Let $(M,g)$ be a riemannian manifold. Show that the topology induced by the metric coincides with the topology of $M$.

Question

I know that I have to show that $M$ with these two notions of topology have the same open. To do this, I have to take elements of the basis of any topology. But I am failing in choose that opens for each case. It is evident that I have to use somehow curves on the manifold, but I really have know idea how to proceed... I appreciate hints/solutions/comments, etc.

P.S. I cannot comment what I have tried so far because I am totally lost in how to do this.... please, be coherent.

Thanks

Answer 1

This is a local question, since manifolds are defined using local diffeomorphism with Euclidean space.

Because of this the idea is, of course, to reduce the question to the local behavior of the metric induced by the Riemannian metric. There are various theorems which enable you to do that. The most basic tool for this is the Gauss lemma, which states that the exponential map is a radial isometry. From this you can conclude that small balls with respect to the induced Riemannian metric are the same as small balls of the space used to describe the topology of $M$ using charts. So you get a basis of the topology with respect to the Riemannian metric by mapping (Euclidean) balls to $M$ using the exponential map, which implies what you are looking for.