Construct a map from unit disk to upper half-plane I want to construct this map in high-dimensional case. Let $D=\{x \in \mathbb{R}^n:|x|^2<1\}$,and $H=\{u\in\mathbb{R}^n:u^n>0\}$. Well, it is quite clear when $n=2$, but I find it is hard for me to generalize it. Could any one help me? Thanks a lot! 
 A: Theoretic considerations
I guess you want to obtain a conformal map. In the case of $n=2$ this would be a Möbius transformation. One observation which generalizes nicely to higher dimensions is the fact that every Möbius transformation can be expressed as a concatenation of an even number of inversions in circles. So in the case of arbitrary dimension, you can still perform an even number of inversions in hyperspheres in order to obtain the higher-dimensional equivalent of a Möbius transformation.
The decomposition of Möbius transformations into inversions is not unique. With a bit of trial and error, I found out that an inversion in the circle with midpoint $(0, 1)$ and radius $\sqrt2$ will map the unit disk to the lower half-plane. So the second inversion in that case would be inversion in the $x_1$ axis, which can be interpreted as a circle of infinite radius.
Generalizing this, I suggest you invert in a hypersphere with midpoint $(0,0,\dots,0,1)$ and radius $\sqrt 2$, then reflect in the hyperplane $x_n=0$.
An explicit formula
The inversion in the sphere followed by the reflection can be written like this:
$$
\begin{pmatrix}x_1\\x_2\\\vdots\\x_{n-1}\\x_n\end{pmatrix} \mapsto
\frac{2}{1-2x_n+\lVert x\rVert^2}
\begin{pmatrix}x_1\\x_2\\\vdots\\x_{n-1}\\1-x_n\end{pmatrix} -
\begin{pmatrix}0\\0\\\vdots\\0\\1\end{pmatrix}
=:
\begin{pmatrix}u_1\\u_2\\\vdots\\u_{n-1}\\u_n\end{pmatrix}
$$
Checking the result
To show that the map has the desired properties, let's have a closer look at the last coordinate of the image vector.
$$
u_n =
\frac{2}{1-2x_n+\lVert x\rVert^2}(1-x_n)-1 =
\frac{2-2x_n-1+2x_n-\lVert x\rVert^2}{1-2x_n+\lVert x\rVert^2} =
\frac{1-\lVert x\rVert^2}{1-2x_n+\lVert x\rVert^2}
$$
As you can see, if $\lVert x\rVert = 1$ then $u_n=0$ so the unit hypershpere gets mapped to the $u_n=0$ plane.
If $\lVert x\rVert<1$, then the resulting coordinates are positive. This is a bit more difficult to see, but since $\lVert x\rVert=1$ is the only zero of the nominator, and $x=(0,0,\dots,1)$ is the only position where the denominator is zero, you can be convinced that the sign of the last coordinate stays the same for the whole interior of the unit ball. So check it for $x=0$, where you obtain $u_n=1$, and you should (hopefully) be convinced.
Non-uniqueness
The above result is not unique. It is characterized by fixing the intersection of the unit hypersphere with the $x_n=0$ hyperplane. You could choose an arbitrary circle in the $x_n=0$ plane and modify the map so that these points are fixed points. Simply translate and scale everything up front, then back afterwards.
