Trying to understand open (closed) subfunctors I am trying to read about functor of points and I am struggling with the definition of open subfunctor.
The definition is the following. A subfunctor $\alpha\colon G\to F$, where $F,G\in \mathsf{Fct}(\mathsf{Rings},\mathsf{Sets})$ is called open (closed) if for any $\psi\colon h_{\mathrm{Spec}(R)}\to F$ the fibered product $G_\psi=G\times_F h_{\mathrm{Spec}(R)}$ yields a map $G_\psi\to h_{\mathrm{Spec}(R)}$ isomorphic to injection from the functor represented by some open (closed) subscheme of $\mathrm{Spec}(R)$.
Can you, please, explain why is it true that open subfunctors of $F=h_{\mathrm{Spec}(S)}$ are exactly the functors $G$ of the form $G(T)=\{\phi\in F(T)\mid \phi^*(I)T=T\}$ for some ideal $I\subset S$? This is the exercise $VI.6$ in Eisenbud-Harris "Geometry of Schemes". Can you, please, explain how to think about these subfunctors?
Thank you very much!
 A: $G$ is an open subfunctor of $F = h_{\operatorname{Spec}(S)}$ means that for any $T,$ $G(T) = \operatorname{Hom}(\operatorname{Spec}(T), U)$ where $U\subseteq\operatorname{Spec}(S)$ is an open subscheme. Thus $\varphi\in G(T)$ implies that $\phi = \iota\circ\varphi\in F(T)$ where $\iota:U\hookrightarrow \operatorname{Spec}(S)$ is the inclusion, and there is an obvious bijective correpondence between such $T$-valued points of $\operatorname{Spec}(S)$ and $T$-valued point of $U.$ 
Thus, we simply need to realize that the condition "$\phi\in F(T)$ factors through $U$" is equivalent with the condition "$\phi^*(I)T = T,$ where $\phi^*:S\to T$ is the dual map, and $I$ is the ideal of the complement of $U.$"
For example, in the case $U = \operatorname{Spec}(S_f)$ for some $f\in S,$ we are asking that $\phi^*$ factors through $S_f$ which is equivalent to $\phi^*(f)\in T^\times$ is invertible. The complement of $U$ is $V(f),$ so what this boils down to is that $\phi^*(I)T = T,$ where $I=(f).$ I leave the general case to you.
A: This question is old, but I would like to present a more synthetical way to define open and closed subfunctors. Some work shows that this yields the same open and closed subfunctors, but I believe that it justifies the definition better. My answer will not explain why the open/closed subfunctors as defined in the question agree with open and closed subschemes, but it will explain how to think about them.
You start with the big Zariski topos $Zar$ and write $Spec(R)$ for the representable sheaves $Spec(R) = Hom_{Ring}(R,-)$. The topos $Zar$ has a generic internal ring $\mathbb A$, which is the forgetful functor to $Set$. Then you define the basic open subfunctors of representables $Spec(R)$ to be those of the form $$D(f) = \{x:Spec(R)|\, \neg f(x) = 0\}$$
for each morphism $f:Spec(R) \to \mathbb A$ in $Zar$. You also define $$V(f) = \{x:Spec(R)|\, f(x) = 0\}$$
Some work (combined with basic knowledge over the internal language of topoi) shows that $D(f)$ is precisely the subobject $Spec(R_f)\to Spec(R)$ and $V(f)$ is the subobject $Spec (R/f)\to Spec R$. Next, you make the following definition.

*

*A subfunctor $S \hookrightarrow Spec(R)$ is closed if and only if it is a meet $S = \bigwedge_i V(f_i)$ of basic closed subunctors in the subobject lattice of $Spec(R)$  in $Zar$. if $I$ is an ideal in $R$, then $\bigwedge_{f\in I} V(f)$ will be precisely the closed subfunctor $V(I)$ described in the question.

*A subfunctor $S \hookrightarrow Spec(R)$ is open if and only if it can be written as a join $\bigvee_i D(f_i)$ in the subobject lattice of $Spec(R)$ in $Zar$.

Warning: The lattice of open subfunctors of the functor $Spec(R)$ will be closed under taking finite meets $\wedge_{Zar}$ and of course under taking infinite joins. In contrast, the closed subfunctors are not closed under taking finite joins $\vee_{Zar}$. The closed subobject $V(x)\vee_{Zar}V(y)$ in $Spec(\mathbb Z[x,y])$ will be strictly smaller than $V(xy)$. The lattice of closed subfunctors has finite joins, but they do not agree with the joins taken in $Zar$.
