If you have say an affine variety defined over $\mathbb{R}$, then its image under a morphism (also defined over $\mathbb{R}$) is a constructible set. But presumably there would be no good reason in general why the image of the set of $\mathbb{R}$-rational points under the morphism would have to be equal to the set of $\mathbb{R}$-rational points of the constructible set in question.
Or is there some useful sufficient condition for this to happen?