= Solution
If all three points are finite, define the <Möbius transformation>
$$
\boxed{g(z)=\frac{\beta-\gamma}{\beta-\alpha}\,\frac{z-\alpha}{z-\gamma}}.
$$
Its numerator vanishes at $\alpha$, its pole is at $\gamma$, and substitution of $\beta$ gives $1$. The three points are distinct, so the numerator and denominator are not proportional and the associated $2\times2$ <matrix> has nonzero <determinant>.
If one point is infinite, the limiting formulas are
$$
\boxed{\begin{array}{c|c}
\text{infinite point}&g(z)\\\hline
\alpha=\infty&(\beta-\gamma)/(z-\gamma)\\
\beta=\infty&(z-\alpha)/(z-\gamma)\\
\gamma=\infty&(z-\alpha)/(\beta-\alpha)
\end{array}}.
$$
In each case the finite values and the limit at infinity give $g(\alpha)=0$, $g(\beta)=1$, $g(\gamma)=\infty$, and the corresponding <determinant> is nonzero. Distinctness permits at most one of the points to be infinite. Hence \b[every ordered triple of distinct points on the Riemann sphere can be normalized to $(0,1,\infty)$].
Back to article page
