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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!

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  • $\begingroup$ There are unicoherent, one-dimensional planar continua which separate the plane. Take a circle with an arc spiraling closer and closer around it. In the paper "A Survey on Unicoherence" by Garcia-Maynez and Illanes, planar continua which are open-unicoherent but not unicoherent, and unicoherent but not open-unicoherent, are exhibited. Still don't have a counterexample or proof in the hereditarily unicoherent case. $\endgroup$ – John Samples Jun 25 '17 at 3:00
  • $\begingroup$ In Q2 you've defined the HEREDITARILY unicoherent continuum instead of simply unicoherent. $\endgroup$ – Wlod AA Jun 25 '17 at 14:27
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    $\begingroup$ Oops, you're right! Fixed. $\endgroup$ – John Samples Jul 1 '17 at 23:56
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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".

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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.

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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').

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  • $\begingroup$ Re (3), attaching an arc to the pseudo-circle is decomposable, hereditarily unicoherent and separates the plane. A hereditarily decomposable, hereditarily unicoherent continuum is called a $\lambda$-dendroid, and Cook proved that these are tree-like: eudml.org/doc/214232 $\endgroup$ – John Samples Jun 21 '18 at 13:02
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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.

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