  Let $f$ be a transcendental entire function, and let $U,V\subset\mathbb{C}$
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 order
to prove that a transcendental entire function cannot have two disjoint
completely invariant domains. (A domain $U\subset \mathbb{C}$ is completely
invariant under $f$ if $f^{-1}(U)=U$.)
  It was recently observed by Julien Duval that there is a flaw in Baker's
argument (which has also been used in later generalisations and extensions of
Baker's result). We show that the answer to the above question is negative; so
this flaw cannot be repaired. Indeed, for the function $f(z)= e^z+z$, there is
a collection of infinitely many pairwise disjoint simply-connected domains,
each with connected preimage. We also answer a long-standing question 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 above
properties 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 of
the original results do hold.
  For the convenience of the research community, we also include a description
of the error in the proof of Baker's paper, and a summary of other papers that
are affected.
