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Let $X$ be a $d$-dimensional integral variety over an algebraically closed field $k$ and let $f\colon \mathbb{P}^n_k --> X$ be a dominant rational map. Why does this imply that there is a dominant rational map $g\colon \mathbb{P}^d_k --> X$?

Supposedly, the restriction of $f$ to a sufficiently "general" subspace $\mathbb{P}^d_k \subset \mathbb{P}^n_k$ should give a desired $g$, but I can't see how to prove that a single such subspace exists.

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    $\begingroup$ The general fiber $Z$ of $f$ is a $(n-d)$-dimensional subvariety of $\mathbb{P}^n$. A general $\mathbb{P}^d\subset \mathbb{P}^n$ will meet $Z$ along a finite subset (if you don't see this, start with $d=n-1$ and proceed by induction). Thus the fiber of $f_{|\mathbb{P}^d}$ at a general point of $X$ is finite. This implies that $f_{|\mathbb{P}^d}$ is dominant. $\endgroup$
    – abx
    Dec 31, 2014 at 9:02
  • $\begingroup$ This is a duplicate of mathoverflow.net/questions/95306/… $\endgroup$
    – user5117
    Dec 31, 2014 at 10:42
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    $\begingroup$ @ArtiePrendergast-Smith — I think the question is not exactly a duplicate, but your answer does answer this question. On the other hand, OP indicate that they knows the rough sketch of your answer, but don't know why the sufficiently "general" subspace exists. I think this justifies this question. $\endgroup$
    – jmc
    Dec 31, 2014 at 12:47
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    $\begingroup$ @abx: Thanks, but could you explain in a bit more detail (that would be very helpful)? Is $Z$ the generic fiber or a general fiber. My stumbling point is that "a general fiber" $Z$ is really infinitely many fibers parametrized by an open of $X$, so even though I could convince myself that each one of them has a finite intersection with most $\mathbb{P}^d$'s, I don't see how to guarantee finite intersection with every one $Z$ over that open. On the other hand, if $Z$ is the generic fiber, then I don't know how to argue finite intersection, because $Z$ then consists of non-closed points. $\endgroup$
    – Lisa S.
    Dec 31, 2014 at 17:09
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    $\begingroup$ @Lisa S. : Just take one fiber (above a closed point $x\in X$) of dimension $n-d$. Then $f_{|\mathbb{P}^d}$ has finite fibers above $x$, hence above a Zariski open subset of $X$. $\endgroup$
    – abx
    Dec 31, 2014 at 17:36

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