Some doubts about the cellular boundary formula

(from Hatcher's book on algebraic topology).

1 , 2

1. What is the identification they refer to when saying "we are identifying the cells $$e_{\alpha}^n, e_{\beta}^{n-1}$$ with generators of the coressponding summands.."?

2. I do not see why the attaching map of $$e_{\alpha}^n$$ which is $$\phi_{\alpha}$$ is involved here and so the summation in the formula is finite.

3. when saying "collapsing the complement of $$e_{\beta}^{n-1}$$ in $$X^{n-1}$$ to a point", how this look as a map (in the diagram)?

4. They mention Generatosr of $$Z$$ summand of a homology/reduced homology group corresponding to a cell.., but did not they explicitly which summand it is?

Thanks in advance for helping me better understand these things!

One way to think about this is by the corresponding map on the quotient spaces $$\tilde{\Phi} : D^n_\alpha / \partial D^n_\alpha \to X^n/X^{n-1}$$ The picture in your head should be that $$D^n_\alpha / \partial D^n_\alpha$$ looks like an $$n$$-sphere and $$X^n/X^{n-1}$$ is a wedge of $$n$$-spheres corresponding to each cell. The map $$\tilde{\Phi}$$ then just sends the sphere $$S^n$$ to the corresponding sphere in the wedge representing the cell $$e_\alpha$$. The key point is that this is something topological, and when we move to the algebraic world of homology, the map induced on homology by $$\tilde{\Phi}$$ tells us how to send the generator of $$H_n(D^n_\alpha/\partial D^n_\alpha) \cong H_n(D^n_\alpha, \partial D^n_\alpha) \cong \mathbb{Z}$$ to one of the generators in $$H_n(X^n/X^{n-1}) \cong H_n(X^n, X^{n-1}) \cong \oplus \mathbb{Z}.$$ The takeaway is that each $$\mathbb{Z}$$ summand in the homology of $$X^n/X^{n-1}$$ corresponds to a cell, and we can work out which one by looking at the image of the map $$H^n(\tilde{\Phi}_\alpha)$$ - it really is induced by something topological.
Now for your third point, we can use a kind of similar intuition. We now think of each sphere in $$X^{n-1} / X^{n-2} = \wedge S^{n-1}$$ as corresponding to an $$(n-1)$$-cell in the complex. For a given $$(n-1)$$-cell $$e^{n-1}_\beta$$, the map he is talking about sends every $$(n-1)$$-cell in the wedge to the point except for $$e^{n-1}_\beta$$. In some sense, this does the "transpose" algebraic thing as the last example: now it induces a map on homology from $$\oplus \mathbb{Z} \to \mathbb{Z}$$, where it sends every generator to zero except the one representing the cell $$e^{n-1}_\beta$$.
• Thab you @richokicked800goals may I ask how the image of $e_{\alpha} ^n$ meets only finitely many cells $e_{\beta} ^{n-1}$. I just do not see it by the given formula!. Feb 2, 2023 at 9:28
• It's part of the definition of a CW complex that the image of the attaching map of an $n$-cell $e_\alpha^n$ intersects with finitely many $(n-1)$-cells $e_\beta^{n-1}$. Feb 2, 2023 at 15:12
• Okay, may I ask in [1] (image) why is $d_1$ is the same as the simplicial boundary map? Isn't $d_1$ take 1-cells $e$ and sends then to $e(1)-e(0)$ (0-cells) ? Feb 2, 2023 at 15:38