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.
Best Answer
From the author: I realized that hint was a little too cryptic. In the corrections to the book (which you should probably download and keep handy), there's a more detailed hint.