CW Complex Decomposition of $S^{n}$

Question

I'm currently delving into the foundational concepts of algebraic topology and came across the definition of CW complexes in Hatcher's "Algebraic Topology" (Chapter 0, p. 5):

(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 $X^{n−1} \cup \cup_{\alpha} D_{\alpha}^n$ of $X^{n−1}$ with a collection of $n$ disks $D_{\alpha}^n$ under the identifications $x ∼ \phi_{\alpha}(x)$ for $x \in \partial (D_{\alpha}^n)$ (...)

I found Hatcher's exposition a bit challenging to grasp. Fortunately, I stumbled upon an alternative definition via a commutative diagram on this post: $$\require{AMScd} \begin{CD} \coprod_\alpha S^{n-1}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^n_\alpha \\ @VV (\phi_\alpha) V @VV V\\ X^{n-1} @>>> X^n \end{CD}$$ which appeared more intuitive. However, I've encountered what seems to be a contradiction on the next page of the book in which the process of constructing ${S^n}$.

For context, the construction of $S^n$ can be decomposed through the following commutative square diagram for $S^{n-1}$ and a sum of $D^n$ discs:

$$\require{AMScd} \begin{CD} \coprod_\alpha S^{n-1}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^n_\alpha \\ @VV (\phi_\alpha) V @VV V\\ {e^0} @>>> S^n \end{CD}$$

This diagram is easy to verify on $S^1$ and $S^2$. However, when applying the recursive definition of CW complex, we get a slightly different diagram:

$$\require{AMScd} \begin{CD} \coprod_\alpha S^{n-1}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^n_\alpha \\ @VV (\phi_\alpha) V @VV V\\ X^{n-1} @>>> S^n \end{CD}$$

Starting with $X^0 = {e^0}$, we have the following for $S^0$ into a coproduct of $D^1$ discs:

$$\require{AMScd} \begin{CD} \coprod_\alpha S^{0}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^1_\alpha \\ @VV (\phi_\alpha) V @VV V\\ X^{0} @>>> X^1 \cong S^1 \end{CD}$$

This leads to $X^1$ being homeomorphic to $S^1$. Yet, continuing this process seems to suggest that $X^2$ would not correspond to $S^2$, given $X^1$ is already $S^1$. A answer regarding this example here pointed towards considering $X^0, X^1, \ldots, X^{n-1}$ as ${e^0}$, but this seems incompatible with having themselves attached $D^1, D^2, \ldots, D^{n-1}$.

Could someone help clarify the proper way to construct $S^n$ as a cell complex within the CW complex framework? Is there a step or concept I'm misunderstanding in the transition from $X^{n-1}$ to $X^n$? Any insights or resources that could help illuminate this issue would be greatly appreciated.

Answer 1

This diagram is easy to verify on $S^1$ and $S^2$.

No. You did not define what $\alpha$ is and what attaching maps (vertical arrows) are. It is impossible to deduce anything from that diagram, it is incomplete in the particular case.

There are infinitely many CW structures on a sphere, two of them are "standard".

The first one is recursive. Take two copies of $D^{n}$ and attach both of them to $S^{n-1}$ by glueing over identity $S^{n-1}\to S^{n-1}$. The resulting space is homeomorphic to $S^n$. And by induction (until $S^0=\{-1,1\}$) we get CW decomposition of $S^n$ having $2n$ cells, two at each dimension.

But there's a simpler CW structure on $S^n$. Just take $X^0=\{*\}$ to be a point, ignore all dimensions between $1$ and $n-1$, and add a single $n$-cell. There's only one possible choice of glueing $S^{n-1}\to X^0$, which corresponds to collapsing boundary of $D^n$ into a point. And this results in the $n$-sphere as well, but this time it has $2$ cells (regardless of $n$).

This leads to $X^1$ being homeomorphic to $S^1$. Yet, continuing this process seems to suggest that $X^2$ would not correspond to $S^2$, given $X^1$ is already $S^1$.

That depends on which CW structure you are actually talking about (which honestly is unclear). $X^m$ will be homeomorphic to $S^m$ in the recursive structure. But for the "two cell" structure each $X^m$ will be a singleton except when $n=m$.

This for example:

For context, the construction of $S^n$ can be decomposed through the following commutative square diagram for $S^{n-1}$ and a sum of $D^n$ discs: $$\require{AMScd} \begin{CD} \coprod_\alpha S^{n-1}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^n_\alpha \\ @VV (\phi_\alpha) V @VV V\\ {e^0} @>>> S^n \end{CD}$$

Is $e^0$ supposed to denote a singleton? Then $\alpha$ is a singleton as well, and this actually tells me that you are talking about "two cells" decomposition, not the recursive one. And in this setup indeed each $X^m=e^0$ for $m<n$. Which perfectly fine.