Prove $\Bbb C^2 \setminus \{0\}/\Bbb C^*$ is homeomorphic to $(\Bbb C^2 \setminus \{0\})/_{\sim f}$, where $f=\frac{qi\bar{q}}{|q|^2}$, $q$ quaternion The equivalence relation $\sim f$ is defined s.t. $q_1 \sim q_2$ iff $f(q_1) = f(q_2)$. I am having problems to start.
First I have problem on understanding $\Bbb C^2 \setminus \{0\}/\Bbb C^*$. I know this is called $\Bbb {CP}^1$ but unfortunately I am familiar with neither expression. Seems this is a $\Bbb R^2$ plane, but how to get this plane? If there is visualization it is better.
Second I know image of $f$ is a 2-sphere, but what does equivalence class $[a]$ look like? What is the shape? I tried to solve $f(a)=p$ for some $p$ on the image but did not succeed.
I guess that they are homeomorphic because if $g: \Bbb C^2 \setminus \{0\} \rightarrow \Bbb C^2 \setminus \{0\}/\Bbb C^*$ is a quotient map then equivalence classes of $\sim g$ and $\sim f$ are same. Is the guessing correct?
 A: For the projective line $\mathbb P^1(\mathbb C)$, it is just as the projective line over any field: the union of $\mathbb C$ and a point at infinity (think of $\mathbb P^1$ as the set of all directions of lines in the plane - the direction is then given by the slope, except for the vertical line, which has infinite slope). The topology is the obvious one (a basis of neighbourhoods of $\infty$ are the complements of bounded sets). So topologically, $\mathbb P^1(\mathbb C)$ is just a real, 2-dimensional sphere. Turning this paragraph into an actual proof should not be too hard.
For the right-hand side, I think your notations are a bit messed here. You really should use directly the Hamilton quaternion algebra $\mathbb H$ instead of $\mathbb C^2$ - these are isomorphic as $\mathbb R$-vector spaces and even as $\mathbb C$-vector spaces, but certainly not as algebras ($\mathbb H$ is a noncommutative division algebra, while $\mathbb C^2$ is a commutative algebra with zero divisors). Since you already know that $f(\mathbb H)$ is a real 2-sphere, and $\mathbb P^1(\mathbb C)$ also is, your problem is solved.
