An ordered group $(G,G^+)$ is an abelian group $G$ together with a subsemigroup $G^+$ containing the identity $0$, having these properties:
$G^+-G^+=G$.
$G^+\cap (-G^+)=\{0\}$.
We have for $x,y\in G$: $x\le y$ if $y-x\in G^+$.
A morphism $f:(G,G^+)\to (H,H^+)$ of ordered abelian groups is a group homomorphism $f:G\to H$ satisfying $f(G^+)\subseteq H^+$.
In my situation, $f$ is an injective morphism of ordered abelian groups (not sure if injectivity is important). Set $f(G)^+ :=f(G)\cap H^+$.
Is $f(G^+)= f(G)^+$ (at least, if $f$ is injective)?
Trivially, $f(G^+)\subseteq f(G)\cap H^+$. But what about $f(G)\cap H^+\subseteq f(G^+)$?
I came as far as this: Let $a,b\in f(G)^+$ with $a\le b$, let $x,y\in G$ with $f(x)=a\le f(y)=b$. Then $b-a=f(y-x)$. But does $b-a$ lift to an element in $G^+$, i.e. is $y-x \in G^+$?
If $-x \in G^+$, then $-f(x) = f(-x) \in H^+$, and therefore $f(x) \in -H^+$. So $f(-G^+) \subseteq -H^+$.