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!
Best Answer
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".