# Proving the representability of a functor that is covered by open subfunctors

I already posted this on math.stackexchange some while ago, but haven't received any answers yet. (https://math.stackexchange.com/questions/3221006/proving-the-representability-of-a-functor-that-is-covered-by-open-subfunctors)

I want to prove Theorem 8.9 from Algebraic Geometry I ( U.Görtz, T.Wedhorn), which reads as follows:

Let $$S$$ be a scheme $$F: Sch/S°\rightarrow Set$$ a functor such that:

1. F is a sheaf for the Zariski topology
2. F has a cover by open subfunctors $$\alpha_i:F_i\rightarrow F$$, such that every $$F_i$$ is representable by a scheme $$X_i$$

Then F is representable.

A cover by open subfunctors means, that for every scheme $$T$$ and for every morphism $$h_T\rightarrow F$$, the pullback $$F_i\times_F h_T$$ is representable, say by $$Y_i$$ and the morphism of schemes $$Y_i\rightarrow T$$ corresponding to the projection $$F_i\times_F h_T\rightarrow h_T$$ is an open immersion. In addition the images of $$Y_i\rightarrow T$$ form an open covering of $$T$$

Let me explain what I have done so far and where I am stuck:

The $$X_i$$ can be glued to a scheme $$X$$. The morphisms $$\tilde{\alpha_i}:h_{X_i}\cong F_i\rightarrow F$$ correspond via the yoneda lemma to elements $$f_i\in F(X_i)$$. Using the sheaf property of F, the $$f_i$$ glue together to an element $$f\in F(X)$$ which gives us a natural transformation $$\alpha: h_X\rightarrow F$$. For a scheme $$T$$ and a morphism $$g\in\mathrm{Hom}(T,X)$$ this is given by $$\alpha(T)(g)=F(g)(f)$$. The last step is to show that this assignment is bijective. I managed to show the surjectivity, but can't find a proof for the injectivity. It would suffice to show that the following diagram is a pullback $$\require{AMScd}$$ $$\begin{CD} h_{X_i} @>>> h_X\\@VidVV @VV\alpha V\\ h_{X_i} @>>\tilde{\alpha_i}> F \end{CD}$$ where the morphism $$h_{X_i}\rightarrow h_X$$ is induced from the open immersion $$X_i\rightarrow X$$. The commutativity of this diagram is clear to me. To proof that this is a pullback we could try the following: Since $$h_{X_i}$$ is an open subfunctor, there is an open subscheme $$U_i$$ of $$X$$, such that the following square is cartesian: $$\begin{CD} h_{U_i} @>>> h_X\\@VVV @VV\alpha V\\ h_{X_i} @>>\tilde{\alpha_i}> F \end{CD}$$ By the commutativity of the first square the open immersion $$X_i\rightarrow X$$ factors through $$U_i\rightarrow X$$, so that $$X_i$$ is an open subscheme of $$U_i$$. But I couldn't find a way to show $$X=U_i$$. Another way would simply be checking that, $$h_{X_i}$$ satisfies the universal property. However, when trying to do so one needs that $$\alpha(T)$$ is injective for every scheme $$T$$.

I also looked at the proof in EGA I (Springer edition 1971), where this is Proposition 4.5.4 in chapter 0. There Grothendieck uses (without comment) that the fiber product $$F_i\times_F h_X$$ is represented by $$X_i$$, which I think is equivalent to saying that the square from above (the one with $$X_i$$) is cartesian.

I am thankful for any thoughts on this.

I found another way to prove this Theorem, which then also shows that this map is injective. Consider two Zariski sheaves $$F,G:\mathrm{Sch}^{\circ}\rightarrow\mathrm{Set}$$ and for each sheaf a cover by open subfunctors $$\alpha_i:F_i\rightarrow F$$ and $$\beta_i:G_i\rightarrow G$$. Suppose we have isomorphisms $$\varphi_i: F_i\rightarrow G_i$$, such that the diagram $$\require{AMScd}$$ $$\begin{CD} F_i\times_FF_j @>>> F_i @>\varphi_i>> G_i\\@VVV @. @VV\beta_iV \\ F_j @>\varphi_j>> G_j @>\beta_j>>G \end{CD}$$ commutes and the induced morphism $$F_i\times_F F_j\rightarrow F_i\times_G F_j$$ is an isomorphism. Then $$F\cong G$$. The proof of this statement is similiar to the situation, where $$F$$ and $$G$$ are sheaves on a topological space and $$\varphi_i:F_{|U_i}\rightarrow G_{|U_i}$$ are isomorphisms of sheaves agreeing on overlaps. We can apply this to the functor $$h_X$$ with open cover $$h_{U_i}\rightarrow h_X$$ and the functor $$F$$ with open cover $$\alpha_i:F_i\rightarrow F$$. One can check that the isomorphisms $$h_{U_i}\cong F_i$$ satisfy the conditions above. In this way we obtain an isomorphism $$h_X\rightarrow F$$. The yoneda lemma implies, that this morphsim is actually the same as the morphism defined in my question.