Is representability of Zariski sheaves local on the base?

Let $F: \mathsf{Sch_{/S}}^{op} \to \mathsf{Set}$ be a Zariski sheaf on the category of $S$-schemes. $F$ being a sheaf means it satisfies the following property:

Sheaf condition: For every $S$-scheme $X$ and every open cover $\{U_j\} \subset Open_S(X)$ of $X$ by open $S$-subschemes. The following diagram is an equalizer:

$$F(X) \rightarrow \prod_{i} F(U_i) {{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i, j} F(U_i \times_X U_j)$$

The following theorem gives a necessary and sufficient condition for $F$ to be representable by an $S$-Scheme:

Theorem: $F$ is representable by an $S$-scheme iff $F$ has an open cover by representable subfunctors.

This is pretty satisfying, but let's try something even bolder.

Is representability local on the base? : Let $F: \mathsf{Sch_{/S}}^{op} \to \mathsf{Set}$ be a Zariski sheaf and $\{U_j\} \subset Open(S)$ an open cover of $S$ satisfying that all pullbacks $F \times_S U_j$ are representable. Must $F$ be representable?

The just glue attitude doesn't seem to work for me here. I've played for hours with pullback cubes and the like without getting anywhere.

Unless I'm misunderstanding something (which I probably do) this version of "locality" is used implicitly in a lot of arguments I've encountered. Is it really true? If so, why?

If not, why not and is there perhaps a different locality principle which does hold?

• Well, $F \times_S U_j$ is an open subsheaf of $F$, so you get the desired open cover... – Zhen Lin Jan 11 '16 at 21:59
• @ZhenLin This is precisely the statement I'm trying to prove for hours now. It should be a play with pullback diagrams, but it gets very confusing. – Saal Hardali Jan 11 '16 at 22:00
• That should be more or less immediate with the standard definition. Do you know the pullback pasting lemma? – Zhen Lin Jan 11 '16 at 22:02
• @ZhenLin No :(. I had the feeling that I'm missing some piece of abstract nonsense... – Saal Hardali Jan 11 '16 at 22:03
• Well, it will be useful for showing that the pullback of an open subsheaf is an open subsheaf. – Zhen Lin Jan 11 '16 at 22:07

Since $F\times_S U_j$ is representable, say by a scheme $X_j$ over $U_j$, then for any $i,j$, $X_j|_{U_i\cap U_j}$ is an open subscheme of $X_j$, and $X_i|_{U_i\cap U_j}$ is an open subscheme of $X_i$. Since they're both pullbacks of $F$ along the same morphism $U_i\cap U_j\rightarrow S$, they are "uniquely" isomorphic, and the associated isomorphisms for triples of indices $i,j,k$ must satisfy the cocycle condition (again by uniqueness of isomorphisms), so now you have a family of schemes $X_i$ and various open subschemes and isomorphisms between the open subschemes, so you're in the classical gluing situation (see, for example, Hartshorne chapter II, Exercise 2.12), from which you can construct a scheme $X$ over $S$, whose functor of points must coincide with $F$ because of "the fully faithfulness of pullback functors of descent data".
• This is what I thought worked until I realized I don't know how to show that $F \times_S U_j \to F$ is an open immersion. I just need to show that for every pullback we get an open immersion. But that's when I get confused by monstrous 3-dimensional diagrams... I asked my advisor today and he told me something along these lines. I'm still trying to fill the gaps. – Saal Hardali Jan 12 '16 at 20:35
• The explanation in my answer doesn't need to show that $F\times_S U_j\rightarrow F$ is an open immersion. Nonetheless, to show that it's an open immersion, it suffices to show that for any $S$-scheme $X$ and morphism $X\rightarrow F$, the induced map $X|_{U_j}\rightarrow F|_{U_j}$ is the pullback of $X\rightarrow F$ along $F|_{U_j}\rightarrow F$. This seems pretty straightforward to prove just by universal properties of fiber products. – oxeimon Jan 12 '16 at 20:40