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At the beginning of Lee's "Introduction to Smooth Manifolds", Lee gives the example of a the square and the circle being homeomorphic as an intuitive motivation for smoothness not being invariant under homeomorphisms. But I read somewhere that the unit square does admit a smooth structure and now I'm confused. I can't come up with a structure for the square, but I can't figure out how to prove that there isn't one. Help?

rhodonedra
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    Let $f$ be a homeomorphism from the circle to the square. Push the smooth structure from the circle onto the square via $f$. Done. The point is that the square doesn't admit a smooth structure when you restrict the smooth structure of $\mathbb{R}^2$ to the square. – James May 30 '15 at 17:19
  • Oh, thanks. I missed that. – rhodonedra May 30 '15 at 17:23
  • They are only "pointy" when you consider it relative to the ambient space $\mathbb{R}^2$. If you consider a square as an abstract topological space $\mathscr{T}=(T,\tau)$ there is no notion of corner. Every point looks like every other point. This is just the observation that a square and a circle are homeomorphic. – James May 30 '15 at 17:26
  • This previous question and its answers may be useful. – user225318 May 30 '15 at 19:47

1 Answers1

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For the sake of having an answer:

  1. Since the unit square is homeomorphic to $S^1$, it has a smooth structure given by transporting a smooth structure along any such homeomorphism, but
  2. No such smooth structure will make the unit square an embedded smooth submanifold of $\mathbb{R}^2$ smooth (because the derivatives of such an embedding can't be nonzero at the corners).
Qiaochu Yuan
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    Slight correction: With the standard smooth structure on $S^1$, there is in fact a smooth topological embedding of $S^1$ into $\mathbb R^2$ whose image is the unit square. (You just have to ensure that all derivatives of the embedding map vanish at the corners.) You can use this embedding to put a smooth structure on the square such that the inclusion map is smooth. But you can't do it in such a way that the square is an embedded smooth submanifold of $\mathbb R^2$. – Jack Lee May 30 '15 at 20:06
  • @Jack: oops. Good point. I've fixed my answer. – Qiaochu Yuan May 31 '15 at 03:12