsc x,0)$, whose colimit is the free abelian group $\mathbb{Z}^{\oplus \mathbb{N}}$. Another interesting example is the sequence of maps $\mathbb{Z}^n \to \mathbb{Z}^{n+1}$, $a \mapsto (a,a)$, but I don't know how the colimit looks like here, concretely.
The question is equivalent to: given a sequential colimit of free abelian groups taken in $\mathbf{Ab}$, does it have a "liberation"? More generally, we might ask which abelian groups admit a "liberation" at all, except for the free abelian groups of course. By what I wrote before, $\mathbb{Z}[1/2]$ admits a "liberation", namely $0$.
