Manifold and the topology of $\Bbb R^n$

i'm starting to study manifold in order to understand Einstein's relativity but there is something I don't understand. A manifold $$M$$ is defined as being locally homeomorphic to $$\Bbb R^n$$. Homeomorphisms can be defined in terms of how they map open sets, namely an homemorphism $$f$$ and its inverse $$f^{-1}$$ have to map open sets to open sets. The question is: Are the topology of $$M$$ and $$\Bbb R^n$$ the same? Because then I put a distance in $$\Bbb R^n$$ and the distance will be locally the same in $$M$$. So since a distance induce a topology I think they must be the same.

Can I say that if $$R^n$$ is a metric space and $$M$$ is a manifold on this metric space then: The metric induce a topology on $$R^n$$, the metric of $$R^n$$ induce a metric on $$M$$ and so a topology?

• A similar question was asked here. – Dietrich Burde Mar 12 at 14:59
• The metrics won't be the same globally: think of a solid donut sitting in 3-space. The metric inside the solid donut between points $P$ and $Q$ is given by measuring the length of the shortest curve joining $P$ and $Q$ (lying a bit here). Imagine that $P$ and $Q$ are on opposite sides of the donut hole. Their path won't be a straight line! – Prototank Mar 12 at 19:30