Let $S$ be a scheme, let $T$ be an $S$-scheme, and let $M$ be a set. Let $M_{S}$ be the disjoint union of $M$ copies of $S$, considered as an $S$-scheme. (Notation from [SGA 3, Exp. I, 1.8].) Then $S$-scheme morphisms $T \to M_{S}$ correspond to locally constant functions $T \to M$, i.e. continuous functions $T \to M$ where $M$ is given the discrete topology. The functor $G_{0} : \operatorname{Set} \to \operatorname{Sch}/S$ sending $M \mapsto M_{S}$ is a sort of "partial right adjoint" to the functor $F : \operatorname{Sch}/S \to \operatorname{Top}$ sending $(T,\mathscr{O}_{T}) \mapsto T$, i.e. taking the underlying topological space of the $S$-scheme.
Can the functor $G_{0}$ be extended to a right adjoint $G : \operatorname{Top} \to \operatorname{Sch}/S$ of $F$?
My naive guess is to take a topological space $X$, give $X_{S} := S \times X$ the product topology and set $\mathscr{O}_{X_{S}} := \pi^{-1}(\mathscr{O}_{S})$ where $\pi : X_{S} \to S$ is the projection. Then $(X_{S},\mathscr{O}_{X_{S}})$ is indeed a locally ringed space and gives the usual construction when $X$ is a discrete space, but in general it is not a scheme. Consider $S = \operatorname{Spec} k$ and $X = \{x_{1},x_{2}\}$ the two-point set with the trivial topology; then the only open subsets of $X_{S}$ as defined above are $\emptyset$ and $X_{S}$ itself, so that $X_{S}$ is not even a sober space.
What if I restrict the target category of $F$ to the category of sober spaces?
The product of sober spaces is sober, so it's no longer immediately clear to me whether the above construction fails.
Best Answer
Another way to see that the functor $\mathrm{Sch} \to \mathrm{Top}$ is not a left adjoint is to see that it does not preserve colimits. In this MO answer, Laurent Moret-Bailley gives an example of a pair of arrows $Z \rightrightarrows X$ in $\mathrm{Sch}$, such that the canonical map from $X$ to the coequalizer $Y$ is not surjective (as a function between the sets of points of the underlying spaces). Since in $\mathrm{Top}$ those canonical maps to the coequalizer are always surjective, this coequalizer cannot be preserved by the forgetful functor.