- A holomorphic mapping \(f : U \to V\) is a local bijection on U if for every \(z \in U\) there exists an open disc \(D \subset U\) centered at z, so that \(f : D \to f ( D )\) is a bijection.
Prove that a holomorphic map \(f : U \to V\) is a local bijection on U if and only if \(f ^ { \prime } ( z ) \neq 0\) for all \(z \in U\)
[Hint: Use Rouch´e’s theorem as in the proof of Proposition 1.1.]
Solution
Suppose first that \(f'(z_0)\ne0\). Write \[ f(z)-f(z_0)=(z-z_0)g(z), \] where \(g\) is holomorphic near \(z_0\) and \(g(z_0)=f'(z_0)\ne0\). Choose a small closed disc \(\overline D\) centered at \(z_0\) on which \(g\) has no zeros and \(f(z)\ne f(z_0)\) on \(\partial D\). Let \[ m=\min_{\partial D}|f(z)-f(z_0)|>0. \] For \(w\) with \(|w-f(z_0)|<m\), Rouché’s theorem shows that \[ f(z)-w \] has the same number of zeros in \(D\) as \(f(z)-f(z_0)\), namely one, counted with multiplicity. Hence every such \(w\) has a unique preimage in \(D\). After shrinking \(D\) if necessary, \(f(D)\) is open and \(f:D\to f(D)\) is bijective. Thus \(f\) is a local bijection at \(z_0\).
Conversely, suppose \(f\) is locally injective at \(z_0\). If \(f'(z_0)=0\), then for some \(m\ge2\), \[ f(z)-f(z_0)=(z-z_0)^m g(z),\qquad g(z_0)\ne0. \] On a sufficiently small disc, \(g\) has a holomorphic \(m\)th root \(h\) with \(h(z_0)\ne0\). Then \[ f(z)-f(z_0)=\bigl((z-z_0)h(z)\bigr)^m. \] The map \(\phi(z)=(z-z_0)h(z)\) has \(\phi'(z_0)=h(z_0)\ne0\), so it maps a neighborhood of \(z_0\) biholomorphically onto a neighborhood of \(0\). But \(w\mapsto w^m\) is not injective on any neighborhood of \(0\) when \(m\ge2\). Hence \(f\) cannot be locally injective, a contradiction. Therefore \(f'(z_0)\ne0\).
Since \(z_0\) was arbitrary, \(f\) is a local bijection on \(U\) exactly when \(f'\) never vanishes.