# Show that the two notions of isometry between Riemannian manifold are equivalent

We now have two kinds of “metrics” on a Riemannian manifold—the Riemannian metric and the distance function. Correspondingly, there are two definitions of “isometry” between Riemannian manifolds—a Riemannian isometry is a diffeomorphism that pulls one Riemannian metric back to the other, and a metric isometry is a homeomorphism that pulls one distance function back to the other. I am trying to prove that both are equivalent.

The implication ($\Rightarrow$) is quite straightforward. I mean, given a metric isometry on the first sense, there are a isometry on the distance sense.

I am having a lot of problems about how to start the converse.

I understand that I have to show that since there exist an homeomorphism between $(M,g)$ and $(N,h)$ two riemannian manifolds such that $f^{\ast}d_{h} = d_g$ then necessarily $f^{\ast}h = g.$

This exercise comes from Lee's book of Riemannian geoemtry, and is given the following hint:

Use the exponential map to show that this homeomorphism is a diffeomorphism.

In fact I don't know how to use this hint to start the exercise.

I do appreciate any hints.