There is no diffeomorphism between one quadrant in the plane and the half plane

Question

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.

Answer 1

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.