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Let $f_1 \colon \mathbb{R}^3 \to \mathbb{R}^3$ be a homeomorphism, and let $K_1 \subseteq \mathbb{R}^3$ be compact. Does there always exist a homeomorphism $f_2 \colon \mathbb{R}^3 \to \mathbb{R}^3$ and a compact $K_2 \subseteq \mathbb{R}^3$ such that

  1. $K_1 \subseteq K_2$
  2. $f_1(K_1) = f_2(K_1)$
  3. $f_2$ is identity on $\mathbb{R}^3 - K_2^\circ$?

(Hypothesis 1 is probably unnecessary in light of (2) and (3), I just thought it might be clearer to include)

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    $\begingroup$ Not necessarily if $f_1$ is orientation-reversing, but if you assume it preserves orientation, I would guess yes. $\endgroup$
    – Wojowu
    Commented Apr 26, 2021 at 10:02
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    $\begingroup$ This is true in all dimensions, provided that $f$ is orientation-preserving, since $f$ will ne isotopic to identity. See my answer here. $\endgroup$
    – Moishe Kohan
    Commented Apr 26, 2021 at 11:43
  • $\begingroup$ Wow, excellent --- thank you very much! This is very helpful. $\endgroup$
    – cloudman123
    Commented Apr 26, 2021 at 17:40

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