Prove that the complex projective line is a union of two preimages of closed unit disks in $\mathbb{C}$.

Question

I am reading Rick Miranda's Algebraic Curve and Riemann Surfaces. In Chapter $1$, he introduced the complex projective line $\mathbb{C}\mathbb{P}^{1}$ as an example of the Riemann surface. I understand most of what he was doing, but I got lost in the end.

Consider the following two sets $$U_{0}=\{[z:w]:z\neq 0\}\ \ \text{and}\ \ U_{1}=\{[z:w]:w\neq 0\}$$ and the maps $$\phi_{0}:U_{0}\longrightarrow\mathbb{C}\ \ \text{defined by} \ \ \phi_{0}([z:w]):=\frac{w}{z}$$ $$\phi_{1}:U_{1}\longrightarrow\mathbb{C}\ \ \text{defined by}\ \ \phi_{1}([z:w]):=\frac{z}{w}.$$

At the end of the discussion, he claims two things:

1. $p=[1:0]$ and $q=[0:1]$ can be separated by $\phi_{0}^{-1}(\mathbb{D})$ and $\phi_{1}^{-1}(\mathbb{D})$, i.e. $p\in \phi_{0}^{-1}(\mathbb{D})$ and $q\in \phi_{1}^{-1}(\mathbb{D})$ and $\phi_{0}^{-1}(\mathbb{D})\cap \phi_{1}^{-1}(\mathbb{D})=\varnothing$.

2. $\mathbb{C}\mathbb{P}^{1}=\phi_{0}^{-1}(\overline{\mathbb{D}})\cup \phi_{1}^{-1}(\overline{\mathbb{D}}),$ where $\mathbb{D}$ is the open unit disk in $\mathbb{C}$.

He described $\mathbb{C}\mathbb{P}^{1}$ as simply the set of all $1-$dimensional subspaces of $\mathbb{C}^{2}$. And every point in $\mathbb{C}\mathbb{P}^{1}$ is written in the form $[z:w]$ such that $z$ and $w$ are not both zero, and this expression $[z:w]$ represents the span of $(z,w)\in\mathbb{C}^{2}$.

He did not explicitly define the projective line using equivalence classes and quotient space, so I do not want to prove his claim using concepts relating to these things.

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I understand that $p=[1:0]$ as a span of $(1,0)\in\mathbb{C}^{2}$ basically describe the $\mathbb{C}^{*}$ and $q=[0:1]$ as a span of $(0,1)\in\mathbb{C}^{2}$ is also $\mathbb{C}^{*}$. (And perhaps I am wrong.) But why they are contained in $\phi_{0}^{-1}(\mathbb{D})$ and $\phi_{1}^{-1}(\mathbb{D})$?

I do not where to start to prove the second claim.. Any idea?

Answer 1

Don't we have $\phi_0([1:0])=0 \in \mathbb D$ and $\phi_1([0:1])=0 \in \mathbb D$?

For the second claim simply use that for all $z \in \mathbb C$ you have $|z| \le 1$ or $|z^{-1}| \le 1$. So $\phi_0([z:w]) \in \overline{\mathbb D}$ or $\phi_1([z:w]) \in \overline{\mathbb D}$.