Embedded surface in $\mathbb{R}^3$ Let $U \subseteq \mathbb{R}^2$ be an open set and let $\sigma : U \rightarrow \mathbb{R}^3$ be a parametrization of an oriented surface $S$ embedded in $\mathbb{R}^3$ whose unit normal in $\sigma (u,v)$ is $N(u,v)$. $\forall \delta \in \mathbb{R}$ we define the map $\tau_\delta : U \rightarrow \mathbb{R}^3$ as
$\tau_\delta (u,v) := \sigma (u,v)+ \delta N(u,v)$
Prove that, if $V \subseteq U$ is an open set with compact closure in $U$, there exist $\epsilon > 0$ such that $\tau_\delta|_V$ is the parametrization of a regular embedded surface $\forall \delta \in (-\epsilon , \epsilon)$.
How can I approach this kind of problem?
 A: Consider the map $F(u,v,t)=\sigma (u,v)+t N(u,v)$. If $\sigma $ is $C^2$ smooth, then $N$, and consequently $F$, is $C^1$ smooth. I claim that the Jacobian  matrix   $DF$ is invertible at every point with $t=0$. Indeed, at such a point $F_u$ and $F_v$ are linearly independent tangent vectors, while $F_t=N(u,v)$, a normal vector. These three are linearly independent, proving the claim. 
Use the inverse function theorem to cover the surface with 3D balls $B_i$ in which $F$ is a diffeomorphism. Choose a finite subcover of   $\overline{V}$. By the Lebesgue number lemma there exists $\delta>0$ such that every subset of $\overline{V}$ with diameter $\le 2\delta$ is contained in some $B_i$. The surface  $\sigma_\delta$ has no self-intersections:  if $\sigma (u,v)+\delta N(u,v)=\sigma (u',v')+\delta N(u',v')$, then $\sigma (u,v)$ and $\sigma (u',v')$ are at distance at most $2\delta$ from each other, contradicting the choice of $\delta$. This and $F$ being a diffeomorphism imply that $\sigma_\delta$ is regular.
A: Let's see if it works...
Let $N_S(\eta)$ be a tubular neighborhood of $S$ (where $\eta:S\rightarrow(0,+\infty)$ is a continuous function) and let $N_{\sigma(\overline V)}(\eta)$ be the restriction of this tubular neighborhood to the set $\sigma(\overline V)$ (observe that $\overline {N_{\sigma(\overline V)}(\eta)}$ is a compact set).
Suppose for contradiction that $\forall\epsilon>0$ exists $\delta\in(-\epsilon,\epsilon)$ such that $\tau_\delta(V)$ has a singular point. We choose $n\in\mathbb{Z}_+$ and set $\epsilon=\frac{1}{n}$, then there is $\delta_n$ with $|\delta_n|<\frac{1}{n}$ such that $\tau_{\delta_n}(V)$ has a singular point, let's call it $p_n$.
Now the sequence $\{p_n\}_{n\in\mathbb{Z}_+}$ is definitively contained in the compact set $\overline {N_{\sigma(\overline V)}(\eta)}$, so there is a subsequence $\{p_{n_k}\}$ which converges in $\overline {N_{\sigma(\overline V)}(\eta)}$ to a certain point $\overline p$; also $\overline p\in\sigma(\overline V)\subseteq\sigma(U)$.
Let $A\subseteq\sigma(U)$ be any open neighborhood of $\overline p$, let $a$ be a point of $A$ and let $I_S(a,\eta(a))$ be the segment $a+(-\eta(a),\eta(a))N(a)$ of lenght $2\eta(a)$ centered in $a$ and normal to $T_PS$. Then in $A$ there are at least two points $x,y$ such that $I_S(x,\eta(x))\cap I_S(y,\eta(y))\neq\emptyset$, contradicting the fact that $N_S(\eta)$ is a tubular neighborhood of $S$.
Do you think it is correct?
