Problem 4. Suppose that \(f : D _ { 1 } ( 0 ) \to \mathbb { C }\) is a one-to-one holomorphic function with \(\Omega = f \left( D _ { 1 } ( 0 ) \right)\) Let \(g : D _ { 1 } ( 0 ) \to \Omega\) be another holomorphic function with \(g ( 0 ) = f ( 0 )\) . Show that for each \(0 \leq r < 1\) \(g \left( D _ { r } ( 0 ) \right) \subset f \left( D _ { r } ( 0 ) \right)\) .
Solution
1. \(h\da f^{-1}\circ g\) is a holomorphic self-map of \(D_1(0)\) with \(h(0)=0\).
An injective holomorphic map has nonvanishing derivative and open image, so \(f^{-1}\colon\Omega\to D_1(0)\) is holomorphic. Then \(h\) is holomorphic, and \(h(0)=f^{-1}(f(0))=0\).
2. \(\abs{h(z)}\le\abs z\) on \(D_1(0)\).
This is the Schwarz lemma applied to \(h\), using step 1.
3. Q.E.D.
For \(\abs z<r\), step 2 gives \(h(z)\in D_r(0)\), so \(g(z)=f(h(z))\in f(D_r(0))\).