The subgroup $H=\mathbb Z^\mathbb N$ of $G_1=\mathbb Z^\mathbb Z$ is mapped to a proper subgroup by translation. By considering the semidirect product $G_1\rtimes\mathbb Z$, you can make translation on $G_1$ an inner automorphism.
The first theorem on p.26 tells you that if $G=\pi(B,b)$, $H=p(\pi(E,e))$, $H_1=p'(\pi(E',e'))$, the unique map $g\colon E\to E'$ satisfying $g(e)=e'$ is not an isomorphism. However, by the first proposition on p.23 there is an
$e'_1\in E'$ such that
$p(\pi(E',e'_1))=p(\pi(E,e))$, and the corresponding $E\to E'$ is an isomorphism.