Timeline for Homeomorphism and boundary of a complementary component
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19 at 6:37 | answer | added | Benjamin Vejnar | timeline score: 1 | |
| Nov 18 at 20:44 | answer | added | Moishe Kohan | timeline score: 1 | |
| Nov 18 at 17:17 | history | edited | D.S. Lipham | CC BY-SA 4.0 |
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| Nov 18 at 17:13 | comment | added | D.S. Lipham | Thanks Moishe and Benjamin. Every homeomorphism of a continuum $X\subset \mathbb R^2$ must map $\partial X$ to itself, does not require that $X$ is a retract, correct? Also, although my original question has a negative answer, I wonder if the following is still true: Let $B$ be the union of all boundaries of complementary components of $X$. Then for every homeomorphism $h$ of $X$, $h[B]=B$. I will edit my original post to include this. | |
| Nov 18 at 7:42 | comment | added | Benjamin Vejnar | Seems that for graphs it can not hold. Not sure, whether the following is related: If a continuum $K\subseteq \mathbb R^2$ is a retract (of the plane or equivalently an absolute retract), then every homeomorphism $h: K\to K$ maps the boundary of $K$ onto itself (invariance of domain) and also every self homeo of the boundary can be extended to a homeomorphism of $K$. This is of interest for two-dimensional continua. We used this in the paper: J. Dudák, B. Vejnar; The complexity of homeomorphism relations on some classes of compacta with bounded topological dimension, Fund. Math. 263 (2023). | |
| Nov 17 at 18:07 | comment | added | Moishe Kohan | On the other hand, if your compact is the Sierpinski carpet (equivalently, 1-dimensional planar Peano continuum without local cut-points), then the answer is positive. | |
| Nov 17 at 17:52 | comment | added | Moishe Kohan | Of course not, think of graphs with 2 vertices and 4 edges. | |
| Nov 17 at 17:35 | history | asked | D.S. Lipham | CC BY-SA 4.0 |