# Why is Maps(X,Y) an open subfunctor of Hilb(X x Y)?

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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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)? –  Peter Arndt Sep 29 '10 at 20:17
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.) –  Arend Bayer Sep 29 '10 at 22:03