How to Prove Zariski Open Covering for Representable Functors in Algebraic Ge...
Problem
Possibility of being Zariski open covering of a functor for some family I am trying to show reductibility to affine case in showing representability of some kind of functor and came across next question. Next are prelimanery definitions ( C.f. Gortz, Wedhorn's A.G. book ( 8.2 ) ). Let $X$ be a scheme and let $g: X\to G$ be a morphism in $\widehat{(Sch)}$ ( recall that we do not distinguish explicitly between a scheme $X$ and the attached functor $h_X$ ). Let $F \times_G X$ be the fiber product in the category $\widehat{(Sch)}$ ( Example 8.3 ; $(F \times_G X) ( T) := F(T) \times_{G(T)} X(T) $ ). Denote by $p : F \times_G X \to F$, $q : F \times_GX \to X$ the two projections. Definition 8.5. A morphism $f: F\to G$ of functors in $\widehat{(Sch)}$ is called representable if for all schemes $X$ and all morphisms $g: X \to G$ in $\widehat{(Sch)}$ the functor $F\times_G X $ is representable. Let $Z$ be a scheme and let $\zeta : Z \xrightarrow{\cong} F \times_G X$ be an isomorphim. By the Yoneda lemma 4.6 the composition of $\zeta$ with the second projection $F \times_G X \to X$ is given by a unique scheme morphism $Z \to X$ which is independent of the choice of $(Z, \zeta)$ up to composition with an isomorphism. Thus the following definition makes sense. Definition 8.6. Let $\mathbf{P}$ be a property of morphisms of schemes such that the composition of a morphism possessing $\mathbf{P}$ with an isomorphism from the right or from the left has again property $\mathbf{P}$. We say that a representable morphism $f: F \to G$ of functors in $\widehat{(Sch)}$ possesses the property $\mathbf{P}$ if for all schemes $X$ and for every morphism $g : X \to G$ the second projection $f_{(X)} : F\times_G X \to X$ possesses $\mathbf{P}$. Definition ( open subfunctor, Zariski open coverings). Let $S$ be a fixed scheme and let $F : ( \mathrm{Sch}/S )^{\mathrm{opp}} \to (\mathrm{Sets})$ be a contravariant functor. An open subfunctor $F'$ of $F$ is a representable morphism $f: F' \to F$ that is an open immersion ( Definition 8.6). By definition this means that for every $S$-scheme $X$ and for every $S$-morphism $g: X \to F$ the second projection $f_{(X)} : F' \times_F X \to X$ is an open immersion of schemes. A family $( f_i : F_i \to F ) _{ i\in I}$ of open subfunctors is called a Zariski open covering of $F$, if for every $S$-scheme $X$ and every $S$-morphism $g:X \to F$ the images of the $(f_i)_{(X)} : F_i \times_F X \to X$ form a covering of $X$. Let me explain what covering of $X$ means as follows. Since each $f_i$ is representable, $F_i \times_F X$ is a representable functor so that there are isomorphisms $\zeta_i : Z_i \xrightarrow{\cong} F_i \times_F X$. By composing with the $(f_i)_{(X)}$, we have scheme morphisms $l_i : Z_i \to X$. Then covering of $X$ means $\bigcup_{i\in I} l_i(Z_i) =X$. Definition ( Zariski sheaves ). Let $F : ( \mathrm{Sch}/S)^{\mathrm{opp}} \to ( \mathrm{(Sets)}$ be a functor. If $j: U \to X$ is an open immersion of $S$-schemes and $\xi \in F(X)$, we write, like for sections of presheaves, simply $\xi|_U$ instead of $F(j)(\xi)$. We say that $F$ is a Zariski sheaf ( on $(\mathrm{Sch}/S)$) if the usual sheaf axioms are satisfied, that is, for every $S$-scheme $X$ and for every open covering $X= \bigcup_{i \in I}U_i$ we have : Given $\xi_i \in F(U_i) $ for all $i\in I$ such that $\xi_{i}|_{(U_i \cap U_j)} = \xi_{j}|_{(U_i \cap U_j)}$ for all $i,j \in I$, there exists a unique element $\xi \in F(X)$ such that $\xi|_{U_i} =\xi_i$ for all $i\in I$. Q. My question is, let $S$ be a fixed scheme and let $F : ( \mathrm{Sch}/S )^{\mathrm{opp}} \to (\mathrm{Sets})$ be a Zariski sheaf with $S = \bigcup_{i\in I} U_i$ an affine open covering. Then $(p_i : F|_{U_i} := F \times_S U_i \to F )_{i\in I}$ foms a Zariski open covering of $F$? Here $p_i$ are first projections. Unwinding definitions, next are true? 1) The fiber product $F|_{U_i} := F \times_S U_i$ is well-defined? ; i.e., does there a natural morphism $F \to S$? 2) Each $p_i$ is open subfuntor ; i.e., $p_i$ is representable ( I can't show this at all ) and open immersion ? Similarly as an open immersion of schemes is stable under base change, we can try to consider that $p_i$ maybe open subfunctor. But is it really works? 3) The family $(p_i)_{i\in I}$ really forms a Zariski open covering of $F$? Above definitions are somewhat complex to grasp.
Step-by-Step Explanation
What We're Solving
You're working on a beautiful and sophisticated problem in algebraic geometry! You want to understand whether, given a Zariski sheaf $F$ on $(\text{Sch}/S)$ and an affine open covering $S = \bigcup_{i \in I} U_i$, the family $(p_i : F|_{U_i} \to F)_{i \in I}$ forms a Zariski open covering of $F$. This involves three key questions about fiber products, representability, and the covering property.
The Approach
This is a fundamental technique in algebraic geometry called "reduction to the affine case." The idea is that if we can show representability locally on affine opens, and these affines give us a nice covering, then we can often glue the local results to get global representability. Think of it like proving something about a manifold by working in coordinate charts!
Your approach is exactly right - we need to carefully verify each part of the definition to make sure our construction works.
Step-by-Step Solution
Let me walk you through each of your three questions:
Question 1: Does $F \times_S U_i$ make sense?
Yes! Here's why: Since $F$ is a functor on $(\text{Sch}/S)$, there's an implicit morphism $F \to S$ built into the structure. More precisely, for any $S$-scheme $X$ with structure morphism $X \to S$, we have $F(X)$ as a set, and the functoriality respects the $S$-structure.
The fiber product $F \times_S U_i$ is well-defined as a functor: for any $S$-scheme $T$, $$(F \times_S U_i)(T) = F(T) \times_{S(T)} U_i(T)$$
This construction respects the categorical framework you're working in.
Question 2: Are the $p_i$ open subfunctors?
This is the trickiest part! Let's break it down:
- Representability of $p_i$: This is NOT automatic and depends on properties of $F$. If $F$ itself has nice representability properties (like being representable or having representable diagonal), then $p_i$ might inherit representability. But in general, you need additional hypotheses.
- Open immersion property: Even if $p_i$ is representable, showing it's an open immersion requires careful analysis. The intuition is that $F|_{U_i}$ should "sit inside" $F$ like $U_i$ sits inside $S$, but the functor setting makes this more subtle.
Question 3: Do they form a Zariski open covering?
If questions 1 and 2 work out, then this becomes a question about whether the images of the induced morphisms cover each scheme $X$ mapping to $F$.
Here's the intuition: Since $F$ is a Zariski sheaf and $S = \bigcup U_i$, the sheaf property should ensure that the local pieces $F|_{U_i}$ "cover" $F$ in the appropriate sense.
The Answer (Framework for Moving Forward)
Your questions reveal that this isn't a straightforward application of definitions - you need additional hypotheses! Here's a structured approach:
- 1. Clarify your assumptions: What do you know about $F$? Is it representable? Quasi-representable? Does it have representable diagonal?
- 2. Check representability systematically: For question 2, try to use specific properties of your functor $F$ to show the $p_i$ are representable.
- 3. Use the sheaf property: The fact that $F$ is a Zariski sheaf should be crucial for question 3 - this is where the "gluing" happens.
- 4. Look for similar results: Check if Görtz-Wedhorn or other references have lemmas about base change preserving representability properties.
Memory Tip
Think of this like covering a curved surface with flat patches (charts). The functor $F$ is like your curved object, the affine opens $U_i$ are your flat patches, and you're checking that the "restricted" functors $F|_{U_i}$ properly cover $F$ just like the $U_i$ cover $S$. The Zariski sheaf property is what makes the "gluing" work!
Keep working through this systematically - you're asking exactly the right questions! The key is often finding the right additional hypotheses that make everything work smoothly.
This explanation was generated by AI. While we work hard to be accurate, mistakes can happen! Always double-check important answers with your teacher or textbook.

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📷 Problem detected:
Solve: 2x + 5 = 13
Step 1:
Subtract 5 from both sides...
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