Are they examples of easy chain complexes...
that have the same homotopy type but are not isomorphic?
That have the same homology groups but haven't got the same homotopy type?


For two homotopic non-isomorphic chain complex, take any simplicial complex with a simplicial contraction, e.g. $EG$ for any finite group $G$. Then the extra degeneracy will give a chain homotopy from $C^{\text{simp}}(EG, \mathbb{Z})$ to the chain complex with $\mathbb{Z}$ in degree $0$ and zero everywhere else. On the other hand $C^{\text{simp}}_1(EG, \mathbb{Z})=\mathbb{Z}[G]$.

For homologous chain complexes which are not homotopic, note that homotopic chain complexes are still homotopic after tensoring with any abelian group. Then the chain complexes $\require{AMScd} \begin{CD} \ldots @>>> 0 @>>> \mathbb{Z} @>\times 2>> \mathbb{Z} @>>> 0 @>>> \ldots\\ @. @| @VV0V @VV\text{mod }2V @| @.\\ \ldots @>>> 0 @>>> 0 @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \ldots \end{CD}$ are homologous, but after tensoring with $\mathbb{Z}/2\mathbb{Z}$ they are not. Hence they cannot be homotopic.


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