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Consider the handle decomposition of the manifold $Y:=Y_N$ where we let $Y_k=Y_{k-1}\cup_{\chi} H^{\gamma_k}$ be a $\dim Y_{k-1}$-manifold with a $\gamma_k$-handle attached $H^{\gamma_k}=D^{\gamma_k}\times D^{\dim Y_{k-1}-\gamma_k}$ along the embedding map $\chi:S^{\gamma_k-1}\times D^{\dim Y_{k-1}-\gamma_k}\to\partial Y_{k-1}$ with $Y_0:=D^m$ for $1\le k\le N$.

Then we collapse each handle $D^k\times D^{n−k}$ to $D^k$ to get a homotopy equivalent CW complex $X$ with the same number of $k$-cells as the manifold has $k$-handles. Thus, the CW can be given via its $(N-1)$-skeleton $X_N$ by attaching $\gamma_N$-cells, i.e. $X_N=X_{N-1}\cup_{} (e_{\gamma_i}^N)_i$. Cellular homology of this complex, $H_*^{CW}(X)$, is the homology of the cellular chain complex $(C_*(X),d_*)$ indexed by the cells of $X$ with differentials $d_n:C_n(X)\to C_{n-1}(X)$. Let $X$ have $n_{\gamma_i}$-many $\gamma_i$-cells (for $0\le i\le N$) with $\gamma_N$ the highest-dimensional cells. Note, the inclusion $i:X_N\hookrightarrow X$ induces an isomorphism $H_k(X_N)\cong H_k(X)$ if $k<N=dim(X)$.

Then can we conclude from the long exact sequence of relative homology that the cellular homology of the CW complex is $$H_k(X_N)=H_k^{CW}(X_N)=\begin{cases} \mathbb{Z}^{n_{\gamma_1}}, & \text{if }k=\gamma_1 \\ &\vdots\\ \mathbb{Z}^{n_{\gamma_N}}, & \text{if }k=\gamma_N \\ \mathbb{Z}_2, & \text{if }k=0 \\ 0, & \text{otherwise} \\ \end{cases}$$?

Attempted Proof: Let $X_k=X_{k-1}\cup_{\chi} H^{\gamma_k}$. Suppose one has a retraction $r_N:X_N\to X_{N-1}$, so $r_N\circ i_N=id_{X_{N-1}}$ where $i_N: X_{N-1}\to X_N$ is the inclusion. We similarly define $r_k:X_k\to X_{k-1}$, $i_k:X_{k-1}\to X_k$ for $1\le k\le N$. The induced map $(i_N)_*:H_n(X_{N-1})\to H_n(X_N)$ is then injective for $(r_N)_*(i_N)_*=id$. It follows that the boundary maps in the long exact sequences for $(X_N,X_{N-1})$ are zero, so the long exact sequences breaks up into short exact sequences $0\to H_n(X_{N-1})\overset{(i_N)_*}\longrightarrow H_n(X_N)\overset{(j_N)_*}\longrightarrow H_n(X_N,X_{N-1})\to 0$ and, in general, $0\to H_n(X_{k-1})\overset{(i_k)_*}\longrightarrow H_n(X_k)\overset{(j_k)_*}\longrightarrow H_n(X_k,X_{k-1})\to 0$, so $H_n(X_k)=H_n(X_{k-1})\oplus H_n(X_k,X_{k-1})$ for $1\le k \le N$. Thus, $$H_n(X_N)=H_n(X_{N-1})\oplus H_n(X_N,X_{N-1})=H_n(X_{N-2})\oplus H_n(X_{N-1},X_{N-2})\oplus H_n(X_N,X_{N-1})=\cdots=H_n(D^m)\oplus H_n(X_1,X_0)\oplus\cdots\oplus H_n(X_N,X_{N-1})=\begin{cases} \mathbb{Z}^{n_{\gamma_1}}, & \text{if }n=\gamma_1 \\ &\vdots\\ \mathbb{Z}^{n_{\gamma_N}}, & \text{if }n=\gamma_N \\ \mathbb{Z}_2, & \text{if }n=0 \\ 0, & \text{otherwise.} \\ \end{cases}$$

Motivation: The canonical computation by Hatcher AT, P 141 for $M_g$ a closed oriented surface of genus $g$ with the CW structure of one $0$-cell, $2g$ $1$-cells, and one $2$-cells admits such a result. The cells are attached by the product of commutators $[a_1,b_1]\dots[a_g,b_g]$. The cellular chain complex of $M_g$ is $0\overset{d_3}\longrightarrow\mathbb{Z}\overset{d_2}\longrightarrow\mathbb{Z}^{2g}\overset{d_1}\longrightarrow\mathbb{Z}\overset{d_0}\longrightarrow 0$ with zero maps $d_1$ and $d_2$. Then the homology groups are $$H_k(M_g)=\begin{cases} \mathbb{Z}, & k=0,2 \\ \mathbb{Z}^{2g}, & k=1\\ 0, & \text{otherwise}. \end{cases}$$ I thought that a similar argument with cellular homology applies.

Any help would be much appreciated. Thanks in advance!

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  • $\begingroup$ I don't understand where your $\mathbb{Z}_2$ comes from, nor do I understand what your hypotheses are exactly; does your question imply that the homology of a manifold in degrees greater than $0$ is always torsion-free? Because that's badly false; any finitely generated abelian group can be $H_1(X)$ for a manifold $X$. $\endgroup$ – Qiaochu Yuan Oct 10 '18 at 5:20
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    $\begingroup$ Where do you get the retraction $r:X_N\rightarrow X_{N-1}$? What you describe in the opening comments is the process of collapsing the cocore $0\times D^{n-k}$, and this is not the same as collapsing the core, which in general cannot be done continuously. The map $r$ is should be a retraction $\text{$X$ with handle attached}=X\cup_{S^{k-1}\times D^{n-k}}(D^k\times D^{n-k})\rightarrow X\cup_{S^{k-1}\times 0}(D^k\times 0)=\text{$X$ with cell attached}.$ $\endgroup$ – Tyrone Oct 10 '18 at 9:09
  • $\begingroup$ Thanks @Tyrone! See 147 of Hatcher. I use this retraction $r_k$ from the $k$-skeleton $X_k$ to the $(k-1)$-skeleton $X_{k-1}$ so that I can chase the exact sequence $0\to H_n(X_{k-1})\to H_n(X_k)\to H_n(X_k,X_{k-1})\to 0$. $\endgroup$ – Sergio Charles Oct 10 '18 at 16:40
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    $\begingroup$ @Multivariablecalculus, I've added an answer with examples. $\endgroup$ – Cheerful Parsnip Oct 10 '18 at 22:09
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    $\begingroup$ @Multivariablecalculus, the retraction does not exist. See the simple counterexample I gave above. Also see the comments I gave before regarding which of the two spheres you are collapsing to go from a handle to cell decomposition. $\endgroup$ – Tyrone Oct 11 '18 at 6:52
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Here is an example. Let's consider the surfaces you could build from a single $0$-handle, $1$-handle and $2$-handle. There are essentially two ways to attach a $1$-handle to a $0$-handle, depending on whether you have a twist or not.

  • If there is no twist, you can visualize it like a basket with a handle. You can then add your $2$-handle to cap off one of the two boundary components, and the result will be homeomorphic to a disk.

  • If there is a twist, then you have a Möbius strip, and this has one boundary component to which we can attach a $2$-handle, giving a projective plane.

The chain complex in both cases is of the form $$0\to \mathbb Z\to\mathbb Z\to\mathbb Z\to 0,$$ but the boundary maps induced by the attaching maps are different.

  • The first chain complex is $0\to\mathbb Z\overset{\mathrm{id}}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0$, which yields homology only in degree $0$.
  • The second chain complex is $0\to\mathbb Z\overset{\cdot 2}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0$, which has $H_1\cong\mathbb Z/2\mathbb Z$.
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  • $\begingroup$ Thanks! But your answer is for a surface with a $0$-handle, $1$-handle and $2$-handle; what I would really like to know is how this can be generalized to $D^m\cup H^{\gamma_1}\cup\dots\cup H^{\gamma_N}$? @CheerfulParsnip $\endgroup$ – Sergio Charles Oct 11 '18 at 2:05
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    $\begingroup$ @Multivariablecalculus: I wanted to show you that the answer depends on how things are attached and does not depend only on the number of cells. I also wanted to give you an example of what the chain complexes would look like. In general, you need to compute the boundary maps $d_i$ by looking at how the cells attach. $\endgroup$ – Cheerful Parsnip Oct 12 '18 at 2:13
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This is not a detailed answer, but a suggestion of a possible area to look for developments.

The book by V.V. Sharko, "Functions on Manifolds. Algebraic and Topological Aspects", (Transl. Math. Monogr. 131. Amer. Math. Soc., 1993), remarks in Chapter VII on: "The need to make use of homotopy systems in order to study Morse functions on non-simply connected closed manifolds or on manifolds with one boundary component ...". Homotopy systems were developed by J.H.C. Whitehead in his 1949 paper "Combinatorial Homotopy II", and are now called crossed complexes. Their homotopically defined example involves operations of fundamental groupoids on relative homotopy groups, rather than chain complexes and homology as in the current standard approach to CW-filtrations.

The book partially titled Nonabelian Algebraic Topology EMS Tract (2011), (which we call NAT) gives computational methods for crossed complexes of filtered spaces, with applications mainly to cellular filtrations, i.e. to CW-complexes.

The usual tensor product of chain complexes generalises to crossed complexes, as detailed in NAT, so that there are potential applications to such handlebody decompositions.

Obtaining such applications is the essential content of Problem $16.1.17$ of NAT.

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