Let $\Bbb{P^1}$ be the complex projective line, with the topology given by the quotient topology of $\Bbb C^2-\{0\}$ under scaling by $\Bbb C^\times$. Meaning as a set we have:
$$\Bbb P^1 = \Bbb{C}^2-\{0\}/\Bbb C^\times.$$
We will denote the orbit of $(a,b)$ under this action by $[a:b]$.
Is it true that for any $U\subset \Bbb C$ open, the set:
$$\{[1:z]\mid z\in U\},$$
is open in $\Bbb P^1$?
I can't seem to show this. I've tried to take the preimage under the quotient map:
$$q^{-1}(\{[1:z]\mid z\in U\})=\{\lambda(1,z)\mid \lambda\in\Bbb C^\times,z\in U\},$$
$$=\bigcup_{\lambda\in\Bbb C^\times}(\{\lambda\}\times\lambda U)$$
and then thinking of the product topology, but I can't seem to finish this off.
Best Answer
Hint : the set $A = \{[1:z] : z \in \Bbb C\}$ is open in $\Bbb P^1$, so open sets in $A$ are also open in $\Bbb P^1$.