Find $\epsilon$ such that $\varphi^{\epsilon} = \sigma + \epsilon(\cos t \mathbf{n} + \sin t \mathbf{b})$ is injective ($\sigma$ biregular curve) .

Question

I’ve been struggling for an entire day trying to approach the third point of this exercise (Ex. 4.23 from ‘Curves and Surfaces’ by M. Abate and F. Tovena).

Let $\sigma:I \to \mathbb{R}^3$ be a biregular curve parametrized by arc length, and assume there is $M>0$ such that $\kappa(s) \leq M \quad \forall s \in I$. For all $\epsilon >0$, let $\varphi^{\epsilon}:I\times (0,2\pi) \to \mathbb{R}^3$ be defined as $\varphi^{\epsilon}(s,t)=\sigma(s) + \epsilon(\cos t \mathbf{n}(s) +\sin t \mathbf{b}(s)) $.

1. Prove that if $\epsilon <1/M$, then $d\varphi^{\epsilon}_x$ is injective for all $x\in I\times(0,2\pi)$.
2. Assume that there is $\epsilon>0$ such that $\varphi^{\epsilon}$ is globally injective and a homeomorphism with its image, so that it is a local parametrization of $S^{\epsilon} = \varphi^{\epsilon}(I \times (0,2\pi))$. Find a normal versor field on $S^{\epsilon}$ and compute the Gaussian and mean curvatures of $S^{\epsilon}$.
3. Prove that for any interval $[a,b] \subseteq I$ there exists $\epsilon>0$ such that the restriction $\varphi^{\epsilon}|_{[a,b]\times(0,2\pi)}$ is globally injective and a homeomorphism with its image.

The first two questions are fairly easy, I just computed the derivatives of $\varphi^{\epsilon}$ using Frenet-Serret formulas. That’s what I found.

$\frac{\partial\varphi^{\epsilon}}{\partial s} = (1-\epsilon \kappa \cos t)\mathbf{t}+ \epsilon \tau[-\sin t \mathbf{n} + \cos t \mathbf{b})$

$\frac{\partial\varphi^{\epsilon}}{\partial t} = \epsilon [-\sin t \mathbf{n} + \cos t \mathbf{b})$

The normal versor field: $N \circ \varphi^{\epsilon} = -(\cos t \mathbf{n} + \sin t \mathbf{b})$

The Gaussian curvature: $K = \frac{\cos t}{\epsilon(\epsilon \kappa \cos t -1)}$

The mean curvature: $H = \frac{\cos t}{2\epsilon(\epsilon \kappa \cos t -1)}$

Now, point 3 is the real deal. I tried to proceed in a similar way one can prove the existence of tubular neighborhoods for plane curves. Fix $0<\epsilon<1/M$, then the implicit function theorem applied at every point of $[a,b]\times(0,2\pi)$ yields an open cover of $S^{*\epsilon}=\varphi^{\epsilon}([a,b] \times (0,2\pi))$. Extract a finite subcover and let $\lambda$ be its Lebesgue number. If $\lambda<2\epsilon$, it should be fairly easy to conclude, but actually that’s a very strong claim, so I think that’s not the right idea.

Any help?

Update. Maybe a proof by contradiction could work. First notice that for all $x=(s,t) \in I \times (0,2\pi) \quad \lim_{\epsilon \to 0} \varphi^{\epsilon}(x)=\sigma(s)$. Then assume by contradiction that for all $\epsilon >0$, $\varphi^{\epsilon}$ is not injective. Then for all $n \in \mathbb{N}$ there are $x_n=(s_n,t_n),x’_n=(s’_n,t’_n) \in (a,b)\times (0,2\pi)$, with $ x_n\neq x’_n$ but $ \varphi^{\epsilon}(x_n)= \varphi^{\epsilon}(x’_n)$. Bolzano-Weierstrass theorem yields two convergent subsequences $(x_{n_k})_k, (x’_{n_k})_k$ converging to $x=(s,t),x’=(s’,t’) \in [a,b]\times [0,2\pi]$ respectively. By continuity of $\varphi$, seen also as a function of $\epsilon$, and since we assumed that $\sigma$ is injective in $[a,b]$, we have $s=s’$. Thus the following claim should complete the proof, shifting the problem on a more local point of view:

CLAIM: There exists $\epsilon_0>0$ (presumably 1/M, recalling point 1) such that for all $\epsilon<\epsilon_0$ there is a neighborhood $U^{\epsilon} \subseteq [a,b]$ of $s$ such that $\varphi^{\epsilon}|_{U^{\epsilon}\times (0,2\pi)}$ is injective.

But again, no clues about how to extend what the implicit function theorem yields to the whole interval $(0,2\pi)$.

I was thinking about the hypothesis $\kappa <M$. More geometrically, the radius of any osculating circle is greater than 1/M. This makes me guess that the right $\epsilon_0$ is $\epsilon_0=1/M$, and that near enough to $s$ I can use something related osculating circles.