Why is the unit square a 2-manifold?

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The unit square [0,1]^2 is a smooth 2-manifold with boundary in \mathbb{R}^2, but it is not a smooth submanifold due to its corners, which lack tangent spaces. Smoothness in manifolds is determined by the transition functions, and the unit square cannot be smoothly mapped around its corners. It is homeomorphic to a manifold with boundary, like the unit disc, but does not inherit a smooth structure from \mathbb{R}^2. The discussion clarifies that while the unit square can be treated as a manifold, it does not meet the criteria for being a smooth manifold in the traditional sense. Understanding these distinctions is crucial for proper classification in topology.
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Sorry for the delay; I've been kind of busy recently.
Yes; I can only think of variations of the fact that a function whose graph contains a corner is not differentiable at the corner, since the tangent space would be generated by the line y=f'(x)x+b.
 

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