MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
What you asks make me think of the topological Stein factorization developed by Jean Malgoire and Christine Voisin. (sciencedirect.com/science/article/pii/0040938381900380) – ACL Feb 20 at 11:29
up vote 21 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.