Doubt in the proof of Cellular Boundary formula from Hatcher's Book Theorem: Cellular Boundary Formula: $d_n(e^{n}_\alpha)=\sum_\beta d_{\alpha\beta}e^{n-1}_{\beta} $ where $d_{\alpha\beta}$ is the degree of the map $S^{n-1}_\alpha \rightarrow X^{n-1}\rightarrow S^{n-1}_\beta$ that is the composition of the attaching map of $e^{n}_\alpha$ with the quotient map collapsing $X^{n-1}- e^{n-1}_\beta $ to a point.
Proof:
The map $\Phi_{\alpha^*}$ takes a chosen generator $[D_{\alpha}^n] \in H_n(D^n_{\alpha},\delta D^n_{\alpha})$ to a generator of the $\mathbb {Z}$ summand of $H_n(X^n,X^{n-1})$ corresponding to $e^n_{\alpha}$... 

How does one see the above first line of the proof? 

 
 A: I suggest you define $[e_\alpha^n]\in H_n(X^n , X^{n-1})$ to be the image of $[D_\alpha^n] \in H_n(D_\alpha^n , \partial D_\alpha^n)$ under $\Phi_\alpha^n$.
However, for the purposes of following this proof, the key observation is actually the following:
The quotient $X^n / X^{n-1}$ is a "bouquet" of $n$-spheres, $\vee_\gamma S_\gamma^n$. Consider the sequence of maps:
$$ H_n(D_\alpha^n , \partial D_\alpha^n) \overset{(\Phi_\alpha^n)_\star}{\to} H_n(X^n , X^{n-1}) \overset{\bar q_\star}{\to} \tilde H_n(X^n/X^{n-1}) \overset{\approx}{\to} \tilde H_n(\vee_\gamma S_\gamma^n) \overset{(q_\beta)_\star}{\to} \tilde H_n(S_\beta^n),$$
where $\bar q$ is the quotienting projection from $X^n$ to $X^n / X^{n-1}$ and $q_\beta$ is the quotienting projection from $\vee_\gamma S_\gamma^n$ onto $S_\beta = \vee_\gamma S_\gamma^n / \vee_{\gamma \neq \beta} S_\gamma^n$.
Then the composition $H_n(D_\alpha^n , \partial D_\alpha^n) \to \tilde H_n(S_\beta^n)$ sends $[D_\alpha^n]$ to a generator of $\tilde H_n(S_\beta^n)$ if $\alpha = \beta$, or to the zero element in $\tilde H_n(S_\beta^n)$ if $\alpha \neq \beta $.
[To see why, observe that when $\alpha = \beta$ the composition $H_n(D_\alpha^n , \partial D_\alpha^n) \to \tilde H_n(S_\beta^n)$ is induced by the quotienting projection from $D_\alpha $ onto $D_\alpha / \partial D_\alpha \cong  S_\beta^n$, and since $(D_\alpha^n , \partial D_\alpha^n)$ is a good pair, the generator of $H_n(D_\alpha^n , \partial D_\alpha^n)$ must map to a generator of $\tilde H_n(S_\beta^n)$. If $\alpha \neq \beta$, then the composition $H_n(D_\alpha^n , \partial D_\alpha^n) \to \tilde H_n(S_\beta^n)$ is induced by the contraction of $D_\alpha$ to a single point in $S_\beta^n$, so the generator of $H_n(D_\alpha^n , \partial D_\alpha^n)$ must map to the zero element in $\tilde H_n(S_\alpha^n)$.]
Thus, the generator $[e_\alpha^n]\in H_n(X^n , X^{n-1})$ can be characterised in two ways:
(i) $[e_\alpha^n] = \Phi_\alpha^n([D_\alpha^n])$;
(ii) $(q_\alpha)_\star \circ  \bar q_\star ([e_\alpha^n]) = [S_\alpha^n]; \ \ $ and $ \ (q_\beta)_\star \circ  \bar q_\star ([e_\alpha^n]) = [0 ] \ $ if $\beta \neq \alpha$.
Now let us complete the proof. Suppose the image of $e_\alpha^n$ under the cellular boundary map can be written as
$$ d_n([e_\alpha^n]) = \sum_\gamma c_{\alpha \gamma} [e_\gamma^{n-1}]$$
for some coefficients $c_{\alpha \gamma}$. Our task is to determine the values of these coefficients $c_{\alpha \gamma}$.
First of all, by my characterisation (i) of the generator $[e_\alpha^n]$, we have $[e_\alpha^n] = (\Phi_\alpha)_\star [D_\alpha^n]$, so we can rewrite the equation as
$$ d_n \circ (\Phi_\alpha^n)_\star ([D_\alpha^n]) = \sum_\gamma c_{\alpha \gamma} [e_\gamma^{n-1}].$$
Next, let $\bar q$ be the quotienting projection $X^{n-1} \to X^{n-1}/X^{n-2}$. The morphism $H_{n-1} (X^{n-1}, X^{n-2}) \overset{\approx}{\to} H_{n-1} (X^{n-1}/X^{n-2}, X^{n-2}/X^{n-2}) \cong \tilde H_{n-1} (X^{n-1}/X^{n-2})$ at the bottom-right corner of Hatcher's diagram is induced by the projection $\bar q$. By my characterisation (ii) of the generator $[e_\gamma^{n-1}]$, we know that $(q_\beta)_\star \circ \bar q ([e_\gamma^{n-1}])$ is equal to $[S_\beta^{n-1}]$ if $\beta = \gamma$ and is equal to $[0]$ if $\beta \neq \gamma$. So we can rewrite the equation as
$$ c_{\alpha \beta} [S_\beta^{n-1}]= (q_\beta)_\star \circ \bar q_\star \circ d_n \circ (\Phi_\alpha^n)_\star ([D_\alpha^n]).$$
Now, the map $d_n = j_{n-1} \circ \partial_n $ by definition. So,
$$ c_{\alpha \beta} [S_\beta^{n-1}]= (q_\beta)_\star \circ \bar q_\star \circ j_{n-1} \circ \partial_n \circ (\Phi_\alpha^n)_\star ([D_\alpha^n]) .$$
Also, the morphisms in the bottom-right square of Hatcher's diagram commute, since both downwards morphisms in the bottom-right square are induced by identity maps on the respective topological spaces and both sideways morphisms are induced by the relevant quotienting projections, and these maps of topological spaces commute. So,
$$ c_{\alpha \beta} [S_\beta^{n-1}]= (q_\beta)_\star \circ q_\star \circ \partial_n \circ (\Phi_\alpha^n)_\star ([D_\alpha^n]).$$
Next, the top-left square in Hatcher's diagram commutes: since $\varphi_\alpha$ is by definition the restriction of the attaching map $\Phi_\alpha^n$ to $\partial D_\alpha^n$, this follows from the natural commuting diagram between the LESs for the pairs $(D_\alpha^n, \partial D_\alpha^n)$ and $(X^n, X^{n-1})$. So,
$$ c_{\alpha \beta} [S_\beta^{n-1}]= (q_\beta)_\star \circ q_\star \circ  (\varphi_\alpha^n)_\star \circ \partial ([D_\alpha^n]).$$
Next, the map $q_\beta \circ q \circ \varphi_\alpha^n $ from $S_\alpha^{n-1} \cong \partial D_\alpha^n $ to $S_\beta^{n-1}$ is the map $\Delta_{\alpha \beta}$, by definition. So,
$$ c_{\alpha \beta} [S_\beta^{n-1}]= (\Delta_{\alpha \beta})_\star \circ \partial ([D_\alpha^n]).$$
Finally, the map $\partial: H_n(D_\alpha^n, \partial D_\alpha^n) \to \tilde H_{n-1}(\partial [D_\alpha^n]) \cong \tilde H_{n-1} (S_\alpha^{n-1})$ is an isomorphism, so $\partial[D_\alpha]$ is a generator for $\tilde H_{n-1} (S_\alpha^{n-1})$. The degree  $d_{\alpha \beta}$ of the map $\Delta_{\alpha \beta} : S_\alpha^{n-1} \cong \partial D_\alpha^n \to X^{n-1} \to S_\beta^{n-1}$ is defined so that $(\Delta_{\alpha \beta})_\star \circ \partial[D_\alpha] = d_{\alpha \beta} [S_\beta^{n-1}]$. So,
$$ c_{\alpha \beta} [S_\beta^{n-1}] = d_{\alpha \beta} [S_\beta^{n-1}],$$
from which we deduce that
$$ c_{\alpha \beta} = d_{\alpha \beta}$$
as required.
