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I'm trying to find an explicit bijection from $\{(x,y):(x^2+y^2=1)\}$ and the real numbers $\mathbb R$.

I've found a couple bijection to go from $\mathbb R$ to the unit circle, but none that go the other direction. Any help would be greatly appreciated.

David K
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Amanda
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  • You won't find a continuous bijection, so look for something discontinuous. If you remove one point from the circle it is homeomorphic to $\mathbb R$ (using stereographic projection), you just have to figure out how to deal with that one stray point. – Gregory Grant May 04 '15 at 01:12
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    A bijection already goes both ways. If $f$ is a bijection from $\mathbb R$ to the circle, then $f^{-1}$ is a bijection from the circle to $\mathbb R$. What bijections have you found? – David K May 04 '15 at 01:12
  • So, if a have a bijection from R to the unit circle, I can find its inverse, and that goes the other direction? – Amanda May 04 '15 at 01:14
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    I've found a very common one, where f(t)=((2t/t^2+1), (t^2-1/t^2+1))=(x,y) and another that deals with (cos(2arctan(t)), sin(2arctan(t)))=(x,y) – Amanda May 04 '15 at 01:17
  • @Amanda please see this page about how to properly format mathematics on this site. It is difficult to read your previous comment since it is ambiguous if you mean $\frac{2t}{t^2+1}\dots$ or $\frac{2t}{t^2}+1\dots$ – JMoravitz May 04 '15 at 01:19
  • @JMoravitz, I can't seem to get the formatting to work, but I do mean the first. Each is entirely contained in the fraction. – Amanda May 04 '15 at 01:30
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    It would be OK to add the formulas of your bijections to the original question via editing. When editing the question you get a preview window so it is easier to see how the math will look. I've started you off by formatting the descriptions of the two sets. – David K May 04 '15 at 01:35
  • The bijections you have are each missing one point of the circle (as Gregory Grant's comment implied they would), so you need to patch that hole and then look for the inverse function. – David K May 04 '15 at 01:40

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As commenters said, there is no nice bijection. Conceptually this is the same story as with bijections between closed and open intervals.

There is a natural bijection from the unit circle minus the point $(-1,0)$ to real line: $(x,y)\mapsto y/(x+1)$. Geometrically this is radial projection:

projection

Then you have to create room for $(-1,0)$. Use Hilbert's hotel: on the real line, send $n$ to $n+1$ for $n=0,1,2,3,\dots$. Then $(-1,0)$ can be mapped to $0$.

Community
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