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Let $\mathbb S^1$ the unit circle centered at the origin and the pole $p=(0,1)$. The stereographic projection is the homeomorphism $\varphi:\mathbb S^1\setminus \{p\}\to \mathbb R^1$. In order to find a formula for $\varphi$, note that the point of the semi-straight line $px$ are of the form $p+t((x,y)-p)$ with $t>0$ and $(x,y)\in \mathbb S^1$. This point is in the line $y=0$ whenever the last coordinate $1+t(y-1)$ is zero. So the $\varphi(x,y)=\frac{x}{1-y}$. I'm having troubles to find a formula of the inverse. It seems just high school geometry, but I couldn't find the inverse.

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Let $x^2+y^2=1$ with $x=\varphi(1-y)$ we find $y=\dfrac{\varphi^2-1}{\varphi^2+1}$.

Edit: with $x=t(1-y)$ and $t^2(1-y)^2+y^2=1$ we find $y=\dfrac{t^2-1}{t^2+1}$. From $x=t(1-y)=t(1-\dfrac{t^2-1}{t^2+1})$ which gives us $\psi:\mathbb{R}\to\mathbb{R}^2$, with $\psi(t)=\left(t(1-\dfrac{t^2-1}{t^2+1}),\dfrac{t^2-1}{t^2+1}\right)$.

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    $\varphi$ is a function from a subset of $\mathbb R^2$, therefore $\varphi (1-y)$ doesn't make sense. – 2017-02-21
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    I meant $\psi:\mathbb{R}\to\mathbb{R}^2$, with $\psi(t)=\left(t(1-\dfrac{t^2-1}{t^2+1}),\dfrac{t^2-1}{t^2+1}\right)$. – 2017-02-21
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    could you please add in your answer how do you get this formula? Thank you very much. – 2017-02-21