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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
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Sep 15 at 10:53 history asked Cihan CC BY-SA 4.0