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.


1 Answer 1


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.


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