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Is a closed rectangle a 2d manifold with boundary? It seems like the corners don't have neighborhoods homeomorphic to the Euclidean half space?

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    $\begingroup$ They are homeomorphic but not diffeomorphic. $\endgroup$ – tom Nov 7 '14 at 22:58
  • $\begingroup$ Do you need them to be diffeomorphic for the rectangle to be a manifold with boundary? $\endgroup$ – David Ding Nov 7 '14 at 23:00
  • $\begingroup$ It depends on what is your definition of manifold. Standard notion of manifold is smooth manifold, so every transition map is diffeomorphism. But you can define only continuous manifold, there you require from transition map to be only homeomorphism. $\endgroup$ – tom Nov 7 '14 at 23:05
  • $\begingroup$ I can define a homeomorphism from a closed rectangle to a closed circular disc. $\endgroup$ – Mark Bennet Nov 7 '14 at 23:08
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    $\begingroup$ A closed rectangle can be seen as a smooth manifold, depending on the charts you use. It won't be an embedded submanifold of $\mathbb{R}^2$, however, because the smooth structure would not be the induced one. $\endgroup$ – Seth Nov 7 '14 at 23:11
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Summary of the comments: it is a topological manifold with boundary. It can be regarded as a smooth manifold with boundary, but then the embedding into $\mathbb{R}^2$ fails to be smooth. It can also be regarded as a smooth manifold with corners.

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