No diffeomorphism that takes unit circle to unit square 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?
 A: 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 means $M$ 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 $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$. That the inclusion map from the square to $\mathbb{R}^2$ (identity map restricted to the square) will NOT be differentiable!
A: 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.
A: 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 $\gamma'(0) \ne 0$.
