# homeomorphic topology of quotient space of $S^1$

I got stuck on the problem about quotient space from General Topology of Stephen Williard. Here is the problem:

Let $\sim$ be the equivalence relation $x \sim y$ iff $x$ and $y$ are diametrically opposite, on $S^1$. Which topology is the quotient space $S^1/\sim$ homeomorphic to?

I tried to build a continuous function $S^1$ such that 2 points which are diametrically have the same images, but I couldn't find it. For each point in $S^1$, we can write it as $(\cos(\phi), \sin(\phi))$, then what is the function which satisfies the previous requirement. Can anyone help me with this? I really appreciate.

• What does diametrically opposite mean? – user99914 Oct 12 '15 at 2:28
• Have you seen projective spaces? en.wikipedia.org/wiki/Projective_space – Prahlad Vaidyanathan Oct 12 '15 at 2:28
• Hint: Consider the map $f:S^1\rightarrow S^1$ (thinking of $S^1\subseteq \mathbb{C}$) with $f(z) = z^2$. – Jason DeVito Oct 12 '15 at 2:33
• @PrahladVaidyanathan: Oh, I haven't heard about that space in topology. Let me check, thanks a lot – le duc quang Oct 12 '15 at 3:26
• @leducquang: Yes, your description of $f$ is the same as mine, but it's harder to see that it transforms diametrically opposed points to the same point. In your notation, you need to check that $f(\cos (x + \pi), \sin(x+\pi)) = f(\cos x, \sin x)$, whereas in mine, you need to check that $f(-z) = f(z)$. – Jason DeVito Oct 12 '15 at 3:39

Take a rubber band; that’s your $S^1$. Now fold it into a figure eight: 8. Finally, fold the $8$ about its horizontal midline to get a double circle. Thus, you’ve gone from O to 8 to doubled o. Check that this brings diametrically opposite points of the original O together: they’re simply on different copies of the o.
• @leducquang: You’re basically going around the circle twice as fast: the point whose polar coordinates are $\langle 1,\theta\rangle$ gets sent to the point whose polar coordinates are $\langle 1,2\theta\rangle$. – Brian M. Scott Oct 12 '15 at 3:28
The space you will get is $\mathbb{RP}^1$, the one dimensional real projective space which is by definition of all lines through the origin in $\mathbb{R}^2$.