Paolo Aluffi says something to the effect of what's in the title in his intro to math notes. Link below.
$\mathbb R^2/\sim$ is a quotient of $\mathbb R^2$ modulo the equivalence relation given by $(x_1, y_1) \sim (x_2, y_2) \iff x_1^2 + y_1^2 = x_2^2 + y_2^2.$ If $[(a, b)] \in \mathbb R^2/\sim$, then $[(a, b)] = \{(x, y): x^2 + y^2 = a^2 + b^2\}$ by $\sim.$ So, $\mathbb R^2/\sim$ is a set of concentric circles centered at the origin with the radius $\sqrt{a^2 + b^2}$. Note, $[(0, r)] = \{(x, y): x^2 + y^2 = r^2\}$ is a circle centered at the origin and so $\mathbb R^2/\sim = \bigcup [(0, r)]$. Thus we want to show a bijection from $\mathbb R^{\ge 0}$ to $\bigcup [(0, r)]$. I think the identity function $\color{red}{r_{(\in \ \mathbb R^{\ge 0})} \to [(0, r)]}$ should work since identity is bijective. What if we replace $\mathbb R^{\ge 0}$ with $\mathbb R$? Since the range of $e^x$ is $(0, \infty)$ and since $e^x$ is bijective, $\color{blue}{\text{$x_{(\in \ \mathbb R)} \to [(0, e^x)]$ should(?) work with a caveat: $e^x$ misses $r = 0$}}.$
My questions:
- Does the function in red above work?
- Is the part in blue above reasonable? If so,
- How can we augment the function in blue above such that it doesn't miss $[(0, 0)]?$
Thanks.
Introduction to Advanced MathematicsCourse notes by Paolo Aluffi
Yes, the function in red works perfectly fine and is clearly bijective. You are also correct that the image of $e^x$ is $(0,\infty)$ and not $[0,\infty)$.
Note that if $f:[0,\infty)\to\mathbb{R}$ was continuous, then $f^{-1}[\mathbb{R}]=[0,\infty)$ is open, which is not true, so you will not be able to find a 'nice' function as a bijection between $\mathbb{R}$ and $[0,\infty)$.