Suppose that $M\subseteq \mathbb R^D$ is $d$-dimensional compact submanifold with $0\in M$, having reach $\tau>0$. Thus, for every $p\in M$, $\exp\_p:B\_{T\_pM}(p,\tau)\rightarrow B\_M(p,\tau)$ is a diffeomorphism. Can we say there is $\phi:B\_{T\_0M}(0,\tau)\rightarrow \mathbb R^{D-d}$ such that $\exp\_0(u)=(u,\phi(u))$?



implicit function theorem on manifold
No, consider for example $M=\{x\in \mathbb{R}^2| |(x\_1,x\_2)-(0,1)|=1\}$. Then $exp\_0(u)=(sin(u),1-cos(u))$.


