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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.

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  • $\begingroup$ What is your definition for the topology on $\mathbb{P^n}$? $\endgroup$
    – Claire
    Jan 25, 2019 at 16:07
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    $\begingroup$ Quotient topology $\endgroup$
    – Dalamar
    Jan 25, 2019 at 16:12

1 Answer 1

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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))) $

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