Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p : [0,1] \to \mathbb{R}^2$ be continuous and injective.
Does there always exists a continuous function $f : [0,1]\times \mathbb{R}^2 \to \mathbb{R}^2$ such that

For all $x\in \mathbb{R}^2$, $f(0,x) = x$
For all $t\in [0,1]$, $x\mapsto f(t,x)$ is a homeomorphism
For all $s\in [0,1]$, $f(1,p(s)) = \langle s,0\rangle$


share|cite|improve this question
up vote 15 down vote accepted

The function $f$ that you want is called an "ambient isotopy," and your question is whether all arcs in the plane are ambient isotopic. The answer is yes, although I'm not sure of a good reference. I believe that one can construct the ambient isotopy by hand using the 2D Schönflies theorem. The idea is to connect the endpoints of your Jordan arc by an arc, using the fact that the complement is path-connected, and apply the Schönflies theorem to the result.

The analogous result for arcs in $\mathbb R^3$ is false. There are "wild" arcs which are not ambiently isotopic to a standard arc. See for example this article. There are also wild spheres in $\mathbb R^3$, such as Alexander's horned sphere.

Edit: I asked this question on MathOverflow, which got more definitive answers here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.