Let $\Omega$ be a linearly (i.e. fully) ordered set, and let $\Lambda_{\Omega}$ be the ordered abelian group consisting of those $(\lambda_\omega)_{\omega\in\Omega}\in\mathbb{R}^{\Omega}$ with well-ordered support. Note that $\mathbb{R}^{\Omega}$ itself has incomparable elements, but restricting to elements of $\Lambda_{\Omega}$ ensures that all pairs of elements are comparable: $(\lambda_\omega)_{\omega\in\Omega}<(\lambda'_\omega)_{\omega\in\Omega}$ if $\lambda_{\omega_0}<\lambda'_{\omega_0}$ where $\omega_0$ is the least element of $\Omega$ for which $\lambda_{\omega_0}\neq\lambda'_{\omega_0}$. A group of the form $\Lambda_{\Omega}$ is called a Hahn group.

For $\lambda,\lambda'\in\Lambda_{\Omega}$ with $\lambda'>0$, write $\lambda\ll\lambda'$ if $\lambda=0$ or the first non-zero entry of $\lambda'$ precedes that of $\lambda$. (This means that $\lambda$ is infinitely smaller than $\lambda'$: that is, $n\lambda<\lambda'$ for all $n\in\mathbb{Z}$.)

Let $\Lambda_0$ be a subgroup of $\Lambda_{\Omega}$ and $\zeta:\Lambda_0\to\Lambda_{\Omega}$ a (not necessarily order-preserving) homomorphism for which $\zeta(\lambda)\ll\lambda$ for all positive $\lambda\in\Lambda_0$. Can $\zeta$ be extended to a homomorphism $\Lambda_{\Omega}\to\Lambda_{\Omega}$ that also satisfies $\zeta(\lambda)\ll\lambda$ for all positive $\lambda\in\Lambda_{\Omega}$?

A positive answer would help answer another question.