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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.

1 Answers 1

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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: $f=z$, $f_n=z+(z^2/n)$.

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    How does that help?2011-04-14
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    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.2011-04-15
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    Thanks for the clarification. I was thinking along the same lines, but I couldn't make this argument work. That's why I asked. Nevertheless I guess you're on the right track.2011-04-15
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    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).2011-04-15
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    Jose27, thank you for that last example of $f_{n}$. I believe that is a counter-example to the problem.2011-04-15
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    pel: Yes, I believe so too, and, just for the record, my argument doesn't break down because one of the zeroes of $f_n$ has to tend to infinity as $n\rightarrow \infty$, so maybe at least a partial converse is true (in compact subsets with simple zeroes).2011-04-15