# 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"?

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*}

• Thanks a lot for the answer. I think you meant 'only identifies points of $\color{blue}{\partial D^n_\alpha}$...' in the second line.
– Koro
Commented Mar 4, 2023 at 12:35
• @Koro: Indeed I did. Thanks for pointing out the typo. Commented Mar 4, 2023 at 13:24
• Could you please explain the last equality? How do you get $X^{n-1}$ in the last line? I think it has something to do with 'as sets' but I'm not quite understanding how. Thanks.
– Koro
Commented Mar 4, 2023 at 13:47
• I have this confusion because I think that 'attaching maps' changes the 'cardinality' of the set obtained by attaching. Here, we are attaching $D^n_\alpha$ via identification map $\phi_\alpha$ and this identification is happening on the set $X^{n-1}\sqcup \partial D^n_\alpha$, not on $X^{n-1}$ so I don't understand how it can be concluded that $X^{n-1}= X^{n-1}\sqcup \partial D^n_\alpha$.
– Koro
Commented Mar 4, 2023 at 13:52
• I didn't claim that $X^{n-1} = X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha}$. Rather, as sets, $X^{n-1} = (X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha})/(x\sim\phi_{\alpha}(x))$. Every point of $\partial D^n_{\alpha}$ is identified with a point of $X^{n-1}$ in the quotient via $\phi_{\alpha}$. In the quotient, $\partial D^n_{\alpha}$ doesn't contribute any new points to the set. Commented Mar 4, 2023 at 14:14

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 $$S^1$$.