Let $X$ be a simplicial set. Define the complex $(C^X_\bullet,D)$ by $$C^X_n=\bigoplus_{X_n} \mathbb{Z}$$ and $$D_n=\sum_{i=0}^n (-1)^i d_i:C_n \to C_{n-1}$$ where the $d_i$'s are the face maps.

I wonder if there is a relation between the homology of $(C^X_\bullet,D)$ and the singular homology of $|X|$, the geometric realization of $X$.

I know that there is a weak equivalence $X \to S|X|$ and that the singular homology of $|X|$ is exactly the homology of $(C^{S|X|},D)$ as defined above. But I cannot conclude yet that there is a homotopy equivalence between the complexes $(C^X,D)$ and $(C^{S|X|},D)$, do I ?


1 Answer 1


You have the following two results in the book Simplicial homotopy theory of Goerss and Jardine:

Corollary III.2.7. Suppose that A is a simplicial abelian group. Then there are isomorphisms $$\pi_n(A,0) \cong H_n(NA) \cong H_n(A),$$ where $H_n(A)$ is the $n$th homology group of the Moore complex associated to $A$. These isomorphisms are natural in simplicial abelian groups $A$.

Proposition III.2.16. The free abelian simplicial group functor $X \to \mathbb{Z}X$ preserves weak equivalences.

In their notation, $\mathbb{Z}X$ is the underlying simplicial abelian group of your $C_\bullet^X$ (it's the level-wise free abelian group on $X$), and $H_*(\mathbb{Z}X) = H_*(C_\bullet^X, D)$ by definition.

Let $\eta : X \xrightarrow{\sim} S|X|$ be the weak equivalence you mentioned. Then the Proposition gives you that $\mathbb{Z}\eta : \mathbb{Z}X \to \mathbb{Z}S|X|$ is a weak equivalence too, so in particular it induces an isomorphism on all the homotopy groups. But since the isomorphisms in the Corollary are natural, you have the following commutative diagram: $$\require{AMScd} \begin{CD} \pi_n(\mathbb{Z}X, 0) @>{\mathbb{Z}\eta_*}>{\cong}> \pi_n(\mathbb{Z}S|X|, 0) \\ @V{\cong}VV @V{\cong}VV \\ H_n(\mathbb{Z}X) @>{\mathbb{Z}\eta_*}>> H_n(\mathbb{Z}S|X|) \end{CD}$$

Thus $H_n(\mathbb{Z}X) \cong H_n(\mathbb{Z}S|X|)$. But by definition, $H_n(\mathbb{Z}X) = H_n(C_\bullet^X, D)$ (with your notation), thus you get: $$H_*(X) \cong H_*^\mathrm{sing}(|X|).$$

  • 1
    $\begingroup$ I will look into this. Thanks a lot ! $\endgroup$
    – Nitrogen
    Apr 19, 2015 at 9:31

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