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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.

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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.

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  • $\begingroup$ A similar reason is that the forgetful functor doesn't preserve epimorphisms. That is, there are epimorphisms of schemes which are not surjective. An example is $\coprod_{x \in \mathbb{C}} \mathrm{Spec}(\mathbb{C}) \to \mathbb{A}^1_{\mathbb{C}}$. $\endgroup$ – HeinrichD Sep 12 '16 at 13:40
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No such right adjoint exists, even restricted to sober spaces. For simplicity let us take $S=\operatorname{Spec} k$ for some field $k$, and consider the space $X$ having two points, one of which is closed. If $G(X)$ existed, then maps $M\to G(X)$ would be in bijection with closed subsets of $M$. It is not hard to show no such $G(X)$ exists. For instance, taking $M$ to be Specs of fields extending $k$, you can see $G(X)$ must only have two points, and in particular it must be affine. You then get a $k$-algebra $A$ with a radical ideal $I$ such that for any $k$-algebra $B$ with a radical ideal $J$, there is a unique map $f:A\to B$ such that $J$ is the radical ideal generated by $f(I)$. Clearly no such $(A,I)$ can exist, since for any cardinal $\kappa$ we can find a $(B,J)$ such that $J$ cannot be generated by fewer than $\kappa$ elements.

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