I'm looking for a map from the square (identifying the 2-Torus) onto the closed unit disc, especially regarding:
- Surjectivity: I want a map from the square onto the closed unit disc,
- Smoothness: ideally $C^\infty$ (at least $C^4$),
- Degree $1$: So the map wraps once around the disc's boundary.
Under the idea of constructing a homeo between the square and the disc, for example, see Prove a square is homeomorphic to a circle.
Let $x = (x_1,x_2) \in [-1,1]^2$. Then define $$ T(x) = \frac{\|x\|_\infty}{\|x\|_2}x $$
This maps the square to the unit disc bijectively (except at $x=0$, where we define $T(0)=0$).
Note that the differentiability problem is given by either the absolute value or by the maximum. But, when you use a softmax-based or smoothed $\infty$-norm approximation, the domain is no longer the square itself, but a kind of rounded square, even if you rescale, the mapping’s preimage of the disc becomes a "soft square", not the true square.. So the topological identification to the torus no longer holds cleanly, and this is exactly what I'm trying to avoid.
I was thinking, instead of approximating $∥(x,y)∥_\infty$ directly, define a custom $C^\infty$ function on the whole square that:
- Tends/equal to $∥(x,y)∥_\infty$ outside a compact set near the center,
- Smoothly rounded inside (like a mollified max),
- Still maps the boundary of the square to radius $1$.
I know what I need to fix, but I've run out of ideas. Do you have any ideas ? Would this be a viable approach? Should I try something radial or complex?
Thank you for your attendance :)!
