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I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to $\mathbb R$) are $\mathbb R$ and $S^1$. Is this true? If so, is there a reasonably elementary proof?

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You will probably want to add a Hausdorffness and connectedness condition. –  Miha Habič Feb 26 '12 at 19:55
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There is the proof at the end of Milnor's "Topology from the differentiable viewpoint" which is "elementary" if my memory isn't playing tricks on me. –  Bruno Stonek Feb 26 '12 at 20:07
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Make sure you have paracompact in your definition of "manifold", too. Otherwise you can have the "long line" for example. –  GEdgar Feb 26 '12 at 20:22
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@GEdgar: Second countable is stronger than paracompact for manifolds (equivalent for connected manifolds). –  Chris Eagle Feb 26 '12 at 20:25
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@draks: knots are homeomorphic to $S^1$. –  Damian Sobota Feb 26 '12 at 21:53

1 Answer 1

up vote 6 down vote accepted

There's a proof outlined in Problems 17-5 and 17-7 of John Lee's "Introduction to Smooth Manifolds" that uses a basic classification of integral curves of vector fields, specifically that a nonconstant maximally defined integral curve is either injective or periodic, which implies (after a small amount of work) that the image of any nonconstant integral curve is diffeomorphic to either $\mathbb{R}$ or $\mathbb{S}^1$. The problem is finished by showing that any 1-manifold is orientable, and thus admits a nonvanishing global vector field, of which you consider a maximally defined integral curve.

I don't think this is the same proof as given in Guillemin and Pollack or in Milnor, and for my money it's quite a bit simpler than both.

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