5
$\begingroup$

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 ?

$\endgroup$
7
$\begingroup$

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: $$H_*(X) \cong H_*^\mathrm{sing}(|X|).$$

$\endgroup$
  • 1
    $\begingroup$ I will look into this. Thanks a lot ! $\endgroup$ – Nitrogen Apr 19 '15 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.