How to compute cellular homology of a handle decomposition $D^m\cup H^{\gamma_1}\cup\dots\cup H^{\gamma_N}$

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!

• A handle decomposition is similar to a cell decomposition. In both cases you can set up a chain complex to compute the homology. – Cheerful Parsnip Sep 18 '18 at 19:54
• 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. – Cheerful Parsnip Sep 21 '18 at 3:17
• 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 $X'$. – user98602 Sep 22 '18 at 21:24
• I strongly suggest trying this computation explicitly for a handle structure on $\Sigma_g$ with only one 0- and 2-handle. – user98602 Sep 22 '18 at 21:25
• @Multivariablecalculus There has been 30 edits in this question (up to now). Please, try to refrain from overediting posts. – Aloizio Macedo Sep 27 '18 at 4:46