[Math] Singular $\simeq$ Cellular homology

algebraic-topologyhomological-algebra

Given an arbitrary CW-complex, are the singular chain complex $S_\ast(X)$ and cellular chain complex $C_\ast(X)$ homotopy equivalent or just quasi-isomorphic (some chain map induces isomorphisms on homologies) or only have isomorphic homologies?

I can't find this in the standard AlgTop books. Any references are welcome.

Related: https://mathoverflow.net/questions/59390/when-is-a-quasi-isomorphism-necessarily-a-homotopy-equivalence

Best Answer

You will find this kind of result in

Blakers, A. "Some relations between homology and homotopy groups". Ann. of Math. (2) 49 (1948) 428--461.

I am pretty sure it is in Massey's book on Singular Homology, from a cubical viewpoint.

Proposition 14.7.1 of Nonabelian Algebraic Topology gives a deformation of the singular cubical complex of a space onto that coming from a filtration, under conditions which are satisfied in the case of a cellular filtration.

Later: Here is the detail of the proposition. For the question you can assume $X_*$ is the skeletal filtration of a CW-complex and $R X_*$ is the cubical set of cellular maps $I^n_* \to X_*$:

Let $X_*$ be a filtered space such that the following conditions $\psi (X_*, m)$ hold for all $m \geqslant 0$:

  1. $\psi (X_*, 0) :$ The map $\pi_0 X_0 \rightarrow \pi_0 X$ induced by inclusion is surjective;

  2. $\psi (X_*, 1) :$ Any path in $X$ joining points of $X_0$ is deformable in $X$ rel end points to a path in $X_1$;

  3. $\psi (X_*, m) (m \geqslant 2 ) :$ For all $\nu \in X_0$ , the map $$\pi_m (X_m , X_{m-1} , \nu ) \rightarrow \pi_m (X, X_{m-1} , \nu )$$ induced by inclusion is surjective.

Then the inclusion $i \colon RX_* \rightarrow KX=S^\square X$ is a homotopy equivalence of cubical sets.

The proof is quite direct by induction because the relative homotopy groups may be defined by maps of cubes, and in cubical sets, homotopies are defined using cubes.

Related Question