Say we have a smooth manifold $M$. Does there always exist a metric function $d : M \times M \to \mathbb{R}$ such that $d$ is smooth and gives rise to the same topology of the manifold $M$ ?

I am inclined to think that the answer is affirmative. One could argue that by strong Whitney embedding theorem , $M$ could be embedded in the Euclidean space. Now the restriction of the ordinary metric on the Euclidean space to the manifold would satisfy the conditions. Am I right, or am I making some mistake ? Thanks !

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    $\begingroup$ I think this mathoverflow thread probably answers your question. $\endgroup$ – ಠ_ಠ Oct 22 '16 at 14:37
  • $\begingroup$ @ಠ_ಠ Only if the OP wants it to come from a Riemannian metric. Otherwise their proof (as applied to the squared distance function) is correct. $\endgroup$ – user98602 Oct 22 '16 at 22:49

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