# Not able to make sense of gluing via the functor of points in EGA

I am trying to use a well known result of Grothendieck to show that if $$S$$ is a scheme, and $$\mathcal{B}$$ is a quasi coherent sheaf of $$\mathcal{O}_{S}$$-algebras, then there is a relative affine spectrum which is a scheme over $$S$$.

Denote by $$\text{Sch}_{/S}$$ the category of schemes over $$S$$. I am trying to use the result which states that for a functor $$F: \text{Sch}_{/S} \longrightarrow \text{Set}$$ to be representable, it is necessary and sufficient that it be a Zariski sheaf which is covered by representable open subfunctors.

Grothendieck proves a corollary of this (EGA, 0.4.5.5) which states that if $$\{S_{i} \}$$ is an affine (or indeed any open) cover of $$S$$ and $$F$$ is a Zariski sheaf, then we need only show that the functors $$F_{i} = F \times_{h_{S}} h_{S_{i}}$$ are representable.

My problem is that I have no idea what this pullback diagram even is. The functor $$h_{S}$$ is (presumably) the functor which sends any $$S$$-scheme $$(f: X \rightarrow S)$$ to the set with one element, namely $$\{ f \}$$. But then the $$h_{S_{i}}$$ certainly are not going to be subfunctors of $$h_{S}$$, and by extension the $$F_{i}$$ will not be subfunctors of $$F$$, let alone open subfunctors.

What am I missing here? What is $$h_{S}$$ actually supposed to be? If I assume that it's an abuse of notation and that $$h_{S}$$ is really the functor that sends a scheme (not an $$S$$-scheme) $$X$$ to the set $$\text{Hom}(X, S)$$, then what is the morphism of functors $$F \rightarrow h_{S}$$ supposed to be?

• Luke, was there something else you needed clarified about this problem? – Alex Youcis Apr 7 at 20:04

I think what you're looking for is the following. Let $$F$$ be a sheaf on $$X$$ and let $$\{U_i\}$$ be an open cover of $$X$$. Then, we know by the definition of a sheaf that the diagram

$$0\to F\to \prod_i (\iota_i)_\ast(F\mid_{U_i})\to \prod_{ij}(\iota_{i,j})_\ast(F\mid_{U_i\cap U_j})\qquad (1)$$

is exact where $$\iota_i:U_i\hookrightarrow X$$ and $$\iota_{ij}:U_i\cap U_j\hookrightarrow X$$.

Suppose now that $$F\mid_{U_i}\cong h_{S_i}$$ for some $$U_i$$-schemes $$S_i$$. The isomorphisms $$(F\mid_{U_i})\mid_{U_j}\cong (F\mid_{U_j})\mid_{U_i}$$ give rise to isomorphisms $$\varphi_{ij}:(S_i)_{U_i\cap U_j}\xrightarrow{\approx} (S_j)\mid_{U_i\cap U_j}$$. We then get a $$U$$-scheme $$S$$ (e.g. see Tag01JA). Then, check that $$h_S$$ sits in the same exact sequence as in $$(1)$$ (e.g. see the description of the functor of points as in Lemma 25.14.1 of loc. cit.).

I guess technically I only addressed the question on the small Zariski site of $$X$$, but you can get the claim on the big Zariski site by just noting that if you have any map $$f:X'\to X$$ you can pull the arguments back to the small Zariski site of $$X'$$ using the open cover $$f^{-1}(U_i)$$ and proceeding in the same way.

• Thank you for the response. I'm still not entirely sure how that would give a natural transformation $F \rightarrow h_{S}$. I should also clarify that $h_{S}$ is really an abuse of notation for $h_{(S, \text{id}_{S})}$, right? – Luke Apr 7 at 2:19
• @Luke To show that they have the same description of points functorially? Also, what does $h_{(S,\id)}$ mean? Be careful, I just realized that I switched notation from your post. In my language it's the fact that we've constructed an $X$-scheme $S$, let's say with structure morphism $f:S\to X$, and thus $h_S$ means $h_{(S,f)}$ if I'm interpreting your question correctly. – Alex Youcis Apr 7 at 2:22
• – Alex Youcis Apr 7 at 2:24