Using Argument Principle to show that a function is univalent I sometimes see sentences like "by the Argument Principle, $f$ is univalent". The way I know the Argument Principle, it has no direct connection to univalent function, and it discusses the connection between $f$'s zeros and poles to the integral of $\frac{f'}{f}$. What am I missing? Is it so obvious I have never seen it explained?
 A: Let $\Omega$ be a domain with smooth boundary. Suppose $f$ is a holomorphic function in a domain containing $\overline{\Omega}$.  Then the following are equivalent: 


*

*$f$  is univalent

*for every $w\in\mathbb C$  the function $f-w$ has at most one zero in $\Omega$ 

*$\displaystyle \frac{1}{2\pi i}\int_{\partial \Omega} \frac{f'(z)}{f(z)-w}\,dz \le 1$ for all $w\in\mathbb C\setminus f(\partial \Omega)$. 


The equivalence of 1 and 2 should be clear. Also, 2 implies 3 by the argument principle. Suppose 2 fails; that is, $f(z_1)=f(z_2)=w$ for some $w\in\mathbb C$. If $w\notin f(\partial \Omega)$, then 3 fails by the argument principle (recall that $f-w$ has no poles). What if $w\in f(\partial \Omega)$? Since $f$ is an open map, there is a neighborhood $V$ of $w$ and disjoint neighborhoods $U_i$ of $z_i$ such that $V\subset f(U_i)$ for $i=1,2$. Since $f(\partial \Omega)$ is a smooth curve, there exists $\zeta\in V\setminus f(\partial \Omega)$. Then 3 fails with $\zeta$ instead of $w$. 
One can relax the smoothness assumptions by interpreting 3 as the change of a continuous branch of $\arg(f-w)$ rather than an actual integral. Then it suffices for $\partial\Omega$ to be a simple closed curve, and for $f$ to extend continuously to $\overline{\Omega}$. (One still  has to avoid $w\in f(\partial\Omega)$ in 3.) If the boundary values of  $f$ trace a simple closed curve once, then $\arg(f-w)$ changes by at most $2\pi$ for every $w\in\mathbb C$, and we conclude that $f$ is univalent. 
