# The Homeomorphy of The Tetrahedron, The Cube, and The Octahedron

My question involves all possible dimensions, but this is the lowest dimension where these are qualitatively distinct shapes, so I phrase it in three dimensions for convenience. The analogs for higher dimensions will be apparent once I describe these sets.

The cube $$\mathbf{I}^3$$ is the product space $$[0,1]\times[0,1]\times[0,1]$$. The tetrahedron is the space $$\mathbf{T}^3=\lbrace (x,y,z)\in\mathbf{I}^3: x+y+z\leq 1\rbrace.$$ And the octahedron is the space $$\mathbf{K}^3=\lbrace (x,y,z)\in\mathbf{R}^3:\lvert x\rvert+\lvert y\rvert+\lvert z\rvert\leq 1\rbrace,$$ where $$\mathbf{R}$$ is the real line and $$\mathbf{R}^3$$ is $$\mathbf{R}\times\mathbf{R}\times\mathbf{R}$$.

The spaces $$\mathbf{T}^3$$, $$\mathbf{I}^3$$ and $$\mathbf{K}^3$$ are purported to be homeomorphic. I have been thinking about it, but I can't come up with a satisfactory homeomorphism between the three.

I can give an example for the type of homeomorphism I desire in the best case scenario: In the general case, for natural $$n\geq 0$$, if the unit disk is defined in the usual manner as $$\mathbf{D}^n=\lbrace (x_1,\cdots,x_n)\in\mathbf{R}^n:x_1^2+\cdots+x_n^2\leq 1\rbrace,$$ then the only one of these that is obviously homeomorphic to the unit disk $$\mathbf{D}^n$$ for me is the octahedron$$^*$$ $$\mathbf{K}^n$$ by the map $$f:\mathbf{K}^n\to\mathbf{D}^n$$ defined on any $$p=(p_1,\cdots,p_n)$$ in $$\mathbf{K}^n$$ as $$f(p)=\left(\sqrt[3]{p_1\sqrt{\lvert p_1\rvert}},\cdots, \sqrt[3]{p_n\sqrt{\lvert p_n\rvert}}\right),$$ with the inverse map $$f^{-1}:\mathbf{D}^n\to\mathbf{K}^n$$ defined on any $$x=(x_1,\cdots,x_n)$$ in $$\mathbf{D}^n$$ as $$f^{-1}(x)=(x_1\lvert x_1\rvert,\cdots,x_n\lvert x_n\rvert).$$ This kind of map is ideal, because it is built up only of the composition of elementary operations, and there is no case checking. In particular, for my predicament, if case checking is required, it would be best as a rule of thumb that all cases should constitute open sets—or, put another way, that the composite map is an appropriate gluing of compatible maps defined on open subsets that cover the whole space—but it's even better if this can just be avoided in the first place.

Apparently, there are similar maps from both $$\mathbf{T}^n$$ and $$\mathbf{I}^n$$ to $$\mathbf{D}^n$$. What are examples of such maps? Or, if it is easier to describe, what are instances of homeomorphisms between $$\mathbf{K}^n$$, $$\mathbf{I}^n$$ and $$\mathbf{T}^n$$?

$$^*$$In $$n$$ dimensions, $$\mathbf{K}^n$$ will actually have $$2^n$$ copies of $$\mathbf{T}^n$$, and within all of these copies there will be only one copy of $$\mathbf{T}^{n-1}$$ that constitutes a "face" or $$n$$ dimensional "room" of $$\mathbf{K}^n$$ as a geometric shape. So, for instance, $$\mathbf{K}^4$$ is most aptly called a hexadecachoron, $$\mathbf{I}^4$$ is an octachoron and $$\mathbf{T}^4$$ is a pentachoron. I suppose for $$n\geq 5$$, the general name for these in $$n$$ dimensions would be the $$2^n$$-tope, the $$2n$$-tope, and the $$n+1$$-tope, respectively.

Optional. The way this question comes to be setup makes me think about the remaining two Platonic solids in the following way: In what way can the dodecahedron and the icosahedron be implicitly described as subspaces of 3-Euclidean space, as were described the tetrahedron, the cube and the octahedron? With that, what are the homeomorphisms between the two and the remaining Platonic solids as described? There are also a few more regular polychorons in four dimensions for which one may ask the exactly analogous question, but the three I have found seem to be the only regular polytopes for dimensions no less than 5, so those cases are sorted.

Edit. The answer given for this question is satisfactory for the two and three dimensional cases.

• Topology cares very little about explicit bijections like these. Normally you just inflate any convex polytop to a sphere and call it a day. Dec 14, 2022 at 16:11