# Are the homology groups for simplicial homology and singular homology with coeeficients the same?

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 $$H_n(X;A)\cong H_n^\Delta(X;A)$$?

• 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. Feb 6, 2020 at 8:48
• I will try and look into this. Feb 6, 2020 at 8:57
• In the present form the question is misleading. Your second functor is not defined on $\mathbf{Top}$, but on the category of $\Delta$-complexes. Feb 6, 2020 at 10:55
• You are right. I have edited the question. Feb 6, 2020 at 11:39

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.