# How to construct the schemes $\operatorname{Hom}_S(X, Y)$ and $\operatorname{Isom}_S(X, Y)$ from the Hilbert scheme?

Let $$S$$ be a noetherian scheme. Grothendieck  constructed for any projective $$S$$-scheme $$X$$ the Hilbert scheme $$\operatorname{Hilb}_{X/S}$$, which represents the functor $$T \mapsto \underline{\operatorname{Hilb}}_{X/S}(T) = \{Y \hookrightarrow X \times_S T \,|\, \text{family of closed subschemes, flat over } T\}.$$ Now if $$X, Y$$ are both projective over $$S$$, then the functor $$T \mapsto \underline{\operatorname{Hom}}(X, Y)(T) = \{X_T \to Y_T\}$$ can be embedded into $$\underline{\operatorname{Hilb}}_{X \times_S Y/S}(T)$$, by sending a morphism $$f: X_T \to Y_T$$ to its graph $$\Gamma_f \subset X_T \times_T Y_T$$. This identifies $$\underline{\operatorname{Hom}}(X,Y)(T)$$ with the subset of schemes $$\Gamma \subset X_T \times_T Y_T$$, such that the restricted projection $$\Gamma \to X$$ is an isomorphism. Grothendieck claims that this defines an open subscheme $$\operatorname{Hom}(X,Y) \subset \operatorname{Hilb}_{X \times Y/S}.$$ How can I see that this is the case?

Similarly, Grothendieck claims that $$\operatorname{Isom}(X,Y) \subset \operatorname{Hom}(X,Y)$$ is an open subset. How can I see this?

 Grothendieck, Techniques de construction et théorèmes d'existence en géométrie algébrique IV: les schémas des Hilbert; Séminaire Bourbaki

By definition of the Hilbert scheme, we have a "universal family", i.e. a closed subscheme $$i : \mathscr{Z} \to \mathrm{Hilb}_{X\times_S Y/S} \times_S (X\times_S Y)$$ such that the composite with the projection $$\mathrm{Hilb}_{X\times_S Y/S} \times_S (X\times_S Y) \to \mathrm{Hilb}_{X\times_S Y/S}$$ is flat.
Let $$\alpha : \mathscr{Z} \to \mathrm{Hilb}_{X\times_S Y/S} \times_S X$$ be the composite of $$i$$ with the projection onto the first two factors. Using Stacks 05XD, we have an open subscheme $$\mathscr{U} \subseteq \mathrm{Hilb}_{X\times_S Y/S}$$ such that any map $$f : W \to \mathrm{Hilb}_{X\times_S Y/S}$$ of $$S$$-schemes factors through $$\mathscr{U}$$ if and only if the basechange $$\alpha_W : \mathscr{Z}_W \to (\mathrm{Hilb}_{X\times_S Y/S} \times_S X)_W$$ is an isomorphism. Unwinding definitions, we find $$\mathscr{U}$$ represents exactly the subfunctor $$\underline{\mathrm{Hom}}(X,Y)$$ you described.