Smooth Mapping between Equidimensional Riemannian Manifolds is Local Isometry iff it is Isometry

Theorem
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be Riemannian manifolds of the same dimension.

Let $\phi : M \to \tilde M$ be a smooth mapping.

Then $\phi$ is a local isometry iff it is an isometry, that is, $\phi^* \tilde g = g$.