Coproduct in the Category of Totally Ordered Groups This is a question that arised from a problem in valuation theory of commutative rings. There I need to construct a totally ordered Abelian group "containing" every member of a given family of totally ordered groups. Therefore my question is:
Are there (infinite) colimits in the category of totally ordered Abelian groups with increasing homomorphisms?
Thank you in advance for your help!
 A: 
There are no colimits in general, indeed there is no coproduct for $\mathbf{Z}$ (ordered group) with itself. 

There's ambiguity in the question whether the category $\mathcal{C}$ is: ordered abelian groups, left-ordered groups, bi-ordered groups, and whether arrows are supposed to preserve $\le$ or $<$. This gives 6 possibilities; anyway the argument works in all cases. Denote $G_+$ elements $\ge e_G$ or $>e_G$ according to whether arrows are supposed to preserve $\le$ or $>$.
Observe that $\mathrm{Hom}_\mathcal{C}(\mathbf{Z},G)\simeq G^+$ for every $G\in \mathcal{C}$, the identification being given by $f\mapsto f(1_\mathbf{Z})$. Let $H$ be a coproduct of two copies of $\mathbf{Z}$. Then we have two elements $g,h\in H$ such that for every $G\in \mathcal{C}$ the map $\Phi:f\mapsto (f(g),f(h))$ from $\mathrm{Hom}_\mathcal{C}(H,G)$ to $G_+^2$, is a bijection.
Applied to $G=\mathbf{Z}$, choose $f=\Phi^{-1}(1,2)$ and $f'=\Phi^{-1}(2,1)$. 
So $f(g^{-1}h)=1$ and $f'(g^{-1}h)=-1$. Since both $f$ and $f'$ are increasing, we obtain $g^{-1}h>1_H$ and $g^{-1}h<1_H$ respectively, contradiction.

(Note: it's not really surprising: in the category of total orders with either $\le$-preserving or $<$-preserving maps, there is no coproduct of the point with itself.)
