I am trying to prove rigourosly that the unit square $[0,1]\times [0,1]\subset \mathbb R^2$ is not a differentiable manifold with boundary (I've searched for a rigouros proof, but haven't found anyone) Corners are obviously the problem, so it is reduced to check that otherwise, there would be some open subset around, for example, $(0,0)$, diffeomorphic to $H = \{(x,y)\in\mathbb R^2 : y \geq 0\}$. How can I prove easily that $Q = \{(x,y)\in\mathbb R^2 : x\geq 0, y\geq 0 \}$ is not diffeomorphic to $H$? Any hint? Thank you in advance.

  • $\begingroup$ What is the image of $\mathbb{R}\times \{0\}$ under a diffeomorphism? $\endgroup$ – Daniel Fischer Nov 24 '16 at 15:33
  • $\begingroup$ math.stackexchange.com/q/1820069/10014 is very related. $\endgroup$ – Najib Idrissi Nov 24 '16 at 15:42
  • $\begingroup$ Please, do not delete posts after someone took the time to answer it. Posts in the site are meant to help both the original poster and future visitors in the site. $\endgroup$ – Pedro Tamaroff Nov 28 '16 at 2:20

Show that if $\sigma$, $\tau:[0,1)$ are two curves smooth curves in a manifold $M$ such that $\sigma(0)=\tau(0)$ and with the two vectors $\sigma'(0)$ and $\tau'(0)$ linearly independent in $T_{\sigma(0)}M$, and $\phi:M\to N$ is a diffeomorphism, then $(\phi\circ\tau)'(0)$ and $(\phi\circ\sigma)'(0)$ are linearly independent vectors in $T_{\phi(\sigma(0))}(N)$$.

Then find two curves in the first quadrant such that you reach a contradition if the first quadrant were diffeomorphic to a half plane.

  • $\begingroup$ Thank you very much. Altough the first part is very clear to me, I don't see how to reach a contradiction, as you explain in the second paragraph $\endgroup$ – user55268 Nov 24 '16 at 15:52
  • $\begingroup$ In the first quadrant there are two curves that start out from the vertex which stare at you in the face! $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '16 at 16:17
  • $\begingroup$ (You will need to use the fact that the vertex of the first quadrant cannot be sent to a point in the interior of the halfplane) $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '16 at 16:18
  • $\begingroup$ Well, taking $\varphi:t\mapsto (t,0)$ and $\tau:t\mapsto (0,t)$, $\phi \circ \sigma$ and $\phi\circ \tau$ are smooth curves in the half plane, but don't see how to reach the contradiction. $\endgroup$ – user55268 Nov 24 '16 at 16:31
  • $\begingroup$ Where can the image of $\phi\circ\sigma$ be in the halfplane? $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '16 at 16:45

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