= Solution
\b[No.] Assign the two free generators to $(1,0)$ and $(0,1)$ in the <lexicographically ordered group> $\mathbb Z^2$. The <universal property> gives a <lattice-ordered group homomorphism> onto this group: its image contains both ordinary group generators. It is not an <Archimedean ordered group>, since
$$
n(0,1)<(1,0)\qquad\text{for every positive integer }n.
$$
Thus a lattice-homomorphic image can contain a nonzero positive infinitesimal even when the original free object has an Archimedean pointwise function representation.
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