Suppose that we have a Riemannian manifold $(M, g)$ that is uniformly locally isometric to a ball in $\mathbb{R}^n$, that is, there exists $r > 0$ such that for every $x \in M$ ball $B(x,r)$ in $M$ is isometric to a ball $B(0,r)$ in $\mathbb{R}^n$. Does it follow that there exists a function $\varphi \colon M \to \mathbb{R}^n$ such that $$ \forall y, z \in M \qquad \| \varphi(y) - \varphi(z) \| \le R \implies d(y, z) = \|\varphi(y) - \varphi(z)\|,$$ (where $d(y,z)$ is the metric on $M$ given by the infimum of lengths of curves that connect $y$ with $z$) and $\varphi[M]$ can tile $\mathbb{R}^n$? I also think that this tiling has to be periodic.
The reasoning behind this question is that every periodic tiling of $\mathbb{R}^n$ done using closed sets $F$ (tiles) such that $\mathrm{cl}(\mathrm{Int}(F)) = F$ gives us such a manifold. (The intersections between neighbouring tiles tell us which parts of the tile should be "glued together" to create the manifold).
