[datavirtue]

The first sentence should have been the ball-is-equal-to-egg explanation with mention of topology. Before that I had no idea what they were talking about.

P.s. I have to assume the rules forbid shapes with surfaces of zero thickness. Otherwise I can just smash a ball into an inner-tube. If the shapes have thickness mandated, what is it? Are the thickness of the surfaces a consideration when morphing from one shape to another? Is the surface thickness negative or positive from zero? All of these questions stem from my experience in 3D modeling where these parameters must be defined.

[brianpan]

Have you heard the quip that in physics a cow and a point are equivalent? This is because the physicist cares only about the motion of the thing.

In topology, a doughnut and a coffee mug are equivalent (a mug has exactly one hole, in the handle where your fingers grab). Because the mathematician doesn't care about how hard, thick, or breakable it is; they only care about how complex the shape is. So throw out thickness, size, elasticity, etc.

[dullcrisp]

There is no thickness (or it’s zero if you like). The deformations have to be continuous mathematical functions, so punching a hole isn’t possible.

The study is about the properties of (higher dimensional) shapes rather than concrete objects. It’s like asking what’s the thickness of a circle.

[wcarron]

Easy, it's 0.38mm.

[datavirtue]

If it's zero I can make a doughnut from a ball without tearing.

[teraflop]

No, you can't. If you'd like an analogy from 3D modeling to see why not:

Any polyhedral mesh has an integer called its "Euler characteristic", which is simply calculated by taking the number of vertices, subtracting the number of edges, and adding the number of faces. (V-E+F)

Obviously, smoothly deforming a surface by moving vertices around doesn't change its Euler characteristic. A bit less obviously, any sequence of local refinements to "patches" of the mesh can't change its Euler characteristic either. (For example, splitting one face into smaller regions that are still connected to their surroundings in the same way.) Anything that you might reasonably call a "smooth" transformation will keep the Euler characteristic unchanged. You can convince yourself of this by experimentation with whatever 3D modeling software you like.

But a spherical mesh has Euler characteristic 2, and a torus mesh has Euler characteristic 0. So no smooth deformation can transform one into the other.

The only way to change the Euler characteristic would be to change the mesh topology itself, which would mean there's at least one pair of faces that are connected by an edge in one mesh and not connected in the other, which means the mesh has been "torn" along that edge.

With a lot of math, you can extend this argument to arbitrary continuous surfaces, not just polygons. If two surfaces have different Euler characteristic, then you cannot find a bidirectional continuous mapping between them. Any such bijection must be discontinuous somewhere, which roughly means that arbitrarily close points are "torn apart" from each other.

[dullcrisp]

You can’t. The informal proof may not be very convincing but it’s that the torus has two circles that remain distinct no matter how you deform the space: the smaller and larger circles in this picture [1].

But on a sphere, every circle can be deformed to any other circle. If the torus were itself the deformation of a sphere, you’d be able to deform it the same way as the sphere to get one circle to the other.

Again though, the version of these objects that mathematicians study is formalized such that this is unambiguous.

[1] https://en.m.wikipedia.org/wiki/File:Tesseract_torus.png

[pbhjpbhj]

Mathematical objects are only loosely analogous to physical objects.

A 2D disk has zero thickness, any movement orthogonal to the plane of the disk will take you off the disk. But the disk can't be distorted into a circle in a continuous way.

[agrounds]

To add on to what dullcrisp said, which is all correct, even spheres with thickness are “the same as” spheres of zero thickness from the perspective of homotopy theory. “Sameness” here means homotopy equivalence [1]. In fact the thin sphere is a deformation retract [2] of the thick one. The deformation pushes each point of the thick sphere along radial lines towards the thin sphere. Being a deformation retract implies the two spaces are homotopy equivalent.

[1] https://en.wikipedia.org/wiki/Homotopy#Homotopy_equivalence

[2] https://en.wikipedia.org/wiki/Retraction_(topology)

[topaz0]

You don't actually have to assume anything. You could ask instead, or read some background.