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Let $X_k=X_{k-1}\cup_{\chi} H^{\gamma_k}$ be a $\dim X_{k-1}$-manifold with a $\gamma_k$-handle $H^{\gamma_k}=D^{\dim X_{k-1}-\gamma_k}\times D^{\gamma_k}$ attached along the embedding map $\chi:S^{\gamma_k-1}\times D^{\dim X_{k-1}-\gamma_k}\to\partial X_{k-1}$ with $X_0:=D^m$ (an $m$-ball), $X:=X_N$ for $1\le k\le N$.

Then, how do we compute the $n$-th homology group of $X$ for all $n\ge 0$ (via cellular homology for CW complexes) in $\mathbb{Q}$-coefficients, i.e. $H_n(X;\mathbb{Q})=H_n(D^m\cup_{\chi} H^{\gamma_1}\cup_{\chi}\dots\cup_{\chi} H^{\gamma_N};\mathbb{Q})$? In short, how do we compute homology of a handle decomposition?

There is that there is a deformation retraction of $M\cup H^k$ onto the core $M\cup_{S^{k−1}}(D^k\times 0)$, so homotopically attaching a $k$-handle is the same as attaching a $k$-cell. Thus, we may use cellular homology of the CW complex $X'$ given by collapsing each handle $D^k\times D^{n−k}$ to $D^k$.

Any help would be much appreciated. Thanks in advance!

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    $\begingroup$ A handle decomposition is similar to a cell decomposition. In both cases you can set up a chain complex to compute the homology. $\endgroup$ – Cheerful Parsnip Sep 18 '18 at 19:54
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    $\begingroup$ It's not exactly cellular homology, it just behaves exactly the same way. The $k$-handles are just thickened $k$-cells and you need to examine the degrees of the attaching maps. $\endgroup$ – Cheerful Parsnip Sep 21 '18 at 3:17
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    $\begingroup$ If you have a handle decomposition you can 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 your manifold has k-handles. Then our vegetable friend's chain complex is the same as the cellular chain complex of $X'$. $\endgroup$ – user98602 Sep 22 '18 at 21:24
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    $\begingroup$ I strongly suggest trying this computation explicitly for a handle structure on $\Sigma_g$ with only one 0- and 2-handle. $\endgroup$ – user98602 Sep 22 '18 at 21:25
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    $\begingroup$ @Multivariablecalculus There has been 30 edits in this question (up to now). Please, try to refrain from overediting posts. $\endgroup$ – Aloizio Macedo♦ Sep 27 '18 at 4:46

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