Compute the cellular homology of $n-$dimensional disk $\mathbb{D}^{n}$ for $n\geq 2$. I am learning some introduction of homology, and I am working on an exercise asking me to compute the homology of $\mathbb{D}^{n}$. I had some attempts but got stuck in the end:

For $X=\mathbb{D}^{n}$, we can write $X=e^{0}\cup e^{n-1}\cup e^{n}$. Then, we have $X^{0}=e^{0}$, $$X^{n-1}=e^{0}\cup_{g_{0}} e^{n-1}\ \text{with the attaching map}\ g_{0}\colon\partial e^{n-1}=\mathbb{S}^{n-2}\longrightarrow e^{0}=\text{a point},$$ $$X^{n}=X^{n-1}\cup_{g_{1}}e^{n}\ \text{with the attaching map}\ g_{1}\colon\partial e^{n}=\mathbb{S}^{n-1}\longrightarrow X^{n-1}.$$
Then, given the cell complex structure, we have $$C_{0}(X)=\mathbb{Z}e^{0}\cong\mathbb{Z},\ C_{k}(X)=0\ \text{for all}\ 1\leq k\leq n-2,$$ $$C_{n-1}(X)=\mathbb{Z}e^{n-1}\cong\mathbb{Z}\ \text{and}\ C_{n}(X)=\mathbb{Z}e^{n}\cong\mathbb{Z},$$ and thus we have the chain complex $$\mathbb{Z}\longrightarrow_{\partial_{n}}\mathbb{Z}\longrightarrow_{\partial_{n-1}}0\longrightarrow \cdots \longrightarrow 0\longrightarrow_{\partial_{1}}\mathbb{Z}.$$
Hence, for all $1\leq k\leq n-2$, we have $H_{k}(\mathbb{D}^{n})=0.$ The problem is to determine $H_{n}(X), H_{n-1}(X)$ and $H_{0}(X)$.
Firstly, for $H_{0}(X)$, $\partial_{1}$ has no choice but only to be $\partial_{1}=0$, and thus $$H_{0}(\mathbb{D}^{2})=\dfrac{\ker(\partial_{0})}{Im(\partial_{1})}=\dfrac{\mathbb{Z}}{0}\cong\mathbb{Z}.$$
Now, since there is one $n-$dimensional cell and one $(n-1)$-dimensional cell, $\partial_{n}$ is a $1\times 1$ matrix whose entry is the degree of $$q\circ g_{1}\colon\partial e^{n}=\mathbb{S}^{n-1}\longrightarrow_{g_{1}}X^{n-1}\longrightarrow_{q}\dfrac{X^{n-1}}{X^{n-2}}.$$
Then I got stuck, what is the space $X^{n-1}/X^{n-2}$? 
For $n\geq 3$, $X^{n-2}=0$, and $X^{n-1}=e^{0}\cup e^{n-1}$, so the quotient is $$\dfrac{e^{0}\cup e^{n-1}}{0}=e^{0}\cup e^{n-1}??$$
For $n=2$, $X^{n-2}=e^{0}$ and $X^{n-1}=X^{1}=e^{0}\cup e^{1}$, so perhaps the quotient is $$\dfrac{e^{0}\cup e^{1}}{e^{0}}=e^{1}?$$
I am really confused about the quotient map $q$. If the above quotient results are true, what is the degree of the composition map?
Thank you!
 A: By construction there are no $k$-cells in the range $[1,n-2]$, so as long as $n \geq 2$ then $X^{n-2} = X^0$ (if $n\leq 1$ then $X^{n-2}$ is either empty or isn't defined, depending on your convention). Therefore, like you suspect, the quotient map $$q\colon X^{n-1} \to X^{n-1}/X^{n-2} = X^{n-1}/X^{0}$$ is a homeomorphism since there is only one $0$-cell (every equivalence class in the quotient has a single element).
The attaching map $g_1$ is typically given by choosing an explicit homeomorphism $g_1\colon S^{n-1} \cong X^{n-1}$, which will have degree $\pm 1$ depending on our choices of orientation and isomorphism. Therefore the composition
$$q\circ g_1 \colon \partial e^n \stackrel{\cong}{\to} X^{n-1} \stackrel{\cong}{\to} X^{n-1}/X^{n-2} $$
is a homeomorphism and will have degree $\pm 1$.

For a general $X$, the quotient $X^{n-1}/X^{n-2}$ will be a bouquet of spheres $\vee S^{n-1}$, one for each $(n-1)$-cell of $X$. The boundary maps for a general CW complex can be more difficult to compute and rely heavily on an explicit description of the cell structure.
