# Understanding Hatcher's definition of CW-complex

In Hatcher's book on Algebraic Topology, the author writes:

(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 $X^{n−1} \cup \cup_{\alpha} D_{\alpha}^n$ of $X^{n−1}$ with a collection of $n$ disks $D_{\alpha}^n$ under the identifications $x ∼ \phi_{\alpha}(x)$ for $x \in \partial (D_{\alpha}^n)$ (...)

A space $X$ constructed in this way is called a cell complex or CW complex.

I am having many problems understanding this definition...

1) What is a $n$-cell? (the author does not define it previously)

2) What does he mean by "attaching $n$ cells $e_{\alpha}^n$ via maps $\phi_{\alpha}:S^{n-1} \to X^{n-1}$"? What is the relation between the $e_{\alpha}^n$ and $\phi_{\alpha}$?

3) How can one realize the given definition with the description - "This means that $X^n$ is the quotient space of the disjoint union $X^{n−1} \cup \cup_{\alpha} D_{\alpha}^n$ of $X^{n−1}$ with a collection of $n$ disks $D_{\alpha}^n$ under the identifications $x ∼ \phi_{\alpha}(x)$ for $x \in \partial (D_{\alpha}^n)$ (...)"

I know this is an important concept to grasp in Algebraic Topology, so I would really like to understand it, but this definition wasn't clear at all to me.

1) A n-cell is a map from $D^n$ to the topological space $X$ you're working with.

2) Attaching a space $X$ to a space $Y$ along a map $\phi : E \to Y$, assuming $E \subseteq X$, means to create the quotient space over the disjoint union of $X$ and $Y$ by identifying a point $e \in E$ with its image $\phi(e)\in Y$. Figuratively, we "stitch" $X$ upon $Y$ by gluing $E \subseteq X$ onto $Y$ with $\phi$ and we obtain the space

$$\frac{X \coprod Y}{ < e \sim \phi(e) >}.$$

So far you've constructed you cellular complex up to $n-1$-cells, obtaining the space $X^{n-1}$. View this as the "skeleton" upon which you will attach a $n$-disk $D^n$ along the boundary of the disk $\partial D^n = S^{n-1}$. Thus you need that map $\phi^n_\alpha : S^{n-1} \to X^{n-1}$ to identify the boundary of the disk as points of the skeleton $X^{n-1}$.

$e^n_\alpha$ is the cell, i.e. the map that send the entirety of the disk $D^n$ to the completed space $X$ (or possibly $X^n$), but not to $X^{n-1}$ as the point in the interior of the $n$-disks are not identified with any point of $X^{n-1}$. However on the boundary of the disk, $e^n_\alpha$ should match $\phi^n_\alpha$ :

$$e^n_\alpha|_{\partial D^n} = \phi^n_{\alpha}.$$

3) I already explained the concept of attaching spaces in 2). From that definition,

$$X^n = \frac{ X^{n−1} \cup \big( \bigcup_{\alpha} D_{\alpha}^n \big) }{< x_\alpha \sim \phi^n_\alpha(x_\alpha) >}$$ holds, with the relation being spanned over every $x_\alpha \in S^{n-1}_\alpha$ for any index $\alpha$.

Note that Hatcher uses another notation than the one I used earlier to introduce the concept, where rather than using the disjoin union $\coprod$, he uses the regular union $\bigcup$ and label each of its disks to be attached with a unique index $\alpha$ so they are still distinct after the union. The topological result is the same.

I would post it as a comment, but I really want to draw a commutative diagram... Categorically speaking, $X^n$ is a pushout in the category of topological spaces

$$\require{AMScd} \begin{CD} \coprod_\alpha S^{n-1}_\alpha @>{\coprod_\alpha i_\alpha}>> \coprod_\alpha D^n_\alpha \\ @VV (\phi_\alpha) V @VV V\\ X^{n-1} @>>> X^n \end{CD}$$

Here the arrow $\coprod_\alpha i_\alpha$ is formed by the standard inclusions of $(n-1)$-spheres as boundaries of $n$-disks, and $\phi_\alpha\colon S^{n-1}_\alpha \to X^{n-1}$ are some maps that you choose. The description of $X^n$ as a disjoint union of $X^{n-1}$ and $\coprod_\alpha D^n_\alpha$ with certain identifications is the description of pushouts for topological spaces.

Then an $n$-cell is each map $D^n_\alpha \to X^n$ given by the pushout above.