Let \(\Omega\subseteq\CC\) be open. A function \(f\colon\Omega\to\CC\) is univalent if \(f\) is holomorphic and injective.
Let \(\Omega\subseteq\CC\) be open and let \(f\colon\Omega\to\CC\) be univalent. Then \(f'(z)\neq0\) for every \(z\in\Omega\), the image \(f(\Omega)\) is open, and \(f\colon\Omega\to f(\Omega)\) is a biholomorphism. Conversely, every biholomorphism from \(\Omega\) onto an open subset of \(\CC\) is univalent.
Suppose \(f'(z_0)=0\) for some \(z_0\in\Omega\). An injective function is not constant on any disc, so on a disc \(D\subseteq\Omega\) about \(z_0\), \(f-f(z_0)\) has a zero of order $n$ at \(z_0\) with \(n\ge2\): \(f(z)-f(z_0)=(z-z_0)^n g(z)\) with \(g\) holomorphic and \(g(z_0)\neq0\). Shrinking \(D\), \(g\) has no zeros on \(D\), so it has a holomorphic \(n\)-th root \(k\) on \(D\), and \(f(z)-f(z_0)=\varphi(z)^n\) with \(\varphi(z)\coloneqq(z-z_0)k(z)\). Since \(\varphi(z_0)=0\) and \(\varphi'(z_0)=k(z_0)\neq0\), the holomorphic inverse function theorem gives a neighborhood \(V\subseteq D\) of \(z_0\) mapped by \(\varphi\) bijectively onto a disc \(D_\varepsilon(0)\). For \(0<\delta<\varepsilon\), the \(n\ge2\) distinct points \(\varphi^{-1}(\delta e^{2\pi ij/n})\), \(j=0,\ldots,n-1\), all have image \(f(z_0)+\delta^n\) under \(f\), contradicting injectivity. Hence \(f'\) has no zeros.
By the open mapping theorem applied on each connected component of \(\Omega\), \(f(\Omega)\) is open. For \(w_0=f(z_0)\), the holomorphic inverse function theorem gives open neighborhoods \(V\ni z_0\) and \(W\ni w_0\) with \(f\colon V\to W\) bijective and holomorphic inverse; since \(f\) is injective, \(f^{-1}\) agrees with this local inverse on \(W\). Thus \(f^{-1}\colon f(\Omega)\to\Omega\) is holomorphic. The converse holds because a biholomorphism is holomorphic and bijective onto its image.