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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"?

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2 Answers 2

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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*}

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  • $\begingroup$ Thanks a lot for the answer. I think you meant 'only identifies points of $\color{blue}{\partial D^n_\alpha}$...' in the second line. $\endgroup$
    – Koro
    Commented Mar 4, 2023 at 12:35
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    $\begingroup$ @Koro: Indeed I did. Thanks for pointing out the typo. $\endgroup$ Commented Mar 4, 2023 at 13:24
  • $\begingroup$ 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. $\endgroup$
    – Koro
    Commented Mar 4, 2023 at 13:47
  • $\begingroup$ 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 $X^{n-1}= X^{n-1}\sqcup \partial D^n_\alpha$. $\endgroup$
    – Koro
    Commented Mar 4, 2023 at 13:52
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    $\begingroup$ 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. $\endgroup$ Commented Mar 4, 2023 at 14:14
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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 $S^1$.

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