No diffeomorphism that takes unit circle to unit square

Question

This is not a homework problem. I am trying to learn my own from John M. Lee's Introduction to Smooth Manifolds.

In Chapter 3, there is the problem 3-4

Let $C \subset \mathbb{R}^2$ be the unit circle, and let $S \subset \mathbb{R}^2$ be the boundary of the square of side 2 centred at origin: $S= \lbrace (x,y) \colon \max(|x|,|y|)=1 \rbrace.$ Show that there is a homeomorphism $F:\mathbb{R}^2 \to \mathbb{R}^2$ such that $F(C)=S$, but there is no diffeomorphism with the same property. [Hint: Consider what $F$ does to the tangent vector to a suitable curve in $C$].

I can construct a homeomorphism (by placing the circle inside the square and then every radial line intersects the square at exactly one point). But, I don't know how to do the rest of the problem or understand the hint.

I do not know how to write out what tangent space should be for the square. If there were a diffeomorphism then $F_\star$ is isomorphism between any two tangent space. If I show that the tangent space on the corner of square has dimension zero, would it solve problem?

Answer 1

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 one-dimensional case, the tangent space (in one of its versions) is given in terms of the derivative $f'(t)\; \text{d}t$, but the derivative is not defined at the corners.

Answer 2

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 $\gamma'(0) \ne 0$.