Let $f$ be a transcendental entire function, and let $U,Vsubsetmathbb{C}$be disjoint simply-connected domains. Must one of $f^{-1}(U)$ and $f^{-1}(V)$be disconnected? In 1970, Baker implicitly gave a positive answer to this question, in orderto prove that a transcendental entire function cannot have two disjointcompletely invariant domains. (A domain $Usubset mathbb{C}$ is completelyinvariant under $f$ if $f^{-1}(U)=U$.) It was recently observed by Julien Duval that there is a flaw in Baker'sargument (which has also been used in later generalisations and extensions ofBaker's result). We show that the answer to the above question is negative; sothis flaw cannot be repaired. Indeed, there is a transcendental entire function$f$ for which there are infinitely many pairwise disjoint simply-connecteddomains $(U_i)_{i=1}^{infty}$, such that each $f^{-1}(U_i)$ is connected. Infact, this is the case for $f(z) = e^z+z$. We also answer a long-standingquestion of Eremenko by giving an example of a transcendental entire function,with infinitely many poles, which has the same property. Furthermore, we show that there exists a function $f$ with the aboveproperties such that additionally the set of singular values $S(f)$ is bounded;in other words, $f$ belongs to the Eremenko-Lyubich class. On the other hand,if $S(f)$ is finite (or if certain additional hypotheses are imposed), many ofthe original results do hold. For the convenience of the research community, we also include a descriptionof the error in the proof of Baker's paper, and a summary of other papers thatare affected.
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机译：让$ F $是超越整功能，并让$ U，V 子集 mathbb {C} $是不相交的简单连接结构域。 $ F ^必须一个{ - 1}（U）和$ $ F ^ { - 1}（V）$断开？ 1970年，贝克暗示给了一个肯定的回答了这个问题，在orderto证明了一个超越整函数不能有两个disjointcompletely不变域。 （A域$ U 子集 mathbb {C} $是下completelyinvariant $ F $如果$ F ^。{ - 1}（U）= U $），它被最近由于连Duval的观察，有一个缺陷在贝克的sargument（这也被后面的概括和扩展ofBaker的结果使用）。我们表明，回答上述问题是否定的; SOTHIS缺陷无法修复。事实上，有一个超越整函数$ F $为其中有无穷多的两两不相交的简约connecteddomains $（U_i）_ {i = 1} ^ { infty} $，使得每个$ F ^ { - 1}（ U_i）$连接。逸岸，这是$ F（Z）= E ^ Z + Z $的情况。我们还通过提供一个超越整函数的例子，有无限多极，它具有相同的属性回答Eremenko的长期standingquestion。此外，我们表明，存在一个函数$ F $与aboveproperties使得另外的集合$ S（F）$是有界的奇异值的;换言之，$ F $属于Eremenko-Lyubich类。在另一方面，如果$ S（F）$是有限的（或某些额外的假设被征收），许多原来国税发结果确实持有。研究界的方便，我们还包括descriptionof在贝克的论文的证明错误，以及其他文件的摘要thatare影响。
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