Coproducts of rigid groups

Let $\varepsilon=(\varepsilon_1,\dots,\varepsilon_m)$ be a tuple consisting of zeros and ones. Suppose that a group $G$ has a normal series of the form
$$ G=G_1\ge G_2\ge\dots\ge G_m\ge G_{m+1}=1, $$
in which $G_i>G_{i+1}$ for $\varepsilon_i=1$, $G_i=G_{i+1}$ for $\varepsilon_i=0$, and all factors $G_i/G_{i+1}$ of the series are Abelian and are torsion free as right $\mathbb Z[G/G_i]$-modules. Such a series, if it exists, is defined by the group $G$ and by the tuple $\varepsilon$ uniquely. We call $G$ with the specified series a rigid $m$-graded group with grading $\varepsilon$. In a free solvable group of derived length $m$, the above-formulated condition is satisfied by a series of derived subgroups. We define the concept of a morphism of rigid $m$-graded groups.
It is proved that the category of rigid $m$-graded groups contains coproducts, and we show how to construct a coproduct $G\circ H$ of two given rigid $m$-graded groups. Also it is stated that if $G$ is a rigid $m$-graded group with grading $(1,1,\dots,1)$, and $F$ is a free solvable group of derived length $m$ with basis $\{x_1,\dots,x_n\}$, then $G\circ F$ is the coordinate group of an affine space $G^n$ in variables $x_1,\dots,x_n$ and this space is irreducible in the Zariski topology.