I am currently studying the theory of elementary univalent functions, and am stuck on how to generalise the famous Koebe 1/4 theorem. I have seen the Koebe 1/4 theorem provided in many books, in the form that if $f:\Bbb{D}\rightarrow\Omega$ be a function of the elementary univalent class $S$ (defined as usual). Then $dist(0,\delta\Omega)\in [\frac{1}{4},1]$.
I now want to extend this, as many books do without proof, to the general case where $f:\Omega\rightarrow\Omega'$ is a univalent map, and $z$ is some point in $\Omega$. Then the inequality that follows reads that $$\frac{1}{4}dist(f(z),\delta\Omega')\leq|f'(z)|dist(z,\delta\Omega)\leq4dist(f(z),\delta\Omega')$$
I am not sure how to relate these two inequalities, following the proof for the original Koebe inequality my instinct is to find some function, using our univalent $f$ so that our new function, call it $\phi$, also belongs to $S$, to that we can use Bieberbach's theorem to create boundsin the same way? I just am really unsure on how to go about this proof, and can't find it explicitly in the literature. Any help, or pointing in the right direction, would be really appreciated. Thank you.
Fix $w \in \Omega$; then $g(z)=\frac{f(dist(w,\delta\Omega)z-w)-f(w)}{|f'(w)|dist(w,\delta\Omega)}$ maps the unit disc in an univalent fashion to some domain $W$, while $g \in S$ by our normalizations, hence by the usual Koebe theorem we have $dist(0,\delta W) \ge 1/4$
Noting that $dist(f(w),\delta\Omega') \ge dist(0,\delta W)|f'(w)|dist(w,\delta\Omega)$ we get the RHS inequality in the OP for $w$ arbitrary in $\Omega$:
$|f'(w)|dist(w,\delta\Omega)\leq 4dist(f(w),\delta\Omega')$
But now apply the result above with $f^{-1}$ at the point $f(w)$ (here obviously we must assume $f$ is onto as otherwise, the result is not true since we can take $\Omega$ the unit disc, $f=z$, $\Omega'$ a huge disc containing zero) and the result follows by a simple computation, again under the required assumptions that $f(\Omega)=\Omega'$ since ${f^{-1}}'(f(w))f'(w)=1$