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Sep 17, 2011 at 18:10 comment added Andreas Blass The Riemannian manifolds in the answer are tacitly assumed to be connected.
Sep 17, 2011 at 16:46 comment added Edmund Harriss I agree that surjectivity is all that is required for the isometry; however the question asks what properties of the metric space make it unecessary to assume that isometries are surjective.
Sep 17, 2011 at 16:37 comment added Noam D. Elkies Much simpler counterexamples: the positive or nonnegative reals, or the natural numbers, with the shift $x \mapsto x+1$; or any infinite space with the discrete metric and any map at all that's injective but not surjective.
Sep 17, 2011 at 16:35 history answered Keivan Karai CC BY-SA 3.0