Let $X$ be an algebraic space locally of finite presentation, and let $\tilde{X}$ denote the restriction of $X$ (as a functor on schemes) to the category of complete local rings. Is it true that the mapping $X \mapsto \tilde{X}$ (of algebraic spaces to functors on complete local rings) is a fully faithful functor?


I.e. can we uniquely determine a morphism $f : X \to Y$ of algebraic spaces locally of finite presentation simply by specifying its values on complete local rings?



Algebraic spaces as functors on complete local rings
There is an exponential map from complex line to itself. This is not algebraic, but it is defined on complete local ring valued points. So the functor does not appear to be full.


