Proof of Theorem H1

 Theorem H1 If tεH(2) is direct, then t(z) = κ(z-c)/(c*z-1), If tεH(2) is indirect, then t(z) = (κ(z-c)/(c*z-1))*. where |c| < 1 and |κ| = 1 in each case. 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.