No diffeomorphism that takes unit circle to unit square

Solution 1:

Here's a somewhat rigorous way to see this. Let $\gamma$ be an arc in $C$ such that $F\circ \gamma(0)$ is the corner (1,1). Then (assuming it goes clockwise) there are some functions $x$ and $y$ such that $F \circ \gamma(t) = (1,y(t))$ for $t< 0$ and $F \circ \gamma(t) = (x(t),1)$ for $t > 0$. Thus $F_*\gamma'(t)$ is $(0,y'(t))$ for $t<0$ and $(x'(t),0)$ for $t > 0$. Taking limits, this means that $F_*\gamma'(0) = 0$ contradicting that $\gamma'(0) \ne 0$.

Solution 2:

The original statement above is NOT true. That is a square CAN be given be a differentiable structure!

This follows from the more general theorem

Let M and N be topological manifolds that are homeomorphic with homeomorphism h. If N is also a smooth(differentiable) manifold then a differentiable structure can be defined on M via the pullback defined by h.

And with this differentiable structure on M, h becomes a diffeomorphism from M to N. That M and N are diffeomorphic.

So the homeomorphism given above between the square and the circle can be used to pullback the usual differentiable structure on the circle to the square and with differentiable structure the homeomorphism becomes a diffeomorphism.

Note that this induced differentiable structure on the square is NOT compatible with the usual differentiable structure on R2=R×R. That the inclusion map from the square to R2(identity map restricted to the square) will NOT be differentiable!

Solution 3:

A diffeomorphism would, among other things, induce an isomorphism between the respective tangent spaces. But look at the corners of the square, i.e., the points {(1,1),(1,0),(0,1),(0,0)} ( in the right coordinate system), and see what happens with the tangent space there. More specifically, the tangent space at the corners is not defined, but it must be the image of the tangent space of some point on the circle where the tangent space is defined. Basically, in the 1-D case, the tangent space ( in one of its versions) is given in terms of the derivative f'(t)dt, but the derivative is not defined at the corners.