Let $Y\subseteq \mathbb{A}^n$ be a closed affine set and let $X:=\overline Y\subseteq \mathbb{P}^n$ be its projective closure. Suppose that there exists a morphism $f:X\to \mathbb{P}^k$ such that $f:X\to f(X)$ is a finite morphism. If $f(Y)\subseteq \mathbb{A}^k$ and $\overline{f(Y)}=f(X)$, is it true that $f:Y\to f(Y)$ is a finite morphism?
Restriction of a finite morphism to an affine closed set
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algebraic-geometry
projective-space
1 Answers
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The answer is no. Take $X = \mathbb{P}^1$ and $Y = X \setminus \{P\} \cong \mathbb{A}^1$ ($P$ is a point). Let $f:X \to P^1$ be a double cover not ramified at $P$, and let $P' \ne P$ be the point such that $f(P) = f(P')$. Then $f(Y) = \mathbb{P}^1$, and $f:Y \to f(Y)$ is not finite.
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0But $f(Y)\not\subset \mathbb{A}^1$ – 2017-01-24
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0OK, then take $Y = f^{-1}(\mathbb{A}^1) \setminus Z$ for some nonempty $Z$. – 2017-01-24
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0I don't see why it's not finite. – 2017-01-24
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0It is quasi-finite, but not finite, since it is not proper. – 2017-01-24