Does the fact that a solid circle is homeomorphic to a solid rhombus imply a solid sphere is homeomorphic to a octahedron?

Question

Let $$R = \{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2+z^2 \le 1\}$$ and $$D^3 = \{(x,y,z)\in\mathbb{R}^3 \mid |x|+|y|+|z| \le 1\}.$$ $R$ can be described as a solid sphere (or a $3$-dimensional disk) and $D^3$ can be described as a octahedron. It's relatively straight forward to show $R \cong D^3$ using the maps $f:D^3 \to R$ given by $$f(x,y,z) = \frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}(x,y,z)$$ and $g:R \to D^3$ given by $$g(x,y,z) = \frac{|x|+|y|+|z|}{\sqrt{x^2+y^2+z^2}}(x,y,z).$$ However, I think this method is quite boring.

Instead, is it possible to use the fact that a solid circle is homeomorphic to a solid rhombus? For each $z \in [-1,1]$, $$\{(x,y)\in \mathbb{R}^2 \mid x^2+y^2 \le 1 - z^2\} \cong \{(x,y)\in\mathbb{R}^2 \mid |x|+|y| \le 1 - |z|\}.$$ Thus, can we just correspond each "slice" of $R$ with a "slice" of $D^3$, and use the fact that each of these "slices" are homeomorphic to show that $R \cong D^3$?

Answer 1

To elaborate on Lord Shark's comment, the problem with this method is that this idea of constructing a "slicewise homeomorphism" also requires that the assignment of a slicewise homeomorphism to every point $z \in [-1,1]$ to be, in some sense, continuous in $z$.

Imagine the same problem, but with a Möbius band and a cylinder of finite height. Similarly to your example, we can slice them up into a bunch of intervals $[-1,1]$ arranged along a circle, so all of these slices are homeomorphic. But the global objects are not homeomorphic!

This is due to the fact that when constructing the Möbius band with this method, we need to continuously rotate the different slices, the intervals, as we go around our circle, while we do not have to do anything to that extent with the cylinder (clearly any such cylinder can just be defined with $S^1 \times [-1,1]$). This is why we cannot make these slicewise homeomorphisms into global ones.

Now, I don't know if you know anything about this, but if you replace the word "slice" by "fibre" at every point, this kind of construction of creating a global geometric object from such slices is what people call a fibre bundle. Indeed, in this viewpoint both the cylinder and the Möbius strip are fibre bundles with generic fibre $[-1,1]$ and base space $S^1$, yet they are not homeomorphic, implying that a fibre bundle is determined by more data than just the choice of a fibre and a basis, i.e. you may still potentially "twist" your choice of fibre at every point.