Prove that the mapping $\psi(x)=x/x_{n+1}$ maps the hemisphere $H\colon=S^n\cap[x_{n+1} < 0]$ onto $\R^n$. Is $\psi:H\rar\R^n$ conformal? Determine the inverse of $\psi$.
The case $n=2$ suffices. The matrices of $D\psi(x)$ and $D\psi(x)^*D\psi(x)$ are given by:
$$
\frac1{x_3}
\left(\begin{array}{ccc}
1&0&-\frac{x_1}{x_3}\\
0&1&-\frac{x_2}{x_3}
\end{array}\right)
\quad\mbox{and}\quad
\frac1{x_3^2}
\left(\begin{array}{ccc}
1&0&-\frac{x_1}{x_3}\\
0&1&-\frac{x_2}{x_3}\\
-\frac{x_1}{x_3}&-\frac{x_2}{x_3}&\frac{x_1^2+x_2^2}{x_3^2}
\end{array}\right)
$$
Thus for $x\in S^2$ and $\la u,x\ra=0$ we obtain for $\la D\psi^*D\psi(x)u,u\ra$ (without the factor $x_3^{-2}$):
\begin{eqnarray*}
&&u_1^2+u_2^2
-2\frac{x_1u_1+x_2u_2}{x_3}u_3
+\frac{x_1^2+x_2^2}{x_3^2}u_3^2\\
&=&u_1^2+u_2^2
+2u_3^2
+\frac{1-x_3^2}{x_3^2}u_3^2\\
&=&u_1^2+u_2^2+u_3^2+\frac{1}{x_3^2}u_3^2
\end{eqnarray*}
i.e. $\psi$ is not conformal!
From $y_j=x_j/x_{n+1}$, $x_{n+1} < 0$ and $\sum x_j^2=1$ we infer that
$$
1-x_{n+1}^2=\sum x_j^2=x_{n+1}^2\sum y_j^2
\quad\mbox{i.e.}\quad
x_{n+1}=-\frac1{1+\Vert y\Vert^2}
$$
and for $j=1,\ldots,n$: $x_j=y_jx_{n+1}$.
