Complements of Simply Connected Subsets of the Plane this is my first question here!  Hopefully it is appropriate.  Let $\mathbb{A}$ be the punctured plane, i.e. the 'standard' annulus.  For compact, connected subsets of the plane (planar continua) $X \subset \mathbb{R}^2$, I know it's not necessarily true that $\mathbb{R}^2 \setminus X \simeq \mathbb{A}$ (homeomorphism), for example taking $X$ as the Warsaw Circle.
I was wondering what sort of additional assumptions we can make so that, in fact, $\mathbb{R}^2 \setminus X \simeq \mathbb{A}$.
1) Is it good enough that $\mathbb{R}^2 \setminus X$ be connected?  This seems true using some classical separation arguments in the sphere. [EDIT: This is true, as shown in an answer below]
2) Is it good enough that $X$ be unicoherent, i.e. for each pair of compact, connected subsets $A, B \subset X$ with $A \cup B = X$, their intersection is connected? [EDIT: See my comment below.  Unicoherence is not sufficient]
3) What if $X$ is hereditarily unicoherent, i.e. all its closed subsets are unicoherent?
It seems to me that "open-unicoherence" where the closedness of the sets $A, B$ is replaced by openness should be enough by Cech Homology, but I was unable to find any results on "open-unicoherence" except in the locally connected setting, which is too permissive for what I'm looking at (dendroids).  I suppose a side-question that would be relevant for me is whether hereditary unicoherence implies (hereditary) open-unicoherence.
4) What conditions on a (hereditarily) unicoherent planar continuum are sufficient for it to be (hereditarily) open-unicoherent and vice-versa when not necessarily in the locally connected case (this case has been heavily explored)
So a more specific question would be a reference request for either a proof or counterexample when $X$ is a planar dendroid, i.e. a path-connected and hereditarily unicoherent planar continuum.
EDIT: The question concerning dendroids is known.  It is more strongly known that planar tree-like continua do not separate the plane (but I can't find a reference; anyone know of one?), and since dendroids are tree-like we can apply the answer to #1.  In Kuratowski Topology II, p. 506 Thm. 4 it is known that if $\mathbb{R}^2 \setminus X$ is connected, then $X$ is unicoherent.  The converse is not given, but neither is a counterexample.
So, still stuck on questions 3 and 4, and a reference for the fact that planar tree-like continua don't separate the plane.
Thanks in advance!
 A: Moore's theorem says that if $\sim$ is an equivalence relation on $\mathbb{S}^2$ such that any equivalence class is closed connected and has connected complement then the quotient space $\mathbb{S}^2/\sim$ is homeomorphic to $\mathbb{S}^2$.
In particular the answer to your first question is "yes".
A: After a year the question may be "cooled"; on the other hand good questions never cool off.
Your question concerns the relationship between continua $X \subset \mathbb{R}^2$ amd their complements $CX = \mathbb{R}^2 -X$.
Shape theory provides an answer. We have $CX \approx CY$ iff $X$ and $Y$ have the same shape. There are countably many shapes of continua $X \subset \mathbb{R}^2$: These are represented by $X_0$ = one-point-space, $X_n =$ wedge of $n$ circles and $X_\infty =$ Hawaiin earring. Therefore $CX \approx \mathbb{A}$ iff $X$ has trivial shape (i.e. the shape of $X_0$).
In particular, $CX \approx \mathbb{A}$ iff $CX$ is connected.
A: Here are some thoughts concerning tree-like continua and dendroids. Let me begin by quoting some facts, although most of this will be known to you.
1) From
Cook, Howard. "Tree-likeness of dendroids and λ-dendroids." Fundamenta Mathematicae 68 (1970): 19-22
we know that hereditarily indecomposable continua are tree-like.
2) $X$ being tree-like means that every open cover of $X$ can be refined by a finite open cover having as nerve a tree. This shows that the Cech expansion of $X$ (which is an inverse system in the homotopy category) has a cofinal subsystem consisting of contractible spaces. Therefore $X$ has trivial shape.
3) We conclude that the complement of a tree-like plane continuum is an annulus.
A: 1) is true by the answer above.  Thus in the other cases it's sufficient to show that they don't separate the plane.
2) There is a unicoherent continuum, namely a circle with an arc spiraling closer and closer to it, that separates the plane.
3) There is a counterexample, the pseudo-circle (careful, googling will reveal several things called a 'pseudo-circle').  Here, the pseudo-circle refers to a circle-like continuum written as a union of circular chains $C_i$ such that $C_{i+1}$ is crooked in $C_i$ in the same sense as the pseudo-arc.  This continuum is hereditarily indecomposable.  Thus the new question would be whether there are decomposable hereditarily unicoherent continua which separate the plane.
4) Seems to be largely open, there is a thesis and a couple of papers by Ganea that I'm going to read which might shed light on this.
A reference for the fact that tree-like continua don't separate the plane is in 'Results and Problems in Fixed Point Theory for Tree-Like Continua', theorem 1.5, by Roman Manka (though not first proven here, it seemed to be 'folklore').
