Show that the only holomorphic map \(f:\DD\to \DD\) two distinct fixed points \(a\neq b\) is the identity.
Solution
Note that without loss of generality we can assume \(a=0\) so \(f(0) = 0\) and \(b\neq 0\). If not, if \(a,b\neq 0\) then let \(F\da \psi_a \circ f \circ \psi_a\), then \(F(0) = 0\) and \(F(b') = b'\) for \(b\da \psi_a(b)\).
Since \(f(0) = 0\), Schwarz applies, so \(\abs{f(z)} = \abs{z}\) with equality attained because \(\abs{f(b)} = \abs{b}\), and \(f(z) = \lambda z\) must be a rotation. Since \(b = f(b) = \lambda b\), we have \(\lambda = 1\).