What are cells in CW complex?

Question

I was studying Hatcher's Algebraic Topology regarding CW complex. I am having trouble in understanding definition of CW Complex.

(1) Start with a discrete set $X^0$, the $0$-cells of $X$.

(2) Inductively, form the $n$-skeleton $X^n$ from $X^{n−1}$ by attaching $n$-cells $e^n_\alpha$ via maps $\varphi_\alpha:S^{n-1}\to X^{n−1}$. This means that $X^n$ is the quotient space of $X^{n−1}\amalg_\alpha D^n_\alpha$ under the identifications $x\sim\varphi_\alpha(x)$ for $x\in\partial D^n_\alpha$. The cell $e^n_\alpha$ is the homeomorphic image of $D^n_\alpha−\partial D^n_\alpha$ under the quotient map.

My confusion is if these $n$-cells $e^n_\alpha$ are actually homeomorphic to open disks then why are we attaching closed disk to $X^{n-1}$?

Answer 1

I will give an example in dimension $1$. So the open $1$-disk is just an interval, without the endpoints. Imagine you already have $X_0$, and we now want to attach the cell $e^1$ to $X_0$. In this dimension, there will be basically two ways to do this:

• "Connect" two different points in $X_0$ with the open disk
• "Connect" the same point with itself, forming a loop

Note that, in both cases, the interior of the cell is homeomorphic to the open disk, since both are an open interval, but in the first case, the closure is a closed $1$-disk, and in the second case, it is a $1$-sphere. So the way you attach the closed disk matter, even if the interior is the same open disk.

Answer 2

Here's two examples for you to ponder.

Take $X$ to be the following subset of $\mathbb R^2$: $$X^0 = \{\underbrace{(0,0)}_{e^0_1}, \underbrace{(1,0)}_{e^0_2}\} $$ and $$X=X^1 = X^0 \cup \bigl( \underbrace{(0,1) \times \{1\}}_{e^1_3}\bigr) $$ Now, I'll point out the obvious: $e^1_3$ is (as you say) actually homeomorphic to the open $1$-disc $(0,1)$.

True or false question #1: $X$ is a CW complex with two $0$-cells namely $e^0_1,e^0_2$, and with one $1$-cell namely $e^1_2$.

Okay, now let's take a slightly different example $Y$, with $Y^0=X^0$, but with $$Y = Y^1 = Y^0 \cup \bigl(\underbrace{(0,1) \times \{0\}}_{e^1_4}\bigr) $$ True or false question #2: $Y$ is a CW complex with two $0$-cells namely $e^0_1,e^0_2$, and with one $1$-cell namely $e^1_4$.

Followup question: The answers to #1 and #2 are different, as I hope you have perceived. But the question is: Why are they different?

Answer 3

For the sake of simplicity, assume we are only attaching a single $n$ cell through an attaching map $\varphi:S^{n-1}\rightarrow X^{n-1}$. Then $X^n$ is defined to be the pushout

$\require{AMScd}$ \begin{CD} S^{n-1} @>\varphi>> X^{n-1}\\ @V i V V @VV V\\ D^n @>> > X^n \end{CD} in Top. Note that the map $S^{n-1}\rightarrow D^n$ is injective, which implies that the map $X^{n-1}\rightarrow X^n$ is also injective. Therefore, one can think about what are the "new" points of $X^n$, that are not in the image of $X^{n-1}$.

Let $\overset{\circ}{D^n}$ be the interior of $D^n$. Note that there is a canonical map $\overset{\circ}{D^n}\rightarrow X^n$ given by $$ \overset{\circ}{D^n}\hookrightarrow D^n\rightarrow X^n~.$$ The map $\overset{\circ}{D^n}\rightarrow X^n$ is always injective, and the image of this map is what is referred to as $e^n$. Furthermore, this determines a map

$$e^n\amalg X^{n-1}\xrightarrow{\simeq}\overset{\circ}{D^n}\amalg X^{n-1}\rightarrow X^n$$ that turns out to be a bijection, although be warned this is not a homeomorphism.

Therefore, the space $X^n$ as a set is in bijection with the set $X^{n-1}\amalg e^n$. So, the points $e^n$ can be thought of as the "new" points of $X^n$.