# 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!

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$G$ 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.
Thanks a lot for your answer! By the way, the general case (the one that you've left for me) is also not difficult. $\phi$ factors through $U$ means that preimage $P=\phi^{-1}(V(I))\subset \operatorname{Spec}(T)$ of the complement $V(I)=\operatorname{Spec}(S)-U$ is empty. But this preimage is the closed subscheme of $\operatorname{Spec}(T)$ defined by the ideal $\phi^*(I)T$. $P$ is empty iff $\phi^*(I)T=T$. – Sasha Patotski Jul 4 '13 at 19:39