Proof
The key is the Algebraic Inversion Theorem from inversive geometry.
This states that, if α and β are inverse with respect to C and tεI(2),
then t(α) and t(β) are inverse with respect to t(C).
If tεH(2), t(C) = C, so
t(α) and t(β) are also inverse with respect to C
Suppose first that tεH(2) is direct.
If t(0) = 0, then, since 0 and ∞ are inverse with respect to
C,
we must
also have t(∞) = ∞.
Now t is inversive and direct, so tεM(2) (Theorem I1).
As t(0) = 0 and t(∞) = ∞, t must have the form t(z) = κz.
As t maps C to C, and,as 1εC, t(1)εC, so |t(1)| = 1.
Thus 1 = |t(1)| = |κ|, so t is of the required form, with c = 0.
If t(0) ≠ 0, then t(c) = 0, for some non-zero c.
As t maps D to D, cεD, so |c| < 1.
The inverse of c with respect to C is 1/c*, so, as t(c) = 0, t(1/c*) = ∞.
Since t is in M(2), it must have the form
t(z) = λ(z-c)/(z-1/c*)
= κ(z-c)/(c*z-1), where κ = λc*.
As t maps C to C, t(1)εC,
so |t(1)| = 1.
Thus, we have 1 = |t(1)| = |κ||1-c|/|c*-1|.
But |c*-1| = |(c*-1)*| = |c-1| = |1-c|,
so |κ| = 1, and t has the required form.
Now suppose that t is indirect.
Then u=r0ot is direct and is in H(2) as t and r0 are.
By the first part u has the form u(z) = κ(z-c)/(c*z-1), with |κ| = 1 and |c| < 1.
Then t = r0-1ou = r0ou has the required form.