# Proving the existence of an isotopy

Let $X \subset \mathbb{R}^2$ and $U \subset X$ be an open neighborhood of an embedded line $[a,b] \subset U$. Now does there always exists an isotopy (in the sense that every map $f_t$ is a homeomorphism) $$f:U \times [0,1] \to U$$ such that $f_0(x)=id(x)$ for all $x \in U, f_t(x)=x$ in a ngb. of the boundary of $U$ and which finally maps $a$ to $b$, i.e. $f_1(a)=b$?

Is the following possible? I think one can construct a vectorfield like $g(x)= (a-b) \cdot h(x)$ with some smooth $h$ such that $h(x)=1$ for all $x \in [a,b]$, $h(x) \in [0,1]$ for all $x \in V$ ($V$ a neighborhood of $[a,b]$) and $h(x)=0~$ for all $x \in U\setminus V$ (which certainly exists due to a partition of unity). Now I'm sure that the flow generated by this vectorfield $g$ will move $a$ to $b$ and is per construction also identity on a ngb. of $U$ as the vectorfield vanishes there. I guess that the flow which one certainly can derive from the ODE $$\frac{d}{dt} \phi_t(x)|_{t=0}=g(\phi_t(x))$$ with initial value $\phi_0(x)=x$ is really the isotopy that I'm looking for: As I stated above it really fullfills the properties claimed but actually I'm not sure why it is an isotopy at all. Is every map $\phi_t$ a homeomorphism of $U$? Thank you!!

Yes - the flow of a compactly supported smooth vector field is always an isotopy with the same support. You can prove this using a well-posedness results for ODEs - existence tells you that $\phi_t$ exists, uniqueness tells you that it's a bijection and continuous dependence on the initial data tells you that it's continuous. The inverse of $\phi_t$ is $\phi_{-t}$, which is continuous by the same argument; and thus $\phi_t$ is a homeomorphism.