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

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    $\begingroup$ Quasi-isomorphic implies homotopy equivalent in this situation, because both chain complexes are degreewise projective. $\endgroup$ – Zhen Lin Jun 29 '14 at 9:33
  • $\begingroup$ Zhen Lin, this is a complete answer. $\endgroup$ – Martin Brandenburg Jun 29 '14 at 11:32
  • $\begingroup$ @MartinBrandenburg One still has to prove that there is indeed a quasi-isomorphism... $\endgroup$ – Zhen Lin Jun 29 '14 at 20:54
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    $\begingroup$ @ZhenLin Over $\mathbb{Z}$ (or any hereditary ring), every object of the derived category of modules is isomorphic to the direct sum of its homology groups, so for complexes of free abelan groups, homotopy equivalent, quasi-isomorphic and "same homology" are all equivalent. $\endgroup$ – Jeremy Rickard Jun 30 '14 at 13:22
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    $\begingroup$ Rational Homotopy Theory Felix&Halperin&Thomas, p.51: "We'll construct the cellular chain complex of a CW complex and show it is chain equivalent to singular chain complex". $\endgroup$ – Leo Oct 13 '14 at 17:20

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.

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    $\begingroup$ Do you know of chain complexes $0\to A\to B\to C\to 0$ and $0\to A'\to B'\to C'\to 0$ (possibly over $\mathbb{Z}$) with isomorphic homologies but the isomorphism is not induced from a chain map? $\endgroup$ – Leo Jun 29 '14 at 21:50

Here is a nice zig-zag of quasi isomorphisms. Let $Sing(X)$ denote the realization of the singular set of X. Let $Song(X)$ denote the realization of the simplicial set of singular simplices that are cellular maps.

We have a chain of maps $X \leftarrow Song(X) \rightarrow Sing(X)$, where it is standard that these are weak equivalences and by design are cellular (where the latter spaces are CW complexes since they are realizations of simplicial sets). Hence, on CW chains these are quasi isomorphisms.

It is easy to see that CW and simplicial homology on the realization of a simplicial set coincide, so after taking CW chains we can extend to the right by an isomorphism of chain complexes getting us the simplicial chains of $Sing(X)$. Simplicial chains on $Sing(X)$ is exactly singular chains on X, so we are done.

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  • $\begingroup$ If you have the time, you could maybe post a similar answer here : math.stackexchange.com/questions/3604071/… . I'm just a bit confused : is it obvious that the complex of CW chains on $Sing(X)$ is the same as the one of simplicial chains ? It seems like there's a similar problem as the one pointed out in the question I'm mentioning, but I might be missing something $\endgroup$ – Maxime Ramzi Jul 10 at 8:48
  • $\begingroup$ @Maxime_Ramzi The chain groups are identical, and the requirement that the boundary of a simplex is injective (with maybe some of the other parts of being a simplicial complex) implies that the boundary map in the CW chains will be plus or minus one, depending on the orientation you pick for the cells. $\endgroup$ – Connor Malin Jul 10 at 13:51

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