I have a few questions about computing cellular homology.
Setup:
Here is an outline of the process of computing cellular homology as far as I understand it:
- Put a CW complex structure on your space $X$.
- Form the cellular chain complex $$\ldots \xrightarrow{d_{n+1}} C_n(X) \xrightarrow{d_n} C_{n-1}(X) \to \ldots \xrightarrow{d_1} C_{0}(X) \to 0$$ where each $C_i(X)$ is the free abelian group generated by the $i$-cells of $X$.
- As the boundary maps are between free modules it suffices to define the maps on the generators and extend by linearity. In particular, we have the cellular boundary formula $$d_n(e_\alpha^n)=\sum_\beta d_{\alpha \beta} e^{n-1}_\beta$$ where $d_{\alpha \beta}$ is the degree of the composition of the attaching map $S_\alpha^{n-1} \to X^{n-1}$ of $e_\alpha^n$ with the quotient map $q \colon X^{n-1} \to X^{n-1}/(X^{n-1}-e_\beta^{n-1})$.
- After computing the cellular boundary maps one knows both the images and the kernels of these maps and so may compute the homology by $H_i(X) \cong \ker (d_i) / \text{im}(d_{i+1})$.
My Question:
My question is how does one in practice compute the degrees $d_{\alpha \beta}$? I will describe my understanding of the process that I have acquired from the examples in Bredon's Topology and Geometry.
The image of the attaching map $S_\alpha^{n-1} \to X^{n-1}$ is a word in $i$-cells for $i \le n-1$. By composing with the quotient map $q$ one is killing all elements in the word except $e_\beta^{n-1}$. This results in a word completely made up of $e_\beta^{n-1}$. Now one simplifies this word into a reduced word. The length of this reduced word is the $d_{\alpha \beta}$ (up to a sign).
Is this understanding correct? In my algebraic topology course this was never really explained explicitly. Thank you very much!