Converse to Hurwitz Theorem This is problem 17 from chapter 14 of Papa Rudin.
Suppose we have a region $\Omega$ (open connected subset of the plane) and some $f_n$ that are holomorphic on $\Omega$ and converge to $f$ uniformly on compact subsets of $\Omega$ where $f$ is some one-to-one function.
Fix $K \subset \Omega$ where $K$ is compact. Is is possible for infinitely many of the $f_n$ to be NOT one-to-one when restricted to $K$?
Thanks in advance!
 A: Sketch of the proof:
the result is correct on compacts (there are non-injective sequences but only on the full open set, once you fix a compact, they become eventually injective there) - the idea is to consider first the case $K$ is the closure of a Jordan domain and assuming (passing to subsequences etc) $f_n(z_n)=f_n(w_n)=a_n \to a$ where $z_n \to y, w_n \to w$, first $f(y)=f(w)$ hence $y=w$ and since $f(w)=a$ consider the winding number of $f-a$ around a small circle centered at $w$ (which is $1$) so the same is eventually true for $f_n-a$
But now choosing a small circle of radius $\epsilon$ around $w$ and $|f(\zeta)-a| \ge 2\delta>0, |\zeta-w|=\epsilon$, by uniform convergence $|f_n(\zeta)-a| \ge \delta>0, n \ge n(\delta), |\zeta-w|=\epsilon$ there, so choosing such an $n$ for which $|a_n-a| < \delta/2, |z_n-w|, |w_n-w| <\epsilon$ we can apply Rouche and $f_n-a_n, f_n-a$ have the same number of zeroes in the disc of center $w$ and radius $\epsilon$ hence $f_n-a$ has winding number at least $2$ and that is a contradiction.
To get from a general compact to one as above, "encircle" it by a nice Jordan curve $J$ that stays in $\Omega$, so it is enough to show that $f_n$ is eventually injective on the interior of $J$
For example $(1-z)^2$ is injective on the open unit disc and $(1-1/n)-z)^2 \to (1-z)^2$ uniformly on compacts but it takes the same value when $z+w=2-2/n$ and that clearly can happen for $|w|,|z|<1$ but for any compact within the unit disc, that relation cannot be fulfilled from some $n$ on
