In Hatcher's book we find, when computing the boundary maps of cellular homology,
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.
My question is:
When we identify $X^{n-1}/(X^{n-1}- e^{n-1}_\beta)$ with $S^{n-1}_\beta$, we are choosing a homeomorphism. Doesn't this choice affect the degree of the map in question?
Essentially (as I see it), we must choose a quotient map from $D^{n-1}$ to $S^{n-1}$ which identifies $S^{n-2}$ to a point.
Best Answer
There is indeed a choice of homeomorphism $D^n/S^{n-1} \cong S^n$ but as others have pointed out this won't affect any homology calculations as long as you consistently use this homeomorphism for each cell.
However, the degree is affected by the choice of homeomorphism, so the cellular boundary formula as found in Hatcher is only right to a sign. In order for the cellular boundary formula to make sense it is necessary that the generators of $H_n(X^n, X^{n-1})$ are of the form $e_{\alpha}^n =\Phi_{\alpha *}^n([D^n])$ for a fixed generator $[D^n]$ of $H_n(D^n, S^{n-1})$. Now in order for the cellular boundary formula to hold we need that these generators fit together according to the sequence of isomorphisms
$H_n(D^n, S^{n-1}) \cong H_{n-1}(S^{n-1}) \cong H_{n-1}(D^{n-1}/S^{n-2}) \cong H_{n-1}(D^{n-1}, S^{n-1})$
where the first isomorphism is the boundary map coming from the long exact sequence for the pair $(D^n, S^{n-1})$ and the second isomorphism comes from the chosen homeomorphism $S^{n-1} \cong D^{n-1}/S^{n-2}$. If the homeomorphisms make the chosen generators incompatible then the formula may fail to hold.
This compatibility ensures the last two isomorphisms followed by $\Phi_{\beta *}^{n-1}$ sends $\partial [D^n]$ to $[D^{n-1}]$ and then to $e_{\beta}^{n-1}$. Thus it sends $\Delta_{\alpha \beta} \partial [D^n] = d_{\alpha \beta} \partial [D^n]$ to $d_{\alpha \beta} e_{\beta}^{n-1}$. The standard diagram chase shows $d_n(e_{\alpha}^n) = \sum_{\beta} (H_{n-1}(S^{n-1})\to H_{n-1}(D^{n-1}/S^{n-2}) \to H_{n-1}(D^{n-1}, S^{n-1}))\Delta_{\alpha \beta} \partial [D^n]$ so the theorem is proved.