Help understanding CW-complex construction.

I thought attachment maps would be from $n$-cell to $n$-cell since it says "by attaching $n$-cells", but in that link they're from $S^{n-1}$ to $(n-1)$-cell? Please shed some light on what they mean and how it's related to attachment maps.

What do they mean by $D^n_\alpha$? Is that the open disk in $R^n$? Does this mean CW-complexes are always subsets of $R^n$?

• Please take screenshots or copy down the information yourself into the question, so that your question will still make sense if external content is removed. Commented Aug 3, 2013 at 19:13

You make the $n$-skeleton $X^n$ by gluing copies of the $n$-disk, $D^n$ to the $X^{n-1}$ skeleton. The gluing occurs on the boundary of the disk, which is $S^{n-1}$, hence you get the attachment maps. Let me give an example,
We want to get the unit disk in $\mathbb{R}^2$ as a CW complex. So we start with $X^0$ which will be a point, call it $x_0$. (in general the number of points at the 0 level is the number of connected components you want to end up with). To form $X^1$ we must attach a 1-cell which is $[0, 1]$ to the point $x_o$. To make this attachment we must have a map $\varphi:S^0\rightarrow X^0$. $S^0=\{0, 1\}$ and $X^0=\{x_0\}$ so the only map is to send both elements to the same point. This gives us $X^1$. Note that $X^1$ is homeomorphic to $S^1$.
Now to create $X^2$ we must attach a copy of $D^2$ to $X^1$. So we need an attachment map from $S^1$, (thinking of it as the boundary of $D^2$ to $X^1$. Since $X^1$ is homeomorphic to $S^1$ we can use this homeomorphism as the attaching map, and we are left with $X^2$. Now we are done.
• Please help me understand this. Is $X^1$ the set $X^0 \sqcup [0,1]$ modulo the equivalence relation $x R \varphi(x)$ for each $x \in S^0$? That is, it's a set of equivalence classes, right? So then what are the other equivalence classes besides $[x_o]$? What equivalence class are the points $(0,1)$ in? Thank in advance.
• $$\{x_0, 0, 1\}$$ is one equivalence class. All the points in (0,1) are in distinct equivalence classes. Commented Mar 5, 2015 at 4:34