From p.5 of Hatcher's Algebraic Topology:

enter image description here

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 $R^n$?

  • $\begingroup$ 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. $\endgroup$ Commented Aug 3, 2013 at 19:13

1 Answer 1


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.

For small dimensional examples such as this it really pays off to draw out, (or even make out of paper!) some examples to get a feel for how this works.

  • $\begingroup$ 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. $\endgroup$
    – Sam
    Commented Mar 4, 2015 at 15:38
  • $\begingroup$ $$\{x_0, 0, 1\}$$ is one equivalence class. All the points in (0,1) are in distinct equivalence classes. $\endgroup$ Commented Mar 5, 2015 at 4:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .