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Let $X$ be a topological space with a $\Delta$-complex structure. We know that $H_n^\Delta(X)\cong H_n(x)$ (Theorem 2.27 Algebraic Topology Hatcher). My question is the following: Does this also hold for other coefficients than $\mathbb{Z}$?

To be more precise:

Let $A$ be an abelian group, and let $H_n(X;A)$ denote the object resulting from applying the following composition of functors to $X$: $$ \textbf{Top}\overset{\text{Singular Chains}}{\longrightarrow}\textbf{Ch}\overset{\square \otimes_\mathbb{Z}A}{\longrightarrow}\textbf{Ch}\overset{H_n}{\longrightarrow}\textbf{Ab}. $$ Let $H_n^\Delta(X;A)$ denote the object resulting from applying the following composition of functors to the chain complex of simplicial chains $\Delta_*(X)$: $$ \textbf{Ch}\overset{\square \otimes_\mathbb{Z}A}{\longrightarrow}\textbf{Ch}\overset{H_n}{\longrightarrow}\textbf{Ab}. $$ Is it true that $H_n(X;A)\cong H_n^\Delta(X;A)$?

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    $\begingroup$ I seem to remember from my algebraic topology course that any two sequences of functors satisfying the Eilenberg-Steenrod axioms (en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms) are naturally isomorphic if they have the same coefficient ring, i.e. if they agree on the one point space. $\endgroup$ Feb 6, 2020 at 8:48
  • $\begingroup$ I will try and look into this. $\endgroup$
    – Frederik
    Feb 6, 2020 at 8:57
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    $\begingroup$ In the present form the question is misleading. Your second functor is not defined on $\mathbf{Top}$, but on the category of $\Delta$-complexes. $\endgroup$
    – Paul Frost
    Feb 6, 2020 at 10:55
  • $\begingroup$ You are right. I have edited the question. $\endgroup$
    – Frederik
    Feb 6, 2020 at 11:39

1 Answer 1

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1) I am 100% sure the proof goes through with coefficients in any abelian group $A$.

2) If you don't want to look at the proof again, it follows for any coefficient group $A$ from the case of $\Bbb Z$, the five-lemma, and the universal coefficient theorem.

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