Exercise:
Prove that every injective order preserving homomorphism $\phi$ of two ordered abelian groups satisfies $a < b \iff \phi(a) < \phi(b)$.
Definition:
A homomorphism $\phi \colon G \to H$ of ordered abelian groups is called order preserving if $a \le b$ implies that $ \phi(a) \le \phi(b)$.
Proof of exercise:
$\implies:$ from Definition we have $a \le b \implies \phi(a) \le \phi(b)$. Homomorphism $\phi$ is injective so we have $a < b \implies \phi(a) < \phi(b)$ (?)
$\Longleftarrow: \phi(a) < \phi(b)$ and $\phi$ is order preserving injection. I am not sure what to do know. Should I use the fact that kernel of injective function has only one element?