Timeline for Can every orientation preserving homeomorphism of a manifold isotoped to be identity on a locally flat embedded disk?
Current License: CC BY-SA 4.0
5 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Sep 18 at 12:22 | comment | added | Maxim Prasolov | Yes. Here is a sketch. By composing $f$ with a stable homeomorphism we can assume that $f$ has a fixed point. The image of some ball $B_0$ around this point lies in a bigger ball $B_1$. By the annulus conjecture the closure of $B_1\setminus f(B_0)$ is homeomorphic to $S^{m-1}\times[0;1].$ Using this and (pre)composing $f$ with stable homeomorphisms we can assume that $f$ maps $B_1$ into itself bijectively. One can construct a homeomorphism $g$ of $M$ such that $f=g$ on $B_1$ and $g=id$ outside some neighborhood of $B_1$ thus $g$ is stable. $fg^{-1}=id$ on $B_1$ thus stable. So $f$ is stable. | |
Sep 16 at 20:47 | comment | added | Cihan | @MaximPrasolov I have only seen the stable homeomorphism conjecture stated for Euclidean spaces. Does the version for a general $M$ follow from that? | |
Sep 15 at 16:26 | comment | added | Maxim Prasolov | Yes, it is true. It follows from the stable homeomorphism conjecture (which is proved). It states that $f$ can be decomposed into homeomorphisms each of which is the identity on some open non-empty subset. By an isotopy we can make them all to be the identity on the same open subset. The new composition is $F$, and $\Phi$ is an embedding of a ball into this open subset. | |
Sep 15 at 14:53 | history | edited | Cihan | CC BY-SA 4.0 |
added 124 characters in body
|
Sep 15 at 10:53 | history | asked | Cihan | CC BY-SA 4.0 |