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Let $M$ be a connected (topological) oriented $m$-manifold (say without boundary), and let $\operatorname{Homeo}^+(M)$ be the group of orientation preserving homeomorphisms $M \to M$.

Is it true that for every $f \in \operatorname{Homeo}^+(M)$, there is a locally flat oriented embedding $\Phi \colon \mathbb{D}^m \hookrightarrow M$ and $F \in \operatorname{Homeo}^+(M)$ such that $F$ is isotopic to $f$ and $F \Phi = \Phi$?

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    $\begingroup$ 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. $\endgroup$
    – Maxim Prasolov
    Commented Sep 15 at 16:26
  • $\begingroup$ @MaximPrasolov I have only seen the stable homeomorphism conjecture stated for Euclidean spaces. Does the version for a general $M$ follow from that? $\endgroup$
    – Cihan
    Commented Sep 16 at 20:47
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    $\begingroup$ 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. $\endgroup$
    – Maxim Prasolov
    Commented Sep 18 at 12:22

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