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I want to prove that the stereographic projection:$f:\{\ (x,y):x^{2}+y^{2} = 1 \}\ \rightarrow \mathbb{R}$ where $f$ is defined as: $$ f(x,y) = \frac{x}{1-y}$$ is bijective to the real number. I have already shown that $f$ is one-to-one, but I am struggling on showing this is onto. I also tried to find an inverse for $f$ but I failed. What is the easiest way to show $f$ is bijective?

Arctic Char
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Charlie
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  • What exactly is the domain you're considering? (I can probably guess, but you really ought to be specific about these things.) As for showing that it's bijective, a drawing is probably not a bad place to start. Do you know the standard geometrical idea behind this map? – Arthur Nov 21 '20 at 01:08
  • yes, I do know the domain, and I do know the standard geometrical idea behind this map. To show this is onto, if you try to connect any point on the x-axis with (0,1), you will always pass the circle. But how do I prove this rigorously? – Charlie Nov 21 '20 at 01:57
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    If you know the geometric picture, then a little geometric reasoning can give you the inverse formula. Which would prove surjectivity. – Arthur Nov 21 '20 at 09:53
  • Here is the $\boldsymbol{\text{case in }\Bbb R^3}$:. – Matcha Latte Sep 25 '21 at 06:07

1 Answers1

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Let $\,S^1:=\{x^2+y^2=1\}\subseteq \Bbb R^2$. Consider $\,f :S^1-\{(0,1)\}\to \Bbb R\,$ defined by $\displaystyle \,f(x,y)=\frac x{1-y}$.

Given any $\,a\in\Bbb R$, we search, if exists, $\,(x,y)\in S^1-\{(0,1)\}\,$ such that

$$ f(x,y)=\frac x{1-y}=a. \tag{*}$$

From this we have, by squaring:

$$ \frac{x^2}{(1-y)^2}=\frac{1-y^2}{(1-y)^2}=\frac{1+y}{1-y}=a^2, $$

and hence

$$ y=\frac{a^2-1}{a^2+1}, \quad\text{ from which equation (*) gives }\quad x=\frac{2a}{a^2+1}.$$

As

$$ \left(\frac{2a}{a^2+1}\right)^2 + \left(\frac{a^2-1}{a^2+1}\right)^2=1 \qquad \text{and} \qquad \frac{a^2-1}{a^2+1}\neq 1, $$

and since you have already proved that $\,f\,$ is injective, it follows that there exists $\,f^{-1}\,$, given by

$$ f^{-1}(a)=\left(\frac{2a}{a^2+1},\; \frac{a^2-1}{a^2+1}\right). $$

gpassante
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