TITLE: Uniform limit of injective analytic functions is injective
QUESTION [17 upvotes]: I'm stuck on the following problem:

Let $f_n$ be a sequence of injective analytic functions on the unit disc $D$ such that $f_n$ converges uniformly to $f$ on compact subsets of $D$.  Show that $f$ is either injective or constant.

Already $f$ is analytic as an almost uniform limit.  If $f'$ is identically zero, then $f$ is constant.  Otherwise $f'$ has isolated zeroes.  If $c$ is such an isolated zero, then $f_n'$ converges uniformly to $f'$ on a small disc near $c$.  By Hurwitz's theorem, for almost all $n$ the functions $f_n'$ should also have a zero in that small disc.  
But an analytic function is locally injective at a point if and only if its derivative at the point is nonzero.  And injective implies locally injective.  So the $f_n'$ shouldn't have any zeros in $D$.
So, we can at least conclude that $f'$ is never zero on $D$.  This shows that $f$ is locally injective.  Is there an easy way to conclude injectivity from local injectivity here?

REPLY [8 votes]: Uniform limit of injective analytic functions is injective, if this limit IS NOT CONSTANT. 
Proof. Suppose that $f(z)=a$ at two points $z_1,z_2$. Surround these two points by a Jordan curve $\gamma$ such that $f(z)\neq a$  on $\gamma$. Then according to the argument principle
$$\frac{1}{2\pi i}\int\frac{f'(z)dz}{f(z)-a}\geq 2.$$
(The right hand side equals to the number of solutions of $f(z)=a$ inside $\gamma$,
counting multiplicity). On the other hand, $f_n(z)\neq a$ on $\gamma$ when $n$ is sufficiently large, and the similar integral with $f_n$ instead of $f$ is
$\leq 1$, because $f_n$ are injective. But this is a contradiction because
$f_n\to f$ uniformly, so integral with $f_n$ must converge to the integral with $f$.