My Math Forum are the rational maps are Zariski-continuous?

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 April 2nd, 2013, 10:35 AM #1 Newbie   Joined: Mar 2013 Posts: 2 Thanks: 0 are the rational maps are Zariski-continuous? How can one show that the a rational map f:V??W is Zariski-continuous? (where V&W are affine varieties, i.e. irreducible closed algebraic sets) Interpret that, by definition we need to show that for every closed subset U?W f^(?1)(U):={P?dom(f):f(P)?U} is closed since U is closed there are polynomials h1,...,hn:W?k such that U={P?W:h1(P)= ... =hn(P)=0} and so we need to show that: f?1(U)={P?dom(f):h1?f(P)= ...= hn?f(P)=0} is closed, how can we show that?

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### rational mapd are continulus

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