TITLE: $f_{n}$ converges uniformly on compact sets to a 1-1 function implies $f_{n}$ 1-1?
QUESTION [6 upvotes]: Say we have a region $D$ and a sequence of functions $f_{n}$ holomorphic in $D$, which converges uniformly on compact sets to a one-to-one function $f$.
Can we say that for each compact set $K \subset D$ there is a number $N(K)$ such that $f_{n}$ is one-to-one for all $n >N(K)$?
Thank you for any help or suggestions.

REPLY [4 votes]: How about using Hurwitz theorem, this gives you a ball around each point in $K$ in which $f_n$ and $f$ have the same zeroes, now use compactness of $K$. 
Edit: False if global injectivity of the $f_n$ is required: $f=z$, $f_n=z+(z^2/n)$.