# Covering by open subfunctors and epimorphisms of sheaves.

I am trying to learn about the functor of points approach to algebraic geometry. Given the category of locally ringed spaces $GSp$ (geometric spaces) we have a functor $$\mathcal{G}: GSp \to Set^{Rng} \\ X \mapsto Hom(\operatorname{Spec}-,X)$$

It is clear to me that if $X \cong \operatorname{Spec}A$ then $\mathcal{G}(\operatorname{Spec}A)\cong Hom(A,-)$. I am having problems to understand what happens when $X$ is a general scheme. First of all, a scheme has an affine cover by $\{ \operatorname{Spec}A_i\}$ which is mapped by $\mathcal{G}$ to a family of open subfunctors of $\mathcal{G}X$ in the sense of this Trying to understand open (closed) subfunctors.

Also since $\{ \operatorname{Spec}A_i\}$ is a covering it is clear that the induced morphism $\bigsqcup \operatorname{Spec}A_i \to X$ is an epimorphism. If the notion of covering by open subfunctors is analogous to the topological one then we should have an epimorphism

$$\bigsqcup Hom(A_i,-) \to \mathcal{G}X.$$

This condition is translated in the book of Demazure by saying that for every field $K$, $\bigsqcup Hom(A_i,K) \to \mathcal{G}X(K)$ is an epimorphism.

I think this condition is essentially the same as the one I stated before but since we are working with sheaves on the Zariski site instead of general presheaves the condition of being a sheaf epimorphism can be checked locally on fields. Am I right? In this case how can I proof the fact that both ideas are equivalent? A general statement for this would be:

Given two sheaves $\mathcal{F},\mathcal{G}$ in the Zariski site is it true that $\mathcal{F} \to \mathcal{G}$ is a sheaf epimorphism $\iff$ $\mathcal{F}(K) \to \mathcal{G}(K)$ is surjective for every field?

• It's true for schemes but not for general sheaves. – Zhen Lin Jun 19 '16 at 7:39

Formally: Consider on the one hand the sheaf ${\mathscr F} := {\mathbb A}^1$, i.e. ${\mathbb A}^1(R) := R$. You can view this as a sheaf of 'regular' functions on $\text{Spec}(-)$ by associating to $f\in {\mathbb A}^1(R)=R$ the function on $\text{Spec}(R)$ mapping the point ${\mathfrak p}$ to the image of $f$ in $k({\mathfrak p})=A_{\mathfrak p}/{\mathfrak p}A_{\mathfrak p}$, i.e. a section of the family ${\mathfrak p}\mapsto k({\mathfrak p})$. On the other hand, considering all such set-theoretic sections also gives rise to a sheaf ${\mathscr G}$, and the just explained construction constitutes a morphism of sheaves ${\mathscr F}\to {\mathscr G}$. On fields, this morphism is an isomorphism, but it is not an epimorphism of sheaves: If that was the case, we would have an epimorphism on all $R$-valued points for local $R$, but this is not true: For example, if $R$ is a DVR with residue field $k$ and quotient field $K$, then ${\mathscr G}(R) = k\times K$, while ${\mathscr F}(R)$ would only map to pairs of the form $(\overline{r},r/1)$.
• Thanks @Hanno for this really interesting answer. Then how can i understand the covering condition with the fields without using that the map $\bigsqcup Hom(A_i,-) \to \mathcal{G}X$ is an epimorphism? – Abellan Jun 18 '16 at 13:27
• @Abellan The difference is that the $\mathrm{Hom}(A_i, -)$ are open subfunctors. For a family $(F_i \to F)_i$ of open subfunctors (all sheaves for the Zariski topology), the following statements are equivalent: (1) For any field $K$, the family $(F_i(K) \to F(K))_i$ is jointly surjective. (2) For any local ring $R$, the family $(F_i(R) \to F(R))_i$ is jointly surjective. (3) The family $(F_i \to F)_i$ is jointly epimorphic in the category of Zariski sheaves. – Ingo Blechschmidt May 9 '17 at 11:38