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So I'm trying to find a bijection from $[1, \infty)$ to the surface of the unit hemisphere $U$ (without a flat bottom) in spherical coordinates. I define this below: $$U = \{(\theta, \psi).\, \theta \in [0,2\pi) \wedge \psi \in [0,\pi/2)\} - \{(\theta, 0).\, \theta \in (0,2\pi)\}$$ I defined $U$ like this because the coordinates $(0,0)$ and $(1,0)$ for example represent the same point on the surface ($\theta$ is the annulus and $\psi$ is the polar angle.)

What I'm thinking of doing is defining a function $f:[1,\infty) \to U$ where I partition the domain such that each element $e\in P \subseteq [1,\infty)$ where $P$ is a partition can be mapped to a point in one of the unique circular cross-sections of U. $P$ should also have the same number of elements as the number of points in this cross section. I also want $f(1) = (0,0)$.

I'm not sure if wording $f$ like this is sufficient to show that it is a bijection. Could I please have some help?

Gary
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    Are you aware that $\mathbb{R}$ and $\mathbb{R}^n$ have the same cardinality? It's far easier to construct a bijection if you know this – Chris Sanders Mar 03 '22 at 09:52
  • @ChrisSanders No, I am not aware of this at all. How interesting. – Kookie Mar 03 '22 at 09:52
  • use the graph parametrization $\mathbf{x}\mapsto(\mathbf{x},\sqrt{1-\lVert\mathbf{x}\rVert^2})$ from the interior of the unit ball in $\mathbb{R}^{n-1}$ to the upper unit hemisphere in $\mathbb{R}^n$. then use Chris Sanders’ comment to get a bijection between the interior of the unit ball and $[0,\infty)$ (in the case where $n \geq 2$) – C Squared Mar 03 '22 at 10:14
  • Thanks @CSquared but I don't really understand what you mean by "graph parameterisation". Could you further elaborate? – Kookie Mar 03 '22 at 10:16
  • forget i said graph parametrization. it’s just a special type of function and i shouldn’t have used that terminology in this context anyway. – C Squared Mar 03 '22 at 10:18
  • @CSquared okay, so what would $\mathbf{x}$ be then? Is it a vector on the 3d-coordinate system? In that case is $n=3$ as well? – Kookie Mar 03 '22 at 10:19
  • @Kookie $\mathbf{x}$ is a vector $(x_1,\dots,x_{n-1})$ in the interior of the unit ball in $\mathbb{R}^{n-1}$ (this was implied in the previous comment). – C Squared Mar 03 '22 at 10:20
  • @CSquared I thought we were working with the surface of half a hemisphere without the circular bottom? We also only have a subset of $\mathbb{R}$ which we want to show the bijection from. – Kookie Mar 03 '22 at 10:22
  • @Kookie that is exactly the image of the map $\mathbf{x}\mapsto (\mathbf{x},\sqrt{1-\lVert\mathbf{x}\rVert^2})$ – C Squared Mar 03 '22 at 10:23
  • Let us continue this discussion in chat. – Kookie Mar 03 '22 at 10:25

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