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$S^1=\{z\in\mathbb{C}\mid |z|=1\}$, let $w\sim z$ iff $w=z\vee w=-z$ (identifying antipodal points). Prove $S^1/\sim$ is homeomorphic to $S^1$. Which function should be used to construct a homeomorphism? I am not good at analysis. Thanks!

Kaa1el
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2 Answers2

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$f:S^1\to S^1,\ e^{it}\mapsto e^{2it}$, where $t\in[0,2\pi]$, is the function you should use. Note that this function itself is continuous because it is induced by the map $t\mapsto e^{2it}$ which gives the same value for $t=0$ and $t=2\pi$. Once seen that it is well-defined, you can of course write it as $f(z)=z^2$.

Stefan Hamcke
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Hint: Use $z\mapsto z^2$. $\,\,\!\!\,\,\!\!\,\,\!\!$

Berci
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  • Then, it is obvious that it is a homeomorphism, right? – Lam18373 Oct 02 '20 at 08:25
  • It's a quotient map $S^1\to S^1$, and it induces a homeomorphism $S^1/\sim\to S^1$. – Berci Oct 02 '20 at 10:20
  • Okay, so if i want to show a homeomorphism between $S^1$ and $RP^1$ , can it be proved directly? Or i must use the above homeomorphism? Then how can i prove that $RP^1$ is homeomorphic to $S^1/{\sim}$ ? – Lam18373 Oct 02 '20 at 12:12
  • It depends how you define $\Bbb RP^1$. – Berci Oct 02 '20 at 12:22
  • $RP^1=R^2{0} /{\sim}$ where ${\sim}$ is defined: $x{\simy}$ iff $y=\lambda x$ and $\lambda\in R\setminus{0}$ (the multiplicative group of nonzero scalars in $R$ ) – Lam18373 Oct 02 '20 at 12:40