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There's a claim that Milnor makes in his book Topology from the Differentiable Viewpoint

Every compact $1$-manifold with boundary always has an even number of boundary points.

I'm not quite sure how to prove this, if every compact $1$-manifold with boundary was homeomorphic to a closed interval $[a, b]$ in $\mathbb{R}^1$, then the proof would follow trivially, but I'm not sure if that's a theorem.

How can I go about proving this claim?

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  • $\begingroup$ Yes, every compact $1$-manifold with nonempty boundary is diffeomorphic to $[0,1]$; various books (including Guillemin and Pollack) prove this in various levels of detail. $\endgroup$ – Lord Shark the Unknown Jun 30 '17 at 5:05
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    $\begingroup$ Your comment needs the word "connected" somewhere @LordSharktheUnknown. $\endgroup$ – Lee Mosher Jun 30 '17 at 17:00
  • $\begingroup$ I'm reading through Topology from the Differentiable Viewpoint again, and in the proof of the Homotopy Lemma on page 21, neither $M$ nor $N$ are assumed to be connected, however when he makes use of this theorem it requires $M$ and $N$ to be connected. Do I just take it that the Homotopy Lemma has those added conditions in it? $\endgroup$ – Perturbative Oct 6 '17 at 18:08
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There is a full proof of the classification of compact $1$-manifolds with boundary on the end of Milnor's book itself (which states that those are finite unions of circles and closed intervals).

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