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Let $X$ and $Y$ be projective schemes. Then we can define the mapping scheme between them, $\rm{Maps}(X,Y)$ as follows:

To any map $f:X\rightarrow Y$ we consider the graph $\Gamma_f$ as a closed subscheme of $X \times Y$. So $\rm{Maps}(X,Y)$ is the set of all subschemes of $X \times Y$ that are graphs of morphisms. (Concretely, a subscheme $Z \subset X \times Y$ is the graph of a morphism iff the projection to $X$ is an isomorphism) Of course this all makes sense in families, so $\rm{Maps}(X,Y)$ is a subfunctor of the Hilbert scheme $\rm{Hilb}(X \times Y)$.

Now at this point, I have seen a number of sources casually claim that $\rm{Maps}(X,Y)$ is actually an $\it{open}$ subfunctor and is hence representable. None of these sources even remark on why this is true? So my question is: why is this true?

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  • $\begingroup$ Which Hilbert scheme do you (and Kollar) mean exactly? I am new to this and only know Hilbert schemes w.r.t. some polynomial... Would you take the Hilbert polynomial of the ideal sheaf of some subspace $X \times \{pt\}$ (assuming that $Y$ has a point)? $\endgroup$
    – Peter Arndt
    Commented Sep 29, 2010 at 20:17
  • $\begingroup$ For the above statement you can just take the union of every Hilbert schemes, i.e. of the Hilbert scheme for every possible Hilbert polynomial. (Similarly, Maps(X, Y) contains all maps, not just maps of some fixed degree.) $\endgroup$
    – Arend Bayer
    Commented Sep 29, 2010 at 22:03

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See Grothendieck, seminaire bourbaki 221 "les schemas de Hilbert", bottom of page 221-19.

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    $\begingroup$ Grothendieck is a bit sketchy there. A more detailed argument fleshing out his sketch is given in Theorem 5.23 in the book FGA Explained (for which the preceding result 5.22(b) provides the crucial openness result). $\endgroup$
    – BCnrd
    Commented Sep 30, 2010 at 0:43
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    $\begingroup$ First time I've seen anyone imply that Grothendieck doesnt write enough. $\endgroup$
    – Richard Borcherds
    Commented Sep 30, 2010 at 1:39
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    $\begingroup$ Dear Richard: I think the FGA's weren't meant to be more than sketches (unlike EGA). I didn't intend to suggest he didn't write enough there; as sketches they are pretty good (but still somewhat of a challenge to fill in; less of a burden these days for beginners since now FGA Explained exists). $\endgroup$
    – BCnrd
    Commented Sep 30, 2010 at 3:02
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I can't give the answer right off, but a reference should be Koll{\'a}r's book "Rational Curves on Algebraic Varieties." Here's where he proves that it's a representable functor, and I believe that the lemma on the next page is what says "open subfunctor", though I might be off.

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  • $\begingroup$ The relevant algebra fact is Proposition I.7.4.1 $\endgroup$
    – mdeland
    Commented Sep 29, 2010 at 22:13

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