**Disclaimer:**I posted the following question on MSE, but since there were no answers. I'm migrating it here.

Let $Homeo_0(\mathbb{R}^n)$ ($Homeo_c(\mathbb{R}^n)$) be the space of all (compactly-supported) orientation-preserving homeomorphisms on $\mathbb{R}^n$ to itself equipped with the topology of compact convergence. Is $Homeo_c(\mathbb{R}^n)$ dense in $Homeo_0(\mathbb{R}^n)$?

I tried to use Kirby's local contractibility result, but I couldn't figure if this is correct.

Note: To avoid confusion, as in the comments, recall that a homeomorphism is said to be compactly supported if it is equal to the identity map outside some compact subset of its domain.
See Definition 1.5 of these nice notes.

  • $\begingroup$ I don't think you can approximate an orientation-reversing homeomorphism by compactly supported homeomorphisms $\endgroup$ Nov 9, 2020 at 17:54
  • $\begingroup$ @DenisNardin Thank you for pointing that out, I polised up the post. $\endgroup$
    – AnnieIM
    Nov 9, 2020 at 17:55
  • 1
    $\begingroup$ Possibly relevant may be papers by Morton Brown (and a coauthor), from about half a century ago. $\endgroup$
    – Wlod AA
    Nov 9, 2020 at 21:07

1 Answer 1


I think this is true. It suffices to prove the

Lemma. Given an orientation-preserving homeomorphism $h$ of $\mathbb{R}^n$ there is a compactly-supported homeomorphsim $h_1$ which agrees with $h$ on the unit ball.

By conjugating by dilations one can find $h_r$'s that agree with $h$ on the ball of radius $r$, and this gives a sequence of compactly-supported homeomorphisms converging pointwise to $h$.

The lemma can be proved using the ideas in the proof of the Annulus Conjecture, but it is simpler here to deduce it from the Annulus Conjecture and the closely-related fact that $Homeo^+(S^{n-1})$ is path-connected.

Proof: Given $h$ consider the sphere $h(S^{n-1}) \subset \mathbb{R}^n$. This is compact so is contained inside a sphere $r S^{n-1}$ for some $r \gg 0$. Both these spheres are locally flat, so by the Annulus Theorem the region between them is homeomorphic to $[0,1] \times S^{n-1}$. Combined with $h\vert_{D^n}$ this gives a homeomorphism $$h': r D^n = D^n \cup [1,r] \times S^{n-1} \to r D^n = h(D^n) \cup \{annulus\}$$ which agrees with $h$ on the unit ball. On the boundary it induces an orientation-preserving homeomorphism of $rS^{n-1}$. Thisis topologically isotopic to the identity, and combined with $h'$ this gives a homeomorphism $$h'' : (r+1)D^n = rD^n \cup [0,1] \times S^{n-1} \to (r+1)D^n = rD^n \cup [0,1] \times S^{n-1}$$ which agrees with $h$ on the unit ball and which is the identity on the boundary: extending by the identity gives the required $h_1$.

  • $\begingroup$ I was thinking about this answer and I wonder. Is it necessary for a subgroup of homeomorphisms to have the fragmentation property in order to be dense in Homeo0(Rn); n≥2? Certaintly, Homeoc(Rn) has it. $\endgroup$
    – AnnieIM
    Nov 17, 2020 at 10:57

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