Let \(\psi_a\) be a Blaschke factor and use the Schwarz lemma to prove that \(\psi_a \circ \psi_a = \id_\DD\).
Solution
First, \(\psi_a\) maps \(\DD\to \DD\). Fix \(z\in S^1\), then \[ \abs{\psi_a(z)} \da \abs{a-z\over 1-\bar{a} z} &= \abs{a-z\over 1-\bar{a} z} \cdot \inverseof{\abs{\bar z}} \\ &= \abs{a-z\over \bar{a} - \bar{z}} \\ &= \abs{a-z \over \bar{a-z} } \\ &\da \abs{w \over \bar w } \\ &= \abs{w \over \abs{w}^2/w }\\ &= \abs{w^2 \over \abs{w}^2} \\ &= 1 .\] \(\psi_a\) is holomorphic on \(\DD\) since it has a simple pole at \(z=1/\bar{a}\), but \(\abs{a}<1\) implies \(\abs{1/\bar{a}} > 1\). Let \(g\da \psi_a\circ \psi_a\), then \(g(0) = 0\) so Schwarz applies. Since \(g(a) = a\) with \(a\neq 0\), the 2nd part of Schwarz also applies since \(\abs{g(a)} = \abs{a}\) and \(g(z) = \lambda z\) is a rotation. Since \(a = g(a) = \lambda a\), this forces \(\lambda = 1\), so \(g\) is the identity.