Is there "essentially only 1" Jordan arc in the plane?

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

?

Solution 1:

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.