A holomorphic mapping \(f: U \rightarrow V\) 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 \(f: D \rightarrow f(D)\) is a bijection. Prove that a holomorphic map \(f: U \rightarrow V\) is a local bijection if and only if \(f^{\prime}(z) \neq 0\) for all \(z \in U\).
- Inverse function theorem: if \(F\in C^1(\RR^n\to \RR^n)\) and \(D_f\) is invertible at \(p\), the \(F\) is invertible in a neighborhood of \(p\), and \(\inverseof{F}\) is \(C^1\).
Solution
\(\impliedby\): Let \(z\in U\) be fixed. Since \(f\) is holomorphic at \(z\) and \(f'(z)\neq 0\), consider \(f(x, y)\) and its Jacobian as a real-valued function: \[ D_f = \matt{u_x}{u_y}{v_x}{v_y} \implies \det(D_f) = u_x v_y - v_x u_y = u_x^2 + v_x^2 = \abs{f'}^2 > 0 ,\] so the derivative matrix is invertible at \(z\). Applying the inverse function theorem yields that \(f\) is a smooth diffeomorphism on some neighborhood \(N\ni p\), and in particular is bijective on \(N\).
\(\not\impliedby\): If \(f'(z) = 0\) for some \(z\), then we claim that \(f\) can not be injective. Equivalently, injectivity of \(f\) implies \(f'\neq 0\). Suppose \(f\) is holomorphic at \(z_0\) but \(f'(z_0)=0\). Write \(h(z) \da f(z) - f(z_0)\), which has a zero \(z_0\) of some order \(k\geq 2\). For a disc \(D\) small enough about \(z_0\) avoiding the other (isolated) zeros of \(h\) and \(f'\), for any \(p\) in a neighborhood of \(z_0\) and contained in \(D\), \[ \int_{\bd D} {f'(\xi) \over f(\xi) - p} \dxi = \size Z(f(z) - p) ,\] using the argument principle and that \((f(\xi) - p)' = f'(\xi)\). But for \(D\) small enough, \(\size Z(f(z) - p) = \size Z(f(z) - f(z_0)) = k\) by Rouché, so there are \(k\) solutions to \(f(z) = p\). Since \((f(z) - p)' \neq 0\) in \(D\), none of these can be repeated roots, so these \(k\) solutions are distinct, forcing \(f\) to be \(k\)-to-one and fail injectivity.
Expanding on the Rouché argument: set \(c \da \inf_{z\in D} \abs{f(z) - w_0}\), then for \(D'\) of radius \(c\), set
\(F(z) \da (f(z) - z_0) - (f(z) - p) = z-p\)
\(G(z) = f(z) - z_0\)
\((F+G)(z) = f(z) - p\)
Then \(F>G\) on \(\bd D'\) will imply \(F, F+G\) have the same number of zeros within \(D'\), and this bound follows from \[ \abs{F(z)} = \abs{z-p} < c \leq \abs{f(z) - p } ,\] where the first inequality is from making the disc small and the second from choosing \(c\) as an inf.