$M$ is a manifold of dimension $m$ embedded in $\mathbb{R}^{m+k}$ ($k > 1$).($M$ is not necessarily compact.)
A normal vector field $v$ on $M$ is a vector field such that $\forall x \in M, v(x) \notin T_xM$.
I want to show that if $v_0,v_1$ are normal vector fields on $M$ and homotopic through normal vector fields, then there is an ambient isotopy $F_t : \mathbb{R}^{m+k} \to \mathbb{R}^{m+k}$ with $t \in [0, 1]$ such that $F_0 = \text{id}$, $F_t(x) = x$ for all $x \in M$, and $dF_1 (v_0) = v_1$.
(It is Exercise 2.2 in Elements of Surgery Theory by Rustam Sadykov)
My friend said I could use the following method:
Consider $x \in M$, there is at least a disk of radius $f(x)$ centered at 0 in $TX$ contained in a tubular neighbourhood. We could choose $f(x)$ to be a continuous positive function.
Let $N(r)$ denote the tubular neighbourhood of radius $r$.
Construct $F_t$ in $N(f(x)/2)$ and then "reflect" $F_t$ to the rest of $N(f(x))$. Then $F_t$ would be identity outside $N(f(x))$.
I have no idea how to formally follow this construction. Is it right? Do I necessarily need compactness?