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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.

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  • $\begingroup$ What does diametrically opposite mean? $\endgroup$ – user99914 Oct 12 '15 at 2:28
  • $\begingroup$ Have you seen projective spaces? en.wikipedia.org/wiki/Projective_space $\endgroup$ – Prahlad Vaidyanathan Oct 12 '15 at 2:28
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    $\begingroup$ Hint: Consider the map $f:S^1\rightarrow S^1$ (thinking of $S^1\subseteq \mathbb{C}$) with $f(z) = z^2$. $\endgroup$ – Jason DeVito Oct 12 '15 at 2:33
  • $\begingroup$ @PrahladVaidyanathan: Oh, I haven't heard about that space in topology. Let me check, thanks a lot $\endgroup$ – le duc quang Oct 12 '15 at 3:26
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    $\begingroup$ @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 $f(-z) = f(z)$. $\endgroup$ – Jason DeVito Oct 12 '15 at 3:39
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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.

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  • $\begingroup$ Wow, impressive. But I think we need to prove it clearly by a quotient map, right? Can you show what the function is for the transformation that you did? $\endgroup$ – le duc quang Oct 12 '15 at 3:25
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    $\begingroup$ @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 $\langle 1,2\theta\rangle$. $\endgroup$ – Brian M. Scott Oct 12 '15 at 3:28
  • $\begingroup$ Oh, I see it now. Thanks a lot! $\endgroup$ – le duc quang Oct 12 '15 at 4:13
  • $\begingroup$ @leducquang: You’re welcome! $\endgroup$ – Brian M. Scott Oct 12 '15 at 4:17
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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 $\mathbb{R}^2$.

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