I know that the sphere $S^n$ has the structure of a cell complex with just two cells, $e^0$ and $e^n,$ where the $n$- cell being attached by the constant map $S^{n-1} \to e^0.$ I also know that another structure of a cell complex on $S^n$ is with two cells in each dimension I do not know the attaching maps in this case though. Could anyone clarify to me what are the attaching maps in this case?
Now, if I want to show that the circle $S^1$ has a CW structure with n $0$-cells and n $1$- cells and I know that the definition of CW complexes is the pushout of the diagram $$X^{(n - 1)} \leftarrow \sqcup \partial D_{\alpha}^n \rightarrow \sqcup D_{\alpha}^n.$$ And I know that the CW complex structure can be determined by determining the characteristic maps and once you know them you can determine the attaching maps by restricting its domain to the boundary. How can I determine the characteristic maps in this case? What is the intuition behind that? What are these characteristic maps?
Any help will be greatly appreciated!
Start with two points. Connect them along two different paths and you get a circle. Fill in the circle twice (say, with an upper hemisphere and a lower hemisphere), and you get a sphere (2-sphere). Filling in the sphere in two independent ways gets you a 3-sphere, though this is harder to picture as symmetrically (maybe you can imagine filling in the interior of the ball and then filling up the exterior including a point at infinity).
In this structure, you make an $S^{n-1}$ inductively, then you attach two cells $e^n_+$ and $e^n_-$ both along the attaching map $\operatorname{id}\colon S^{n-1}\to S^{n-1}$, and the result is homeomorphic to $S^n$.
Thinking of $S^n$ as a subset of $\mathbb{R}^{n+1}$, you can think of the two characteristic maps as the embedding of the upper and lower hemispheres (the sets where the final coordinate is positive or negative). The characteristic maps of the non-top-dimensional cells are just the embedding of the lower or upper hemisphere of the equator, or of the equator's equator, and so on.
If you want to be super explicit with $S^2\subset\mathbb{R}^3$ as an example, the upper hemisphere is embedded as $D^2 \to S^2$ by $(x, y) \mapsto (x, y, \sqrt{1-x^2-y^2})$, and the lower hemisphere cell is embedded as $(x, y) \mapsto (x, y, -\sqrt{1-x^2-y^2})$. The two $1$-cells are embedded with maps $[-1, 1] \to \mathbb{R}^3$ given by $x \mapsto (x, \sqrt{1-x^2}, 0)$ and $x \mapsto (x, -\sqrt{1-x^2}, 0)$. The $0$-cells are the points $(\pm 1, 0, 0)$.