I was wondering if there is a simple argument showing that the complex projective line defined as $\mathbb{CP^1} = \big(\mathbb{C}^2 \setminus \{0\}\big)/{\mathbb{C}^{\times}}$ is hausdorff when equipped with the quotient topology.
So far I was picturing this scenario by analogy with $\big(\mathbb{R}^3 \setminus 0\big)/\mathbb{R}^{\times}$ and the 2-sphere therein. Imagining open, disjoint double cones surrounding distinct lines the asseriton seems clear. I also know that it suffices to show that the action of $\mathbb{C}^{\times}$ on $\mathbb{C}^2 \setminus \{0\}$ is proper (which I also don't see). But since I want to avoid defining continuous and proper group actions I was hoping for a more elementary approach.
Optimally I'd like to see a way of directly establishing that $\mathbb{CP^1}$, defined this way, is homeomorphic to the 2-sphere $\mathbb{S^2}$. However, I'm looking for an answer which is short and preferably doesn't use (continuous) group actions. Maybe there is no way around that?
Best Answer
You can indeed work by analogy with the real case. To do so, you must think of $\mathbb{CP}^1$ as the quotient of the complex sphere by the aciton of the unit length complex numbers: $$\mathbb S^3\times \mathbb S^1\to\mathbb S^3, ((z,z'),\lambda)\mapsto (z\lambda,z'\lambda)$$ There is an obvious identification $$\mathbb S^3/\mathbb S^1\simeq\mathbb{CP}^1$$ and actually, we can define two maps $$\begin{array}{ccc} \mathbb S^3 & \leftrightarrow & \Bbb C^2\setminus\lbrace 0\rbrace \\ p & \rightarrow & p \\ \frac{(z,z')}{\vert (z,z')\vert} & \leftarrow & (z,z') \end{array}$$ These two (obviously continuous) maps are compatible with the respective group action (said another way, they pass to the quotient), thus, by definition of the quotient topology, they define a pair of continuous inverse bijections i.e. they induce homeomorphisms $$\mathbb S^3/\mathbb S^1\rightarrow\mathbb{CP}^1, \mathbb S^3/\mathbb S^1\leftarrow\mathbb{CP}^1$$ We will be done once we show $\mathbb S^3/\mathbb S^1$ to be Hausdorff. For this I refer you to a general
This shows that the left quotient space is Hausdorff, and since it is homeomorphic to the one you are interested in, so is $\Bbb{CP}^1$.