I'm using the following definition of a (smooth) manifold:

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It's from J.Munkres "Analysis on Manifolds". This is an exercise taken from this book: Is the unit square $[0,1]\times [0,1]$ a $2$-manifold in $\mathbb{R}^2$? I searched for answers on the web and they all came out to be different: for example see https://www.physicsforums.com/threads/why-is-the-unit-square-a-2-manifold.566654/ or Is the unit square a submanifold/manifold?. Some people say it's a topological manifold with boundary, others say it's not a manifold with boundary. Which is the correct answer? I know there's a big difference between a smooth manifold and a topological manifold; however some people say that $(0,0)$ has no neighborhood homeomorphic to an open ball in $\mathbb{R}^2$ or an open in $\mathbb{H}^2$, so this means that it cannot be a topological manifold, right? It cannot be neither a smooth manifold because of the "corners", right? Keeping corners out of the picture for a moment, is the tangent vector well defined inside the square? If I took a point inside the square I would be able to draw countless tangent vectors however...that's the point I can't understand; can you please give me a complete answer? Maybe giving a formal proof (I mean not just saying because of corners, or such other things, but proving really the tangent vector is not defined there). One last thing, I think this problem Cartesian product between manifolds could turn out to be useful, however it has no answer yet.

  • $\begingroup$ Look for manifold with corners. math.stackexchange.com/questions/801361/…. $\endgroup$ – hvedrung Jul 3 '15 at 11:26
  • $\begingroup$ The two links you cite are not inconsistent: they both say that the unit square can be given the structure of a smooth manifold with boundary (an alternative way of seeing this is to note that it is homeomorphic with the unit disk). The second link correctly says that the unit square is not a smooth submanifold of $\mathbb{R}^2$ (with its usual smooth structure). The point is that the "corners" are not topologically definable features of the unit square itself. $\endgroup$ – Rob Arthan Jul 7 '15 at 16:41

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