Consider a $d$-dimensional smooth manifold $M$ in $\mathbb{R}^D$ ($d < D$). Denote $B_r(c)$ as an open Euclidean norm ball in $\mathbb{R}^D$ with center $c$ and radius $r$. For a positive real number $\Delta < r$ and $c \in M$, let $K = B_r(c) \setminus B_{r - \Delta}(c)$ be the region between these two balls. We set $U = M \bigcap B_r(c) \neq \emptyset$ and $V = M \bigcap K \neq \emptyset$.

Suppose the reach of the manifold is upper bounded by $\tau$. We pick $r < \frac{\tau}{2}$, and let $\phi$ be the projection onto the tangent space at $c$. As shown in http://people.cs.uchicago.edu/~niyogi/papersps/NiySmaWeiHom.pdf, the derivative of the projection is nonsingular. Then $(U, \phi)$ is a chart for $M$.

Let $x \in V$. We are interested in bounding the Euclidean distance between $x$ and $\phi^{-1}\left(\overline{\phi(U)}\right)$, where $\overline{\phi(U)}$ denotes the "boundary" of $\phi(U)$, i.e., $\textrm{cl}(\phi(U)) \setminus \textrm{int}(\phi(U))$ (cl for closure and int for interior).

I am new to the differential geometry and have stuck on this problem for quite a while. My intuition is that since the curvature of the manifold is restricted, by making $\Delta$ small, the Euclidean distance can be bounded by a function of $\Delta$ and $r$. Unfortunately, I cannot come up with any rigorous argument, nor any counterexamples.


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