# Does the functor Sch to Top have a right adjoint?

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.

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.
• 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 $\coprod_{x \in \mathbb{C}} \mathrm{Spec}(\mathbb{C}) \to \mathbb{A}^1_{\mathbb{C}}$. – HeinrichD Sep 12 '16 at 13:40
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.