Tubular Neighborhood in local coordinates I am trying to verify if I understand the concept of a tubular neighborhood correctly. Any comments on my line of thoughts are very much appreciated.
Let $\mathcal{M}$ be a $d-$dimensional manifold embedded in $\mathbb{R}^D$ through $f:\mathcal{U}\subset \mathbb{R}^d \to \mathcal{M}$. Then, for every point $n$ in the normal space of $\mathcal{M}$ there exists a $u\in \mathcal{U}$ and $v\in \mathbb{R}^{D-d}$ such that
$$n = f(u) + \bot J_f(u)v =:\varphi(u,v)$$
where $\bot J_f(u)$ denotes the matrix consisting of column vectors forming a basis of the normal space $N_x$ in $x$. These column vectors, together with the column vectors of the Jacobian $J_f(u)$, form a basis of $\mathbb{R}^D$ (justifying my notation).
Now, a tubular neighborhood of $\mathcal{M}$ is the image of the mapping $\varphi$ if $v$ is restricted to be sufficiently small, i.e. $||v||<\varepsilon$ for some $\varepsilon >0$, such that $\varphi$ is bijective.
Question: Assuming that this is correct, under which conditions is the tubular neighborhood diffeomorphic to $\mathbb{R}^D$?
Intuition: This is true whenever $\varphi$ is sufficiently smooth and $\varepsilon$ is sufficiently small.
 A: Following the advice of Deane, I use the inverse function theorem to show that indeed the tubular neighborhood is diffeomorphic to $\mathbb{R}^D$ under certain conditions.
Let $\mathcal{B}_{\varepsilon}^{D-d}(0)$ be the open ball in $\mathbb{R}^{D-d}$. Let $\varepsilon$ be sufficiently small such that the image of
$$ \mathcal{U} \times \mathcal{B}_{\varepsilon}^{D-d}(0)$$
under $\varphi$ is a tubular neighborhood of $\mathcal{M}$.
From the inverse function theorem, we know the following:
If the Jacobian determinant of $\varphi$ is non-zero at $(u,0)$, then there exists an open neighborhood $U_u$ of  $(u,0)$ such that
$$\varphi:U_u\to \varphi(U_u)$$
is a diffeomorphism. Without loss of generality, $U_u$ is diffeomorphic to $\mathbb{R}^D$, and thus $\varphi(U_u)$ is diffeomorphic to $\mathbb{R}^D$.
If we can show that this is indeed true for all points $(u,0)$ then we can simply unify all neighborhoods $U_u$, intersect this union with $\mathcal{U} \times \mathcal{B}_{\varepsilon}^{D-d}(0)$, and $\varphi$ restricted to this intersection will yield a tubular neighborhood which is diffeomorphic to $\mathbb{R}^D$
What remains to show is:
The Jacobian determinant of $\varphi$ is non-zero at every point (u,0).
Proof: The Jacobian of $\varphi$, $J_{\varphi}$ is a square matrix. Thus, $\det J_{\varphi}(u,0) \neq 0$ if and only if $\det J_{\varphi}(u,0)^T J_{\varphi}(u,0) \neq 0$. The Gram determinant of $\varphi$ is given by:
\begin{align}\label{eq:}
\det \left( J_{\varphi}(u,0)^T J_{\varphi}(u,0) \right)  
&= \det
\left[
\begin{array}
JJ_f(u)^T J_f(u) & J_f(u)^T \cdot \bot J_f(u)\\ 
\bot J_f(u)^T \cdot J _f(u)^T   & \bot J_f(u)^T \bot J_f(u)
\end{array}
\right] \\
&= \det
\left[
\begin{array}
JJ_f(u)^T J_f(u) & 0_{d\times {D-d}}\\ 
 0_{{D-d}\times d}   & \bot J_f(u)^T \bot J_f(u)
\end{array}
\right]
\\
&= \det J_f(u)^T J_f(u) \cdot \det \bot J_f(u)^T \bot J_f(u)
\end{align}
where we have exploited the fact that the column vectors of $J_f(u)$ and $\bot J_f(u)$ are orthogonal.
Since $f$ is an embedding for $\mathcal{M}$, $\det J_f(u)^T J_f(u) \neq 0$. As the column vectors of $\bot J_f(u)^T$ span the normal space, we may assume without loss of generality that $\det \bot J_f(u)^T \bot J_f(u) =1 $. This ends the proof.
Remark: Note that the inverse function theorem requires $\varphi$ to be in $C^1$. Therefore, $f$ needs to be in $C^2$.
