suppose $f:X\rightarrow Y$ is a morphism between two schemes over scheme $S.$ Do we have the morphism between their hilbert schemes, i.e. is there a natural morphism $Hilb(X/S)\rightarrow Hilb(Y/S)$ over $S$? What about the Douady spaces instead of Hilbert schemes in the analytic spaces case?



functoriality of hilbert scheme
I am just posting my comments above as an answer, so that this question does not remain unanswered. 


Yes, that is also true for the Douady spaces. This has nothing to do with the construction or proof of existence of the Hilbert schemes / Douady spaces. Already for the (contravariant) Hilbert functor $$ \underline{\text{Hilb}}\_{Y/S}: \textbf{Schemes}\_S \to \textbf{Sets},$$ fo the composition of this function with the faitful functor, $$\Phi: \textbf{Schemes}\_{S'} \to \textbf{Schemes}\_S,$$ there is a natural isomorphism of $\underline{\text{Hilb}}\_{Y/S}\circ \Phi$ with the Hilbert functor $$ \underline{\text{Hilb}}\_{Y\times\_S S'/S'}: \textbf{Schemes}\_{S'} \to \textbf{Sets}.$$ Thus, whenever both of these functors are representable, then there is an isomorphism as above. 


