4
$\begingroup$

I was looking into particular cases for the Poincaré-Bendixson theorem and I came across a topological problem about simply connectivity.

If $\gamma$ is a Jordan curve in ${\Bbb S}^2$ then using Jordan-Schoenflies, we have that ${\Bbb S}^2\setminus \gamma = U\sqcup V$ with $U$ and $V$ being simply connected (s.c.). Moreover, as the sphere minus $\gamma$ is homeomorphic to the sphere minus the equator, we also get that $\overline{U}$ and $\overline{V}$ are s.c.

  1. Let $U$ be a s.c. open subset in ${\Bbb S}^2$. We note that $\overline{U}$ need not be s.c. (take an open annulus where you make a transversal cut). On the other hand, the complement $F=S^2\setminus U$ appears to me to be s.c.

  2. ${\rm Int} F$ need not be connected, but a connected component of ${\rm Int} F$ appears to me again to be s.c.

Do the above two claims hold in general? Or have I missed some obvious counter-examples?

$\endgroup$
3
$\begingroup$

$F=S^2\backslash U$ need not be path connected, e.g. $F$ could be homeomorphic to the closed topologists sine curve. However, every path component of $F$ must be simply connected.

By identifying $U$ with the open unit disk (Riemann mapping), you can realize the compact set $F=S^2\backslash U$ as an intersection $\bigcap_{n\in\mathbb{N}}V_n$ where each $V_n$ is homeomorphic to the closed unit disk. This implies that $F$ has trivial shape, i.e. is cell-like.

The path-components of a cell-like continuum don't have to be contractible, e.g. if $F$ is the Knaster buckethandle continuum. However, it is known that every cell-like subset of a 2-dimensional manifold is simply connected. See Corollary 6 of:

H. Fischer, A.Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676.

Various parts this paper could allow you to verify a positive answer to both of your questions without appealing to shape theory. For instance, Lemma 13 is a particularly handy result that would imply that once you know each path-component of $F$ is simply connected, then every component of $int(F)$ is simply connected.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for your comprehensive answer. Exactly what I was hoping for. And nice twist with the sine curve being s.c. but in a somewhat pathological way. I'll look into Fischer and Zastrow. $\endgroup$ – H. H. Rugh Apr 30 '20 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.