Cornered sphere homemorphic to unit sphere

Question

A buddy of mine asked me this question, to which I found it intuitively obvious, but was unable to come up with a proper proof.

Consider the so-called cornered sphere, defined by $x^4+y^4+z^4=1$ in contrast with the standard unit sphere $x^2+y^2+z^2=1$. It is pretty obvious when you make a picture that the cornered sphere is homeomorphic to the sphere proper. However, it seems hard to prove bceause the sort of natural map between the sets is not injective. That is to say, if you use the map $f(x,y,z) = (x^2, y^2, z^2)$ is not a homeomorphism.

But surely an explicit map exists?

Answer 1

Hint: Try $f: \text{cornered sphere} \to \mathbb{S}^2(1)$ given by $$f(x,y,z) = (x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}(x,y,z).$$ This map projects the cornered sphere into the unit sphere, radially.

Answer 2

You want a more linear map. Note for that, that on every ray through the origin, there is precisely one point of each of those spheres. That defines you already a map.

Another way of saying this, is that both spheres are unit norm balls. You can get a map from the cornered one to the normal one by $z \mapsto \frac z {||z||}$ where $||-||$ is the standard norm.

Answer 3

Try $(x,y,z)\mapsto (x|x|,y|y|,z|z|)$. Note that this is invertible by dividing each term by positive square root of absolute value of each coordinate when the coordinate is not 0, and sending 0 to 0 otherwise. Since $\sqrt{|x|x||}=\sqrt{|x^2|}=|x|$, and dividing $x|x|$ by this gives $x$. Since both this function and its inverse are continuous, it is a homeomorphism.