Equivalence between categories of affine schemes over $X$ and representable functors $\operatorname{Points}(x) \to \operatorname{Sets}$ I am reading Strickland - Formal schemes and formal groups. For a functor $X\colon \operatorname{Rings}\to \operatorname{Sets}$, he defines (2.14) the category of $\operatorname{Points}(X)$ in the following way:
Objects: $(R,x)$ with $x \in X(R)$.
Morphisms: ring morphisms $f\colon R\to S$ s.t $X(f)(x)=y$. 
In remark 2.15 he claims that the category $\mathcal{X}_X$ of affine schemes $Y$ over an affine scheme $X$ and the category of representable functors $Y'\colon \operatorname{Points}(X)\to \operatorname{Sets}$ are equivalent using $Y(S)= \coprod_{z \in X(S)}  Y'(S,z)$.  I can't really understand how the equivalence follows. Would appreciate any help.
P.S. I understood the equivalence of (a) and (c)—this is usually referred to as the description of a scheme using its functor of points. 
 A: You want to see that the category $\mathcal{X}_X$ of affine schemes over $X$ (or affine $X$-schemes) is equivalent to the category of representable functors $Y':\text{Points}(X)\to \text{Set}$. 
First, given an affine $X$-scheme $f:Y\to X$, and a point $(x,R)$ with $x\in X(R)$ and $R$ a ring, we associate the set $$f(R)^{-1}(\{x\})=\{ y\in Y(R) \ | \ f(R)(y)=x\},$$ where $f(R):Y(R)\to X(R)$ is the associated map on the $R$-points (usually called just $f$ by abuse of notation). So the functor say $f'$ (bad notation, by sure) we associate is on the objects of $\text{Points}(X)$ defined by 
$$f'((R,x)):=f(R)^{-1}(\{x\}).$$
(Note that an $X$-scheme is better defined by the morphism $f$ that by the scheme $Y$). 
On the contrary, that a functor $Y':\text{Points}(X)\to \text{Set}$ is representable it means (my guess) that there exists a ring $S$ and a point $y\in X(S)$ such that $$Y'(R,x)=\{h:R\to S \ | \ X(h)(x)=y\}.$$
Note that a point $y\in X(S)$ is the same that a map $y:\text{Spec}(S)\to X$. We associate then to $Y'$ such affine $X$-scheme $y$. 
A: Thank you @Xarles for the answer and the corrections and @Lspice for the editting. I think I got it now. For an affine $X$-scheme $f:Y\to X$ with $Y=Spec(R)$ for some ring R, we have the map $f:Spec(R)\to X$. Following the Yoneda lemma, the map $f$ corresponds to an $x_f\in X(R)$. We thus have a point $(R,x_f)$ and can hence associate to it the contravariant functor $Hom_{Points(X)}(\_,(R,x_f)):=  Y'$, which is represented by this point.
This is one direction $Y\to Y'$. 
For the other direction, let $Y': Points(X)\to Set$ be a representable functor with $Y'= Hom_{Points(X)}(\_,(S,y))$ for some point $(S,y)$. We thus have an induced element $y\in X(S)$, which again by Yoneda, corresponds to a map $f_y:Spec(S)\to X$. Alltogether it gives the equivalence. The thing is, I don't fully understand how the identification $Y(S)= \coprod_{z \in X(S)}  Y'(S,z)$ fits in my proof. Any ideas?
