Complex projective line hausdorff as quotient space 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?
 A: 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

Lemma: Let $X$ be a Hausdorff space, $G$ a compact topological group (not necessarily Hausdorff), and suppose $G$ acts on $X$. Then the quotient space $X/G$ is Hausdorff.

This shows that the left quotient space is Hausdorff, and since it is homeomorphic to the one you are interested in, so is $\Bbb{CP}^1$.
A: There are homeomorphic embeddings of $\mathbb{C}$ into $\mathbb{CP}^1$, given by $f(z) = (z:1)$ and $g(z)=(1:z)$. Since $\mathbb{C}$ is Hausdorff and since the images of $f$ and $g$ are $\mathbb{CP}^1$ without $(1:0)$ and $(0:1)$, respectively, you only need to find disjoint neighborhoods of these two points, which is easy. Depending on the topological tools you have at your disposal, this also shows that $\mathbb{CP}^1$ is homeomorphic to $S^2$.
