How to prove that the map is open?

Question

I am trying to prove that $\phi$ is homeomorphism,where $U=\{[1,u,v]|u,v\in \mathbb{R}\}\subset{\mathbb{R}P^{2}}$,and $\phi$:$U\rightarrow\mathbb{R}^{2}$ given by $\phi([1,u,v])=(u,v)$,$\mathbb{R}P^{2}$ is the usual quotient topological space made from $\mathbb{R}^{3}\backslash\{(0,0,0)\}$;in order to show that $\mathbb{R}P^{2}$ is a differentiable manifold,I need to show that $\phi$ is open,but I cannot formulate a formal proof.Can anyone help me?Much appreciate!

Answer 1

Hint: Let $\pi:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}P^2$ be the standard quotient map. Since $\mathbb{R}P^2$ is equipped with the quotient topology induced by $\pi$, a subset $V$ of $\mathbb{R}P^2$ is open if and only if $\pi^{-1}(V)$ is open in $\mathbb{R}^3\setminus\{0\}$. Firstly, $\pi^{-1}(U)$ is open in $\mathbb{R}^3\setminus\{0\}$; secondly, to show $\phi$ is open, it suffices to show $$\phi\circ\pi:\pi^{-1}(U)\to\mathbb{R}^2$$ is an open map.