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It seems that chain complexes have all information of simplicial complexes. If we have an isomorphism between chain complexes which are induced by simplicial complexes, can we conclude that two simplicial complexes are isomorphic?

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    $\begingroup$ I think yes. We can reconstruct the complex starting from vertices (which correspond to rank of $C_0$), adding higher simplexes based on boundary maps. $\endgroup$
    – freakish
    Dec 15, 2022 at 17:13
  • $\begingroup$ @MarianoSuárez-Álvarez yes, I was too hasty. I wonder however, if the chain isomorphism arises from graph morphism, then does this imply that the graph morphism is a graph isomorphism? $\endgroup$
    – freakish
    Dec 16, 2022 at 9:16

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No. For instance, consider two non-isomorphic trees with the same number of vertices. Both will have a chain complex of the form $0\to\mathbb{Z}^{n-1}\to\mathbb{Z}^n\to 0$ where the map $\mathbb{Z}^{n-1}\to\mathbb{Z}^n$ is injective and its cokernel is free of rank $1$. Any two such chain complexes are isomorphic, since the images of the standard basis elements of $\mathbb{Z}^{n-1}$ together with a lift of a generator of the cokernel will form a basis for $\mathbb{Z}^n$, so you can choose a basis for $\mathbb{Z}^n$ such that the map $\mathbb{Z}^{n-1}\to\mathbb{Z}^n$ is just the inclusion given by the first $n-1$ basis vectors.

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  • $\begingroup$ Thanks. In this example, I think isomorphisms of chains are like 12 + 23 -> 13 (12 is an edge with vertices 1, 2). Then can we say that if we have an isomorphism of chain complexes with a coefficient, then chain complexes with any coefficient are isomorphic? $\endgroup$
    – Link
    Dec 16, 2022 at 0:18
  • $\begingroup$ Well if the chain complexes with coefficients in $\mathbb{Z}$ are isomorphic then so are the chain complexes with any other coefficients, since those are obtained by just tensoring with the coefficient group. $\endgroup$ Dec 16, 2022 at 0:30

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