I know that there is no diffeomorphism from the unit circle, $S^1$ to the square of side length 2 centered at 0.
However, can we construct a bijective map from $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that is also smooth and maps $S^1$ smoothly onto the square.
A friend suggested we could to this, by mapping each quarter arc onto a side of the square. Then use smooth bump functions so that the derivative transitions smoothly to 0 at each corner.
However, we are unsure if $f$ constructed in in this way is smooth at the origin.
Best Answer
As discussed in other MSE posts, there is no diffeomorphism of $\mathbb{R}^2$ which sends a circle to a square.
There is however noninjective smooth maps of $\mathbb{R}^2$ to itself which do the trick. It is more difficult to come up with an injective smooth map of $\mathbb{R}^2$ to itself which sends the circle to the square, but it is also possible. The arguments I have for both statements rest on the following already not so trivial fact:
For any small enough $\epsilon > 0$, there exists an ambient diffeomorphism of $\mathbb{R}^2$ which sends a unit circle to a smooth embedded closed curve $C$ which coincides with a square everywhere outside the four disjoint balls of radius $\epsilon$ centred on the corners of the square.
Sketch of proof. Fix a unit circle and consider a square in which it is inscribed. Smooth the corners of the square in your favorite way, so as to obtain a curve $C$ as in the above statement; note that $C$ can be taken to be convex. There is thus a radial interpolation between $C = C_0$ and the circle (call it now $C_1$) through smooth embedded closed curves $C_t$. The 'velocity vectors' $v_t := (dC_t/dt)(t)$ at time $t$ are defined on the curve $C_t$; since the $C_t$ are smoothly embedded, $v_t$ can be smoothly extended to an ambient vector field $w_t$ which is smooth both in $t$ and in the $\mathbb{R}^2$-variables. Using bump functions, $w_t$ can be chosen to have support in a time-independent large ball. The integral curves of the time-dependent vector field $w_t$ are thus defined and determine an ambient diffeomorphism of $\mathbb{R}^2$ which sends $C_0$ to $C_1$. QED
To prove that there is a noninjective smooth ambient map which sends a unit circle to a square, it thus suffices to show that there is a noninjective smooth ambient map which sends the above curve $C$ to a square. It is possible to successively shrink the balls of radius $\epsilon$ to their center points via (noninjective) smooth ambient maps, thus sending $C$ to a square.
To prove that there is an injective smooth ambient map which sends a unit circle to a square, it thus suffices to show that there is an injective smooth ambient map which sends the above curve $C$ to a square. It is however useful to make the choice of $C$ a little more specific. First of all, given any $\epsilon/2 > \delta > 0$, consider a smooth even map $f : \mathbb{R} \to [0, \infty) : u \mapsto f(u)$ whose derivatives of all orders vanish at $u=0$, which is increasing for $u > 0$ and which agrees with $|u|$ for $|u| > \delta$. Consequently, the function $g(u) := \mathrm{sign}(u) \cdot f(u)$ is a smooth bijective function which coincides with $u$ for $|u|>\delta$. Now pick any corner of the square and choose cartesian coordinates $(x,y)$ on $\mathbb{R}^2$ centred at that corner so that the square is locally the set of points $(x, |x|)$. One can smoothen this corner so that the resulting curve is locally given by the set of points $(x, f(x))$. Doing this to every corner yields a curve $C$ as above which has the same symmetry group as the square it approximates. Then choose new cartesian coordinates, still denoted $(x,y)$ for convenience, centred on the barycenter of $C$ and such that the corners of the square lie on either the $x$-axis or the $y$-axis. Consider now the smooth bijective map $G(x,y) = (g(x), g(y))$. One can check that the set $G(C)$ is the square.