By extending the map in Q6 to $\mathbb{C}$, show that $P^1(\mathbb{C}) \cong \mathbb{C} \cup \{ \infty \}\cong S^2$.

By extending the map in Q6 to $\mathbb{C}$, show that $P^1(\mathbb{C}) \cong \mathbb{C} \cup \{ \infty \}\cong S^2$.

sol) Similarly to $Q6$, Let define $\theta : P^1 (\mathbb{C}) \to \mathbb{C} \cup \{ \infty \}$ such that $\theta([x,y]) = x/y$ for $y\neq 0$ and $\theta([x,0]) = \infty$. In the same way it can be shown that $\theta$ is a homeomorphism. Note that $S^2$ is homeomorphic to $H^2 = \{ (x,y,z) | z\geq 0 , x^2+y^2+z^2=1\}$. Consider $N=(0,0,1)$. Let $(u,v,0)$ be an intersection of plane $z=0$ and a straight line connection $N$ and $(x,y,z)$. Ratio of a length between $N$ and $(x,y,z)$ and a length between $N$ and $(u,v,0)$ is $1-z$. When $(x,y)$ and $(u,v)$ are represented through polar coordinates, $(x,y)$ and $(u,v)$ have same degree. Therefore $u= \frac{x}{1-z}, v=\frac{y}{1-z}$. Geometrically, the mapping $(u,v)=(u(x,y,z), v(x,y,z))$ is a homeomorphism, therefore, $S^2 \cong \mathbb{C} \cup \{ \infty \}$

출처- 기하학1 (GIST-강현석교수님 강의) Homework 1 문제 중 일부



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