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.



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


