Two ways of inducing a metric/topology on a manifold.

Question

Consider $\Bbb R^3$ equipped with the usual euclidean metric and topology. And consider the subset $S^2 := \{ x\in\Bbb R^3\,|\,d(x,0)=1\}\subset\Bbb R^3$ . Suppose we wanted to make $S^2$ in to a metric space on its own, it seems to me that there are two distinct ways of doing so. The first way would be to define a metric $\,\overline d:\, S^2\times S^2 \rightarrow \Bbb R\,,\; \overline d(x,y):=d(x,y)\;$ (it is easy to show that this forms a metric on $S^2$). The problem I have with this definition is that it does not really resemble the distance we intuitivly think of when thinking of the sphere (for example the distance between north-and south-pole is 2 instead of $\pi$). The other obvious way would be to define the metric $\widehat d $ by considering all paths on the sphere between to points and defining the distance as the infimum of the lenghts of those paths. This also defines a metric on $S^2$ and seems to be the more intuitive way of doing so (distance between north-and south-pole is $\pi$) . Since the two metrics are not equal, my question is: Do $\overline d$ and $\widehat d$ induce the same topology on $S^2$ ? And if so, is this generally true for manifolds that are embedded in $\Bbb R^n$ ?

Answer 1

Yes they do.

By definition, the topology of an embedded manifold must be the restriction of the topology of the containing space, so that it must be compatible with the metric $\overline{d}$.

For a Riemannian manifold in general, the infimum of path lengths between two points defines a metric that is compatible with its topology, and this is precisely $\widehat{d}$ in your case.