Injective holomorphic maps have no critical points, so their inverses are holomorphic.Suggest a slogan
Let \(U,V\subseteq\CC\) be open and let \(f\colon U\to V\) be holomorphic and injective. Then \(f'(z)\neq0\) for every \(z\in U\), and the inverse \(f^{-1}\colon f(U)\to U\) is holomorphic. In particular, the inverse of a conformal map is conformal.
A nowhere-vanishing derivative does not imply injectivity: \(f(z)=e^z\) satisfies \(f'(z)=e^z\neq0\) on \(\CC\), and \(f(0)=f(2\pi i)\).
See [@SS03, Chapter 8, Proposition 1.1].