cell complex structure of circle

Question

I came to know that the cell complex structure of the circle $S^1$ is $e^0∪e^1$. But in my point of view it should be $D^0∪D^1$ as $e^0$ is nothing but an empty set. Can anybody make me understand where am I wrong?

Answer 1

The are two equivalent approaches to build CW-complexes. One is based on open cells, the other on closed cells. Open cells are pairwise disjoint whereas closed cells are not pairwise disjoint (unless we only have $0$-cells). In both cases the $k$-skeleton $X^k$ of a CW-complex $X$ is defined as the union of all cells with dimension $\le k$. For each $n$-cell $\gamma^n_\alpha$ we have a characteristic map $\phi^n_\alpha : D^n \to X^{n-1}$. If $\gamma^n_\alpha = e^n_\alpha$ denotes an open cell, then $e^n_\alpha = \phi^n_\alpha(D^n \setminus \partial D^n)$, if $\gamma^n_\alpha = \bar{e}^n_\alpha$ denotes a closed cell, then $\bar{e}^n_\alpha = \phi^n_\alpha(D^n)$. In the first case $e^n_\alpha \cap X^{n-1} = \emptyset$ and in second case $\bar{e}^n_\alpha \cap X^{n-1} \ne \emptyset$.

It seems that the "open-cell-approach" is more popular, see for example the Appendix of Hatcher's "Algebraic Topology".

Anyway, in both approaches $0$-cells are single point spaces and the $0$-skeleton $X^0$ is a discrete space. This comes from the fact that $D^0 = \{ * \}$ and $\partial D^0 = \emptyset$.

Hence for $X = S^1$ we obtain an open cell decomposition $e^0 = \{ 1\}, e^1 = S^1 \setminus \{ 1\} \approx (-1,1)$ and a closed decomposition $\bar{e}^0 = \{ 1\}, \bar{e}^1 = S^1$. The characteristic map for the $1$-cell is $\phi^1 : [-1,1] \to \{ 1\}, \phi^1(t) = e^{\pi i t}$.