recently I am in great puzzle when studying algebraic topology using Hatcher's book. Questions are as follows:
- How to understand the "cell" intuitionally? I used to understand it as an open disk, but soon I found that it could also have boundaries.
- Is there any connection between dimension and topology space? For example, should a 0-cell be in 0-dimension and so on?
- What is the product of two topological space A and B? According to Hatcher, the product of two spaces is actually the product of the cells in A and B, but what is the product of cells? For example, a torus is S1 x S1, and S1 is composed of e0 and e1. So I guess the answer should be a set of e0 x e0, e1 x e1, e1 x e0, e0 x e1. What is e0 x e0 or the other like? For example, e0 is a point, so e0 x e0 should be also a point, and the dimension is higher?
I assume you are refering to CW complex.
Well, that depends if we are talking about open cells or closed cells. Open cells are very easy to understand - they are simply open balls (up to homeomorphism).
Closed cells on the other hand can be quite complicated. By the definition a closed cell is the image of a closed ball under an attaching map. So it can be a closed ball itself. But it can be for a example a sphere or any other monstrosity generated by an attachment map.
Note that an open cell is the "inside" of the associated closed cell.
For example in the case of the sphere $S^1\subseteq\mathbb{C}$ in the complex plane. We can introduce the following CW structure: one $0$-cell $\{1\}$ and one closed $1$-cell $S^1$. The attaching map glues ends of $1$-dimensional ball (i.e. the closed interval) with the single $0$-cell. There never are open $0$-cells. There is one open $1$-cell, namely $S^1\backslash\{1\}$. It is the "inside" of $S^1$. Note that the closed cell $S^1$ is not homeomorphic to a closed ball.
You can take a different CW structure on $S^1$: two $0$-cells $\{1\}, \{-1\}$ and two closed $1$-cells $\{z\in S^1\ |\ \text{im}(z)\geq 0\}$ and $\{z\in S^1\ |\ \text{im}(z)\leq 0\}$ (here $\text{im}$ denotes the imaginary part so those sets are the upper and the lower halves of the sphere). Now both closed cells are homeomorphic to closed balls. Attaching maps glue one end of the closed interval to $\{-1\}$ and the other to $\{1\}$ producing our cells.
Yes, $n$-cell is $n$-dimensional by definition. The dimension of the entire CW complex is defined as the maximum among dimensions of its cells.
Product as in the Cartesian product. This is what it is in terms of sets. The topology is then defined as the product topology. It is "inherited" from both components. Cells are topological spaces so it makes perfect sense to talk about their product.
Yes, product of singletons is again a singleton. And no, the dimension does not increase. The formula for dimension is quite simple:
$$\dim(X\times Y)=\dim(X)+\dim(Y)$$
and the dimension of a point (or finite set of points) is $0$.
For example take a plane $\mathbb{R}^2$ and a line $\mathbb{R}$. They have dimensions $2$ and $1$ respectively. But their product $\mathbb{R}^2\times\mathbb{R}=\mathbb{R}^3$ is the $3$-dimensional space.