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Let f: X -> Y be universally open surjective morphism of finite algebraic schemes.

Let Y' -> Y be a base extension and let f':x' -> Y' be the extended morphism. Then the image of every component of X' is a component of Y'.

Why? Is it ok because "f is open"?

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1 Answer 1

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A morphism between finite schemes is finite itself (EGA II, 6.1.5 (v)), so $f'$ is finite as well, hence closed (EGA II, 6.1.10). Now apply EGA IV, 2.3.5 (ii).

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