As we know the inverse Cayley map can be expressed as $f(z)=i\frac{z+1}{1-z}$, i.e. a biholomorphism from the complex unit disk to the upper half complex plane. I have algebraically rewritten this map to a $\mathbb{R}^2$ correspondence, and I find that the map $$\frac{1}{\left(1-x\right)^2 +y^2}(-2y,1-x^2-y^2)$$ would be a real analytic bijective from the unit disk to the upper half plane in $\mathbb{R}^2$. Then, to find the inverse of this map I have done a similar thing by algebraically rewriting the Cayley map as a $\mathbb{R}^2$ mapping. Then I get $$\frac{1}{x^2 +(y+1)^2} (x^2+y^2-1,-2x).$$ But when I try to compose these two maps I don't get the identity map. I can't find why this does not work, I must have done something wrong?
I have also tried finding the inverse using elementary algebra but without luck. Does anyone have an idea?
To check your calculations: if we denote $$ X=-\frac{2y}{(1-x)^2+y^2},\quad Y=\frac{1-x^2-y^2}{(1-x)^2+y^2} $$ then we get with some algebra \begin{align} (Y+1)^2+X^2&=\frac{4((1-x)^2+y^2)}{((1-x)^2+y^2)^2}=\frac{4}{(1-x)^2+y^2},\\ X^2+Y^2-1&=\frac{4x((1-x)^2+y^2)}{((1-x)^2+y^2)^2}= \frac{4x}{(1-x)^2+y^2},\\ -2X&=\frac{4y}{(1-x)^2+y^2} \end{align} and everything combines just fine to end up with $(x,y)$.