Complex projective line 1

Question

I have $\mathbb{C}P^1$ defined by equivalence classes $(z^{1} : z^{2})$ with respect to $(z^{1} : z^{2}) = (\lambda z^{1} : \lambda z^{2})$. Then $U_{k} = \{(z^{1} : z^{2}) | z^{k} \ne 0 \}$ for $k = 1 , 2$. Then I define $\phi_{k}: \mathbb{R}^2 \rightarrow U_{k}$ by $\phi_{1}(x_{(1)}, y_{(1)}) = (1 : x_{(1)} + i y_{(1)})$ and similarly $\phi_{2}(x_{(2)}, y_{(2)}) = (x_{(2)} + i y_{(2)} : 1)$ , but I have no idea how to calculate $\phi_{k} ^{-1}$ for $k = 1, 2$ and I feel like it's probably really obvious? (I know that there's an easier mapping from $\mathbb{C}$ to the projective space but I'm interested in this map in this instance.

Answer 1

The mapping you describe is precisely the mapping $\mathbb{C}\rightarrow U_1, z\mapsto (1:z)$ resp. $\mathbb{C}\rightarrow U_2,z\mapsto (z:1)$ when identifying $\mathbb{C}\cong \mathbb{R}^2,z\mapsto (\mathrm{Re}(z),\mathrm{Im}(z))$. The inverse is pretty straightforward: Let $(z_1:z_2)\in U_1$, then $z_1\neq 0$, so $(z_1:z_2)=(1,\frac{z_2}{z_1})$ and we therefore define $\phi_1^{-1}: U_1\rightarrow \mathbb{C},(z_1:z_2)\mapsto \frac{z_2}{z_1}$. Clearly this is the inverse of $\phi_1$. The same works for $\phi_2$.