Understanding CW complex construction Hatcher's construction of a CW complex is as follows (see page $5$ of Hatcher's Algebraic Topology):
(1) Start with a discrete set $X^0$ , whose points are regarded as $0$ cells.
(2) Inductively, form the $n$ skeleton $X^n$ from $X^{n−1}$ by attaching $n$ cells $e_{\alpha}^n$ via maps $\phi_{\alpha}:S^{n-1} \to X^{n-1}$. This means that $X^n$ is the quotient space of the disjoint union $X^{n−1} \sqcup_{\alpha} D_{\alpha}^n$ of $X^{n−1}$ with a collection of $n$ disks $D_{\alpha}^n$ under the identifications $x \sim \phi_{\alpha}(x)$ for $x \in \partial
(D_{\alpha}^n)$. Thus as a set $X^n=X^{n-1}\sqcup_{\alpha} e^n_\alpha,$ where $e^n_\alpha$ is an open $n-$ disk.
A space $X$ constructed in this way is called a cell complex or CW complex.
1) I don't understand why $X^n=X^{n-1}\sqcup_{\alpha} e^n_\alpha$ as a set.
We have: $X^n=(X^{n-1}\sqcup_\alpha D^n_\alpha)/\sim$, where $\sim$ is the equivalence relation $x \sim \phi(x)$ for every $x\in \partial D^n_\alpha$. How to go from here to "$X^n=X^{n-1}\sqcup_{\alpha} e^n_\alpha$ as a set"?
 A: Note that $e^n_{\alpha}$ is an open $n$-disk, and $D^n_{\alpha}$ is a closed $n$-disk, so $D^n_{\alpha} = e^n_{\alpha}\sqcup\partial D^n_{\alpha}$. The identification $\phi_{\alpha} : \partial D^n_{\alpha} \to X^{n-1}$ only identifies points of $\partial D^n_{\alpha}$ with points of $X^{n-1}$, the points of $e^n_{\alpha}$ do not get identified. So, as sets, we have
\begin{align*}
X^n &= (X^{n-1}\sqcup_{\alpha} D^n_{\alpha})/(x\sim \phi_{\alpha}(x))\\ 
&= (X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha}\sqcup_{\alpha}e^n_{\alpha})/(x\sim\phi_{\alpha}(x))\\ 
&= (X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha})/(x\sim\phi_{\alpha}(x))\sqcup_{\alpha} e^n_{\alpha}\\ 
&= X^{n-1}\sqcup_{\alpha} e^n_{\alpha}.
\end{align*}
A: The reason is that all the boundaries of the disks $D_{\alpha}^n$ are identified with points of $X_{n-1}$. So you only have the points of $X_{n-1}$ and the interior points of $D_{\alpha}^n$, i.e., $e_{\alpha}^n$.
Think of a particular example to understand better. Take $X_0=\{p\}$ consisting on a single point. Now construct $X_1$ by appending $D^1=[-1,1]$, with identification map $\phi$ sending $-1$ and $1$ to $p$. Then
$$
X_1=\{p\} \sqcup (-1,1)
$$
By the way, the obtained topology is that of $S^1$.
