Prove that a map is open in the Zariski topology

Say $$k$$ is an algebraically closed field and define the equivalence relation on $$k^{n+1}$$ given by $$x \sim y \iff x=\lambda y$$ for some $$\lambda \in \mathbb{k}^{\times}$$. Clearly $$\mathbb{P}^{n} = k^{n+1}/\sim$$. Let the map $$q:k^{n+1} \setminus 0 \longrightarrow \mathbb{P}^{n}$$ be the quotient map that sends $$x$$ to its equivalence class. I have to show that it is an open map (for the Zariski topology). Any hint is appreciated.

• What is your definition for the topology on $\mathbb{P^n}$? – lulu Jan 25 '19 at 16:07
• Quotient topology – Dalamar Jan 25 '19 at 16:12

Hint:

1. $$q^{-1}(q(U))=\bigcup_{\lambda\in k^{\times}}\lambda U$$

2. Multiplication by $$\lambda\neq 0$$ is a homeomorphism as for basic opens $$\lambda D(f)=D(f(\frac{1}{\lambda}(\cdot)))$$