Why the space S1 and S1/Z_2 is topologically identical?

Question

I am a physicist studying liquid crystals. My research is bit related to topology but I don't have much knowledge of it. Recently I read from a the book Soft matter physics: An introduction that topologically, $S^1$ and $S^1/Z_2$ is topologically identical. The book doesn't give explain to this statement and I have some problem of understanding.

1. Is the space of $S^1/Z_2$ a semi-circle? If so why it is identical to $S^1$?
2. Is $S^1/Z_N$ also topologically identical to $S^1$?

Can anyone give me some hints? References are also warmly welcome.

Answer 1

Yes, $\mathbb{S}_1 / \mathbb{Z}_2$ is homeomorphic to $\mathbb{S}_1$. The manifold $\mathbb{S}_1 / \mathbb{Z}_2$ is a semi circle with extremities identified, so topologically a circle. This is also called the real projective line.

Start from the circle $\mathbb{S}_1 = \{(x,y):x^2+y^2=1\}$, and let $\mathbb{Z}_2$ act by central symmetry. Quotienting here is equivalent to identifying opposite points, and the data of two opposite points on this circle is exactly the data of a line going through the origin. Hence, $\mathbb{S}_1 / \mathbb{Z}_2$ is the space of lines going through the origin, also called the projective line.

Now, if you start from A fixed line $L_0$, and let it rotate, then you will get back $L_0$, but only after half a turn (when the orientation of the line is reversed). So you get back to your initial point, but only after a half turn. The space you get is still a circle, just twice as small as $\mathbb{S}_1$.

Formally, you can write an homeomorphism from $\mathbb{S}_1$ to $\mathbb{S}_1 / \mathbb{Z}_2$ by $e^{i \theta} \mapsto \{ e^{i \theta/2}, e^{i (\theta/2+\pi)} \}$.

More generally, $\mathbb{S}_1 / \mathbb{Z}_N$ is still topologically a circle for all $N \geq 1$, although you lose the interpretation by lines. I'll let you write an homeomorphism.

Finally, the canonical covering map $\mathbb{S}_1 \mapsto \mathbb{S}_1 / \mathbb{Z}_N$ is $N$-to-one. What this map does is take the circle $\mathbb{S}_1$, and wrap it $N$ times around the smaller circle $\mathbb{S}_1 / \mathbb{Z}_N$ (something you can do with strings or elastics, if you wish).

Answer 2

If you take a set of represantatives for your $S^1/Z_2$ you can chose it as a semi circle, but with one end open and one closed. A homeomorphism to $S^1$ can be given by sending $[z]$ to $z^2$.

If you look at what this is doing geometrically, you stretch your semi-circle along the circle until the open end links up with the closed one.

By the same argument (with $z^n$) the answer to your second question is yes as well (assuming you take the equivalence relation I think, which is: regarding Z_2 as the group of $n$-th roots of unity, acting via multiplication)