# Is the quarter disk diffeomorphic to the half disk?

I am having trouble with the following easy question: consider a quarter of an open disk. Is there a diffeomorphism between it and half of an open disk? I think not since this would make the square $[0,1]\times[0,1]$ a manifold with boundary.

The obvious map I could think of which takes a point and doubles the angle it makes with the $x$-axis, that is, $\begin{bmatrix} \cos(2(\arccos(\frac{x}{\sqrt{x^2+y^2}})& -\sin(2(\arccos(\frac{x}{\sqrt{x^2+y^2}})\\ \sin(2(\arccos(\frac{x}{\sqrt{x^2+y^2}})& \cos(2(\arccos(\frac{x}{\sqrt{x^2+y^2}})\end{bmatrix}$ fails to be continuous at the origin.

But, i don't know what the problem would be in general in finding such a map. Thanks.

added Quarter of an open disk is the open disk centred at the origin intersection the unit square.

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No: you can't create (or remove) a sharp corner using a diffeomorphism. As you say, if they were diffeomorphic, then the quarter disc would be a manifold with boundary, so it suffices to show this is not the case. (I did this once, but my differential geometry skills are a bit rusty which is why I'm only posting this as a comment.) – Clive Newstead Jan 4 '13 at 12:33
By the way, an easier statement of your "obvious map" is the function $z\mapsto z^2$ defined on $\mathbb C$ (a.k.a. $\langle x,y\rangle\mapsto\langle x^2-y^2,2xy\rangle$). This also does some squashing of points, but this version is differentiable everywhere, although the Jacobian is $0$ at the origin. – Mario Carneiro Jan 4 '13 at 12:36
The vague idea I have is that suppose it was a smooth manifold then how does one go about defining a tangent space at the corner... But I don't know how to make this rigorous. – Apoha Jan 4 '13 at 12:40
What exactly is "a quarter of an open disk"? Are the straight parts of the boundary included in the set or not? If not, I don't see any problems with the origin. – Hans Lundmark Jan 4 '13 at 12:42
@HansLundmark Yes, the boundaries are included in the quarter disk. – Apoha Jan 4 '13 at 12:45

Hint: Use the fact that the tangent space at any point of the half disk contains some vector $v$ together with its opposite $-v$; the "tangent space" at the corner of the quarter disk does not. Conclude that there cannot be a linear map between the two tangent spaces of full rank.