Need help understanding cell complexes

Question

I need some help understanding cell complexes and properties behind them. I'm starting Hatcher and can't seem to grasp the simplest ideas here.

From how I understand it: Cell complexes present themselves as a new way to construct spaces by gluing n dimensional disks to all sorts of surfaces at a point along the boundary of the disk. This surface level understanding doesn't help me answer questions such as how a connected cell complex is equivalent to it being path connected which is equivalent to saying the 1-skeletons (arcs if I'm not wrong) of the cell are connected? I understand how these definitions work independently but I can't seem to connect this idea to cell complexes. I just get this mental block on what to do, I'm not sure how to work on this abstractly. Any response will help, from advice on how to think about this or even explanations on how cell complexes work.

Apologies for large text, I wrote this from my phone.

Answer 1

Maybe an example will clarify things. Consider this $2$-dimensional CW complex embedded in $\mathbb{R}^2$:

So it is a line segment together with a filled ellipse. Well, actually is not a CW complex yet, it is only a topological space. We have to define a cell structure on it. There are multiple ways to do that (infinitely many in fact), one possibility is this:

1. two $0$ cells, more precisely endpoints of the line segment
2. two $1$ cells, the line segment on the left and the boundary of the ellipse
3. one $2$ cell, the interior of the ellipse.

With that the $0$-skeleton (i.e. the union of all cells of dimension up to $0$) is simply this:

while the $1$-skeleton (i.e. the union of all cells of dimension up to $1$) is this:

and the $2$-skeleton (i.e. the union of all cells of dimension up to $2$) equals to the space itself.

Note how we glue the $2$-cell to the $1$-skeleton. It is boundary to boundary, not "at a point" as you've suggested in the question. We have lots of freedom on how we glue $n$-cells to the $n-1$-skeleton.

which is equivalent to saying the 1-skeletons (arcs if I'm not wrong) of the cell are connected?

Not arcs. $1$-skeleton (note: singular, without "s" at the end). Every CW complex comes with a structure of cells. Given that structure, the $n$-skeleton is the union of all cells of dimension up to $n$. For a given CW structure and given $n\in\mathbb{N}$ there is exactly one $n$-skeleton. Although they are not necessarily distinct for different $n$, since a CW complex of dimension $m$ does not have to have cells of all dimensions lower than $m$. E.g. the sphere of dimension $m$ can be given CW structure consisting of a single $0$-cell and a single $m$-cell, meaning the $n$-skeleton is equal to $0$-skeleton for any $0\leq n< m$. In particular note that the $n$-skeleton does not have to be of dimension $n$ (although $n$ is the upper bound of its dimension) which is somewhat counterintuitive.

Anyway the theorem says that path connectedness of a CW complex is equivalent to the path connectedness of its $1$-skeleton. The point is that cells themselves are always path connected. But in order to go from one cell to another we can always do that through $1$-skeleton, as long as it is path connected. Hopefully the example above gives a good enough intuition.

The fact that a CW complex is connected if and only if it is path connected is harder to visualize. Perhaps because connected but not path-connected spaces are strange. Here the intuition is that CW complexes are not pathological, unlike some other spaces, e.g. the topologist's sine curve. I suppose you can simply accept the fact and move on.