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!
Best Answer
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.