## 26.15 A representability criterion

In this section we reformulate the glueing lemma of Section 26.14 in terms of functors. We recall some of the material from Categories, Section 4.3. Recall that given a scheme $X$ we can define a functor

$h_ X : \mathit{Sch}^{opp} \longrightarrow \textit{Sets}, \quad T \longmapsto \mathop{Mor}\nolimits (T, X).$

This is called the functor of points of $X$.

Let $F$ be a contravariant functor from the category of schemes to the category of sets. In a formula

$F : \mathit{Sch}^{opp} \longrightarrow \textit{Sets}.$

We will use the same terminology as in Sites, Section 7.2. Namely, given a scheme $T$, an element $\xi \in F(T)$, and a morphism $f : T' \to T$ we will denote $f^*\xi$ the element $F(f)(\xi )$, and sometimes we will even use the notation $\xi |_{T'}$

Definition 26.15.1. (See Categories, Definition 4.3.6.) Let $F$ be a contravariant functor from the category of schemes to the category of sets (as above). We say that $F$ is representable by a scheme or representable if there exists a scheme $X$ such that $h_ X \cong F$.

Suppose that $F$ is representable by the scheme $X$ and that $s : h_ X \to F$ is an isomorphism. By Categories, Yoneda Lemma 4.3.5 the pair $(X, s : h_ X \to F)$ is unique up to unique isomorphism if it exists. Moreover, the Yoneda lemma says that given any contravariant functor $F$ as above and any scheme $Y$, we have a bijection

$\mathop{Mor}\nolimits _{\text{Fun}(\mathit{Sch}^{opp}, \textit{Sets})} (h_ Y, F) \longrightarrow F(Y), \quad s \longmapsto s(\text{id}_ Y).$

Here is the reverse construction. Given any $\xi \in F(Y)$ the transformation of functors $s_\xi : h_ Y \to F$ associates to any morphism $f : T \to Y$ the element $f^*\xi \in F(T)$.

In particular, in the case that $F$ is representable, there exists a scheme $X$ and an element $\xi \in F(X)$ such that the corresponding morphism $h_ X \to F$ is an isomorphism. In this case we also say the pair $(X, \xi )$ represents $F$. The element $\xi \in F(X)$ is often called the “universal family” for reasons that will become more clear when we talk about algebraic stacks (insert future reference here). For the moment we simply observe that the fact that if the pair $(X, \xi )$ represents $F$, then every element $\xi ' \in F(T)$ for any $T$ is of the form $\xi ' = f^*\xi$ for a unique morphism $f : T \to X$.

Example 26.15.2. Consider the rule which associates to every scheme $T$ the set $F(T) = \Gamma (T, \mathcal{O}_ T)$. We can turn this into a contravariant functor by using for a morphism $f : T' \to T$ the pullback map $f^\sharp : \Gamma (T, \mathcal{O}_ T) \to \Gamma (T', \mathcal{O}_{T'})$. Given a ring $R$ and an element $t \in R$ there exists a unique ring homomorphism $\mathbf{Z}[x] \to R$ which maps $x$ to $t$. Thus, using Lemma 26.6.4, we see that

$\mathop{Mor}\nolimits (T, \mathop{\mathrm{Spec}}(\mathbf{Z}[x])) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}[x], \Gamma (T, \mathcal{O}_ T)) = \Gamma (T, \mathcal{O}_ T).$

This does indeed give an isomorphism $h_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])} \to F$. What is the “universal family” $\xi$? To get it we have to apply the identifications above to $\text{id}_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])}$. Clearly under the identifications above this gives that $\xi = x \in \Gamma (\mathop{\mathrm{Spec}}(\mathbf{Z}[x]), \mathcal{O}_{\mathop{\mathrm{Spec}}(\mathbf{Z}[x])}) = \mathbf{Z}[x]$ as expected.

Definition 26.15.3. Let $F$ be a contravariant functor on the category of schemes with values in sets.

1. We say that $F$ satisfies the sheaf property for the Zariski topology if for every scheme $T$ and every open covering $T = \bigcup _{i \in I} U_ i$, and for any collection of elements $\xi _ i \in F(U_ i)$ such that $\xi _ i|_{U_ i \cap U_ j} = \xi _ j|_{U_ i \cap U_ j}$ there exists a unique element $\xi \in F(T)$ such that $\xi _ i = \xi |_{U_ i}$ in $F(U_ i)$.

2. A subfunctor $H \subset F$ is a rule that associates to every scheme $T$ a subset $H(T) \subset F(T)$ such that the maps $F(f) : F(T) \to F(T')$ maps $H(T)$ into $H(T')$ for all morphisms of schemes $f : T' \to T$.

3. Let $H \subset F$ be a subfunctor. We say that $H \subset F$ is representable by open immersions if for all pairs $(T, \xi )$, where $T$ is a scheme and $\xi \in F(T)$ there exists an open subscheme $U_\xi \subset T$ with the following property:

• A morphism $f : T' \to T$ factors through $U_\xi$ if and only if $f^*\xi \in H(T')$.

4. Let $I$ be a set. For each $i \in I$ let $H_ i \subset F$ be a subfunctor. We say that the collection $(H_ i)_{i \in I}$ covers $F$ if and only if for every $\xi \in F(T)$ there exists an open covering $T = \bigcup U_ i$ such that $\xi |_{U_ i} \in H_ i(U_ i)$.

In condition (4), if $H_ i \subset F$ is representable by open immersions for all $i$, then to check $(H_ i)_{i \in I}$ covers $F$, it suffices to check $F(T) = \bigcup H_ i(T)$ whenever $T$ is the spectrum of a field.

Lemma 26.15.4. Let $F$ be a contravariant functor on the category of schemes with values in the category of sets. Suppose that

1. $F$ satisfies the sheaf property for the Zariski topology,

2. there exists a set $I$ and a collection of subfunctors $F_ i \subset F$ such that

1. each $F_ i$ is representable,

2. each $F_ i \subset F$ is representable by open immersions, and

3. the collection $(F_ i)_{i \in I}$ covers $F$.

Then $F$ is representable.

Proof. Let $X_ i$ be a scheme representing $F_ i$ and let $\xi _ i \in F_ i(X_ i) \subset F(X_ i)$ be the “universal family”. Because $F_ j \subset F$ is representable by open immersions, there exists an open $U_{ij} \subset X_ i$ such that $T \to X_ i$ factors through $U_{ij}$ if and only if $\xi _ i|_ T \in F_ j(T)$. In particular $\xi _ i|_{U_{ij}} \in F_ j(U_{ij})$ and therefore we obtain a canonical morphism $\varphi _{ij} : U_{ij} \to X_ j$ such that $\varphi _{ij}^*\xi _ j = \xi _ i|_{U_{ij}}$. By definition of $U_{ji}$ this implies that $\varphi _{ij}$ factors through $U_{ji}$. Since $(\varphi _{ij} \circ \varphi _{ji})^*\xi _ j =\varphi _{ji}^*(\varphi _{ij}^*\xi _ j) = \varphi _{ji}^*\xi _ i = \xi _ j$ we conclude that $\varphi _{ij} \circ \varphi _{ji} = \text{id}_{U_{ji}}$ because the pair $(X_ j, \xi _ j)$ represents $F_ j$. In particular the maps $\varphi _{ij} : U_{ij} \to U_{ji}$ are isomorphisms of schemes. Next we have to show that $\varphi _{ij}^{-1}(U_{ji} \cap U_{jk}) = U_{ij} \cap U_{ik}$. This is true because (a) $U_{ji} \cap U_{jk}$ is the largest open of $U_{ji}$ such that $\xi _ j$ restricts to an element of $F_ k$, (b) $U_{ij} \cap U_{ik}$ is the largest open of $U_{ij}$ such that $\xi _ i$ restricts to an element of $F_ k$, and (c) $\varphi _{ij}^*\xi _ j = \xi _ i$. Moreover, the cocycle condition in Section 26.14 follows because both $\varphi _{jk}|_{U_{ji} \cap U_{jk}} \circ \varphi _{ij}|_{U_{ij} \cap U_{ik}}$ and $\varphi _{ik}|_{U_{ij} \cap U_{ik}}$ pullback $\xi _ k$ to the element $\xi _ i$. Thus we may apply Lemma 26.14.2 to obtain a scheme $X$ with an open covering $X = \bigcup U_ i$ and isomorphisms $\varphi _ i : X_ i \to U_ i$ with properties as in Lemma 26.14.1. Let $\xi _ i' = (\varphi _ i^{-1})^* \xi _ i$. The conditions of Lemma 26.14.1 imply that $\xi _ i'|_{U_ i \cap U_ j} = \xi _ j'|_{U_ i \cap U_ j}$. Therefore, by the condition that $F$ satisfies the sheaf condition in the Zariski topology we see that there exists an element $\xi ' \in F(X)$ such that $\xi _ i = \varphi _ i^*\xi '|_{U_ i}$ for all $i$. Since $\varphi _ i$ is an isomorphism we also get that $(U_ i, \xi '|_{U_ i})$ represents the functor $F_ i$.

We claim that the pair $(X, \xi ')$ represents the functor $F$. To show this, let $T$ be a scheme and let $\xi \in F(T)$. We will construct a unique morphism $g : T \to X$ such that $g^*\xi ' = \xi$. Namely, by the condition that the subfunctors $F_ i$ cover $F$ there exists an open covering $T = \bigcup V_ i$ such that for each $i$ the restriction $\xi |_{V_ i} \in F_ i(V_ i)$. Moreover, since each of the inclusions $F_ i \subset F$ are representable by open immersions we may assume that each $V_ i \subset T$ is maximal open with this property. Because, $(U_ i, \xi '_{U_ i})$ represents the functor $F_ i$ we get a unique morphism $g_ i : V_ i \to U_ i$ such that $g_ i^*\xi '|_{U_ i} = \xi |_{V_ i}$. On the overlaps $V_ i \cap V_ j$ the morphisms $g_ i$ and $g_ j$ agree, for example because they both pull back $\xi '|_{U_ i \cap U_ j} \in F_ i(U_ i \cap U_ j)$ to the same element. Thus the morphisms $g_ i$ glue to a unique morphism from $T \to X$ as desired. $\square$

Remark 26.15.5. Suppose the functor $F$ is defined on all locally ringed spaces, and if conditions of Lemma 26.15.4 are replaced by the following:

1. $F$ satisfies the sheaf property on the category of locally ringed spaces,

2. there exists a set $I$ and a collection of subfunctors $F_ i \subset F$ such that

1. each $F_ i$ is representable by a scheme,

2. each $F_ i \subset F$ is representable by open immersions on the category of locally ringed spaces, and

3. the collection $(F_ i)_{i \in I}$ covers $F$ as a functor on the category of locally ringed spaces.

We leave it to the reader to spell this out further. Then the end result is that the functor $F$ is representable in the category of locally ringed spaces and that the representing object is a scheme.

Comment #4010 by Andrei Bengus-Lasnier on

In the last paragraph of the proof, I don't see why we use the open immersion property to make the $V_i$ maximal. One does not need that in order to prove they agree on overlaps.

Comment #4119 by on

The reason for this is that in the statement of Lemma 26.14.1 when we talk about morphisms from a scheme into the glueing $X$, there is a condition on the intersection of $V_i$ and $V_j$. Now it is true that this condition isn't needed to get a morphism... but that is not what the lemma says. Sigh!

So I am going to leave this as is for now. Somebody else can edit this and say it in a better way perhaps.

Comment #5002 by Xavier on

It seems that setting each $V_i$ maximal makes it easier to verify the uniqueness of $g$, since for another $f$ with $f^* \xi' = \xi$ one automatically have $f^{-1}(U_i) \subset V_i$.

Comment #5014 by Alexios Terezakis on

In the second paragraph of the last proof, third line it's $F$ instead of $T$

Comment #5238 by on

@#5002: Thanks for this comment!

@#5014: Thanks and fixed here.

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