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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?

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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 $S^2$.

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Yes! I thought of that, too. You need to check that these embeddings are homeomorphic (which is possible since division is open). If $U = U_1(0)$, one can take $g(U)$ and $f(U)$ as neighbourhoods of the north and south poles which are disjoint because $g^{-1}\circ f\colon z \mapsto \tfrac{1}{z}$. Why did I put a bounty on that? I completely forgot about it! But who knows – maybe there will be even more elementary answers. – k.stm Oct 8 '12 at 16:56

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 $\Bbb{CP}^1$.

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Thanks, this is helpful. But as far as I can tell the Lemma requires the notion of continuous and proper group actions which I wanted to evade. – k.stm Sep 23 '12 at 6:18
The lemma only requires continuity of the group action $G\times X\to X,~(g,x)\mapsto gx$, and is easy to establish. There is no assumption of properness. – Olivier Bégassat Sep 23 '12 at 10:28
Oh, you're right. Thanks again. Still, I wanted something more elementary if possible. I also edited my original question to make that clearer, the edit is pending. – k.stm Sep 23 '12 at 11:14

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