E r plus one is literally equal to the homology of e r

For each r≥0 put $G_r=\{(r,a):a\in \mathbb{Z}\mathrm{/}2\}$ with $(r,a)+(r,b)=(r,a+b)$. Put $G_r$ at (0,0), zero elsewhere and let every differential be zero. The group laws follow immediately by transport from [F2]. Model its homology quotient as the set of singleton cosets $\{\{(r,a)\}:a\in \mathbb{Z}\mathrm{/}2\}$.

Define ${\alpha }_r(\{(r,a)\})=(r+1,a)$. It is additive and has inverse $(r+1,a)\mapsto \{(r,a)\}$, giving the exact data of [F1]. These homology elements are singleton sets, whereas ordered pairs (in the standard set model $(x,y)=\{\{x\},\{x,y\}\}$) at r+1 have two elements as soon as r+1 differs from a. This occurs for some a since ℤ/2 has two distinct elements. Therefore this chosen model of $H(G_r)$ is not literally $G_{r+1}$, although ${\alpha }_r$ is an isomorphism.