# $f_{n}$ converges uniformly on compact sets to a 1-1 function implies $f_{n}$ 1-1?

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.

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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: $f=z$, $f_n=z+(z^2/n)$.
I was thinking along these lines: Pick any $y\in \mathbb{C}$ and define $g_n(z)=f_n(z)-y$ and $g(z)=f(z)-y$ then if $y\in F(D)$ we pick a small disk around this preimage such that $g_n$ has the same number of zeroes in this disk. Add disks that are a positive distance away from the preimage and cover $K$, apply Hurwitz again in this disk and we get, after obtaining a finite subcover, that the only preimages of $y$ under $f_n$ are in the first disk. Now I notice that this would be enough if $g$ had a zero of order 1 (when we let $g_n,g$ vary with $y$), which need not happen. – Jose27 Apr 15 '11 at 0:38
Theo: Since I've got you attention, what about $f=z$, $f_n=z+(z^2/n)$, it satisfies the hypothesis in $D=\mathbb{C}$ (at least as far as I can tell), but clearly $f_n$ is not injective (even in certain compact subsets whch makes me doubt my argument). – Jose27 Apr 15 '11 at 0:49
Jose27, thank you for that last example of $f_{n}$. I believe that is a counter-example to the problem. – pel Apr 15 '11 at 2:01