# How to reduce to affine case to determine whether a given functor is representable

[Definition] These are the contents of Gortz and Wedhorn , Algebraic Geometry.

1. $$\widehat{(Sch)}$$ is the category of functors $$(Sch)^{opp} \rightarrow (Sets)$$

2. For scheme $$X$$, define the functor $$h_X := Hom(\_\_ , X)$$ and identify $$X$$ with $$h_X$$ in $$\widehat{(Sch)}$$

3. A morphism $$f : F \rightarrow G$$ of functors in $$\widehat{(Sch)}$$ is called representable if for all schemes $$X$$ and all morphisms $$g : X \rightarrow G$$ in $$\widehat{(Sch)}$$ the functor $$F \; {\times_G} X$$ is representable

4. An open subfunctor $$F'$$ of $$F$$ is a representable morphism $$f:F' \rightarrow F$$ that is an open immersion.

5. A family $$(f_i:F_i \rightarrow F)_{i \in I}$$ of open subfunctors is called a Zariski open covering of $$F$$ if for every $$S$$-scheme $$X$$ and every $$S$$-morphism $$g:X \rightarrow F$$ the images of the $$(f_i)_{(X)}$$ form a covering of $$X$$, where $$(f_i)_{(X)} : F_i \; \times_F X \rightarrow X$$ is the second projection map of fiber product.

6. [Theorem 8.9] Let $$S$$ be a scheme and $$F:(Sch/S)^{opp} \rightarrow (Sets)$$ be a functor such that
(i) $$F$$ is a sheaf for the Zariski topology
(ii) $$F$$ has a Zariski open covering $$(f_i:F_i \rightarrow F)_{i \in I}$$ by representable functors $$F_i$$.
Then $$F$$ is representable.

7. [Theorem 11.1] Let $$X$$ be a scheme and let $$\mathscr{R}$$ be a quasi-coherent $$\mathscr{O}_{X}$$-algebra. Then there exsist an $$X$$-scheme $$\text{Spec}(\mathscr{R})$$ which is represents the functor $$F: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_X-Alg)}(\mathscr{R},f_{*}\mathscr{O}_T)$$

[Question]

In the proof of the theorem 11.1 , they say that " To show that $$F$$ is representable, by theorem 8.9 we may assume that $$X=\text{Spec}{A}$$ is affine."

I wonder how I can exactly take a family of open subfunctors $$(f_i:F_i \rightarrow F)_{i \in I}$$ so that we can assume $$X$$ is affine.

Let $$X = \bigcup U_i$$ and $$U_i = \text{Spec}A_i$$. $$F_i: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_{U_i}-Alg)}(\mathscr{R}\vert_{U_i},f_{*}\mathscr{O}_T)$$

Then, are these $$F_i$$ open subfunctors of $$F$$ ?

I can't prove this $$(f_i:F_i \rightarrow F)_{i \in I}$$ is a Zariski open covering of $$F$$

• I have a question. Does every $F_i$ representable? If so, why? To apply the theorem 8.9, it seems to need the representability of $F_i$. Is it true? Commented Dec 5, 2021 at 9:12

Here's a rewritten answer for better comprehension. I also apologise for the delay in responding, I was very busy for various reasons.

The answer below is long, but fills in most of the details.

Objective- We want to show that given a scheme $$X$$ and $$\mathcal{B}$$ a quasi-coherent $$\mathcal{O}_X$$-algebra, we want to show that the contravariant functor on $$\operatorname{Sch}_{X}$$ which sends any $$X$$-scheme $$f\colon T\to X$$ to the set $$F(T)\colon= \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},f_*\mathcal{O}_T)$$ is in fact representable.

I will leave the verification that this functor is a Zariski sheaf as an exercise.

Observation 1- Let's note that this functor is affinely representable. This means that if $$X=\operatorname{Spec}(A)$$ for some ring $$A$$ and $$B$$ an algebra over $$A$$, then the functor on $$\operatorname{Sch}_X\to \operatorname{Sets}$$ which sends $$\big(f\colon T\to X \big)\mapsto \operatorname{Hom}_{A-\text{Alg}}(B,\Gamma(T,\mathcal{O}_T))$$ is representable. In this case it is representable by $$\operatorname{Spec}(B)$$.

This is Exercise $$2.4$$ in Hartshorne in the aboslute case (i.e. when $$A=\mathbf{Z}$$) and Prop $$3.4$$ in Gortz-Wedhorn in the relative case. You can also probably find it somewhere in the Stacks Project.

Observation 2- Now let $$X$$ be a general scheme and let $$j\colon U \hookrightarrow X$$ be an affine open subscheme of $$X$$.

Let us try to understand the category $$\operatorname{Sch}_U$$ in terms of $$\operatorname{Sch}_X$$. Note that there is a functor $$j_!\colon \operatorname{Sch}_U\to \operatorname{Sch}_X$$ which sends every $$U$$-scheme $$g\colon S\to U$$ to an $$X$$-scheme via the composition $$j\circ g\colon S\to U\hookrightarrow X.$$

Conversely, every $$X$$ scheme $$f\colon T\to X$$ for which the structure morphism factors as $$T\to U\hookrightarrow X$$ is canonically a $$U$$-scheme. Thus $$j_!\colon \operatorname{Sch}_U\to \operatorname{Sch}_X$$ embeds $$\operatorname{Sch}_U$$ as a full subcategory of $$\operatorname{Sch}_X$$.

Observation 3- Now note that since $$\mathcal{B}$$ is a quasi-coherent $$\mathcal{O}_X$$-algebra, it's restriction $$\mathcal{B}|_U$$ to $$U$$ is a quasi-coherent $$\mathcal{O}_U$$-algebra. Since $$U$$ is affine $$\mathcal{B}|_{U}=\Gamma(U,\mathcal{B})$$ which is an algebra over $$\Gamma(U,\mathcal{O}_U)$$.

So consider the functor $$F_U(-)\colon \operatorname{Sch}^{op}_U\to \operatorname{Sets}$$ which sends $$\big(g\colon S\to U \big)\to \operatorname{Hom}_{\mathcal{O}_U-\text{Alg}}(\mathcal{B}|_{U},g_*\mathcal{O}_S).$$

By Observation $$1$$, this functor is representable by $$\operatorname{Spec}(\Gamma(U,\mathcal{B}))$$ since $$U$$ is affine.

Composing $$F_U$$ with $$j_!$$ of Observation $$2$$, we see that $$F_U\circ j_!\colon \operatorname{Sch}^{op}_U\to \operatorname{Sch}^{op}_X\to \operatorname{Sets}$$ is a functor on the full subcategory $$\operatorname{Sch}_U$$ of $$\operatorname{Sch}_X$$ and we extend it to a functor on all of $$\operatorname{Sch}_X$$ by sending everything not in $$\operatorname{Sch}_U$$ to $$\emptyset$$.

This defines a functor on $$\operatorname{Sch}_X$$ which we again denote by abuse of notation as $$F_U\colon \operatorname{Sch}^{op}_X\to \operatorname{Set}.$$

Observation 4-

There is an adujunction morphism $$\eta\colon \mathcal{B}\to j_*j^{*}\mathcal{B}=j_*\mathcal{B}|_U.$$ On an open set $$V\subset X$$, the morphism is the restriction morphism from $$\eta_V\colon \Gamma(V,\mathcal{B})\to \Gamma(U\cap V,\mathcal{B})$$. This is a morphism of algebras since $$\mathcal{B}$$ is a sheaf of algebras. In particular, for any $$\mathcal{O}_U$$ algebra $$\mathcal{A}$$, and a morphism of $$\varphi \colon \mathcal{B}|_U\to \mathcal{A}$$ of $$\mathcal{O}_U$$ algebras, the composition $$\mathcal{B}\xrightarrow{\eta} j_*\mathcal{B}|_U\xrightarrow{j_*\varphi}j_*\mathcal{A}$$ is a morphism of $$\mathcal{O}_X$$-algebras, and so you get an induced morphism

$$\eta^*\colon\operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(j_*\mathcal{B}|_U,j_*\mathcal{A} )\to \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},j_*\mathcal{A})$$ by pre-composition. Note that this is functorial in $$\mathcal{A}$$ i.e. $$\eta^*$$ is a natural transformation between two functors in the variable $$\mathcal{A}$$.

But note that any $$\mathcal{O}_X$$-algebra morphism $$j_*\mathcal{B}|_U\to j_*\mathcal{A}$$ is just an $$\mathcal{O}_U$$-algebra morphism since $$j\colon U\hookrightarrow X$$ is an open embedding. So in fact the morphism we get is $$\eta^*\colon \operatorname{Hom}_{\mathcal{O}_U-\text{Alg}}(\mathcal{B}|_U,\mathcal{A} )\to \operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},j_*\mathcal{A}).$$

Observation 5- Note that the morphism $$\eta^*$$ is a monomorphism of functors.

Indeed suppose $$\alpha,\beta\colon \mathcal{B}|_U\to \mathcal{A}$$ are morphisms of $$\mathcal{O}_U$$-algebras which satisfy $$\alpha\circ \eta=\beta\circ \eta$$, then in particular at stalks for all $$x\in U$$ they must satisfy $$\alpha_x=\beta_x$$ and so $$\alpha=\beta$$.

Thus we see that $$\eta^*$$ as defined in Observation $$4$$ is a monomorphism.

Observation 6- Now using the definition of $$F_U$$ in Observation $$3$$ and $$F$$ as in Objective, we get a monomorphism of functors $$\eta^*\colon F_U\to F.$$

Indeed for any scheme $$f\colon T\to X$$, we have two conditions. Either $$f$$ factors through $$U$$ or it doesn't. In case it does, then what we discussed above shows that $$\eta^*_T\colon F_U(T)\to F(T)$$ is an injection. In case it doesn't, then $$F_U(T)$$ is $$\emptyset$$ and so tautologically a subfunctor of $$F(T)$$.

Observation 7- Let $$g\colon S\to X$$ be any $$X$$-scheme. An $$S$$-valued point of $$F$$ is an object of $$F(S)=\operatorname{Hom}_{\mathcal{O}_X-\text{Alg}}(\mathcal{B},g_*\mathcal{O}_S)$$ i.e. a morphism $$\varphi\colon \mathcal{B}\to g_*\mathcal{O}_S$$. By Yoneda this is the data of a morphism $$\varphi^*\colon h_S\to F$$.

So we get a morphism of functors $$\eta^*\colon F_U\to F$$ and $$\varphi^*\colon h_S\to F$$ whose fiber product is $$F_U\times_F h_S$$ in the category of presheaves on $$\operatorname{Sch}_X$$.

Observation 8- Since limits can be computed pointwise, let $$f\colon T\to X$$ be an object of $$\operatorname{Sch}_X$$ and so $$F_U\times_F h_S(T)=F_U(T)\times_{F(T)}h_S(T).$$

Assuming $$F_U(T)\neq \emptyset$$, the object on the right consists of pairs $$(\alpha, h)$$ where $$\alpha\colon \mathcal{B}|_U\to f_*\mathcal{O}_T$$ and $$h\colon T\to S$$ an $$X$$ morphism (i.e. $$g\circ h=f$$). These pairs have to satisfy the condition that they agree over $$F(T)$$ i.e. they define the same $$T$$ point of $$F$$ via the natural maps $$\varphi^*\colon h_S\to F$$ and $$\eta^*\colon F_U\to F$$.

More explicitly, this is the condition that the composition $$\mathcal{B}\xrightarrow{\varphi} g_*\mathcal{O}_S\xrightarrow{F(h)}f_*\mathcal{O}_T$$ agrees with the composition $$\mathcal{B}\xrightarrow{\eta}\mathcal{B}|_U\xrightarrow{\alpha}f_*\mathcal{O}_T.$$

Observation 9- We need to show now that the functor $$F_U\times_F h_S$$ is representable by an open subscheme of $$S$$.

It is enough to consider the open subscheme $$V\hookrightarrow S$$ so that $$h\colon T\to S$$ factors as $$T\to V\hookrightarrow S$$ if and only if $$f\colon T\to X$$ factors as $$T\to U\hookrightarrow X$$. Explicitly this subscheme is the preimage $$V\colon=g^{-1}U$$ where $$g\colon S\to X$$ was the structure morphism.

Indeed any $$T$$ point of $$h_V$$ i.e. a morphism $$f \colon T\to V$$ would define a $$T$$ point of $$h_S$$ by composition with the open immersion $$V\hookrightarrow S$$. Moreover, it defines a $$T$$ point of $$F_U$$ by definition of $$V$$. I leave it as an exercise in unwinding definitions to check that these $$T$$-points agree on $$F(T)$$.

Thus the pullback is representable by $$V$$ as defined above.

Observation 10- If $$\bigcup_i U_i$$ is a Zariski cover of $$X$$ by affine opens, then for each $$i$$ the functor $$F_i(-) \colon=\operatorname{Hom}_{\mathcal{O}_{U_i}-\text{Alg}}(\mathcal{B}|_{U_i}, )$$ satisfies the conditions above.

The $$V_i\colon= g^{-1}U_i$$ constructed as above cover $$S$$ for each $$X$$-scheme $$g\colon S\to X$$. Thus these functors form a Zariski open covering of the functor $$F$$ defined as in Objective.

• Why the morphisms of functors $\alpha_i^*:F_i\to F$ you constructed above are monomorphisms? equivalently why for every $f: T \to X$ we have injection $F_i(T) \subset F(T)$? Commented Sep 23, 2021 at 22:10
• also I not understand why for every $f: T \to X$ we have $F_i(T)\neq \emptyset$ iff $f$ factors through $U_i$? Commented Sep 23, 2021 at 22:11
• Hmm, I just saw this right now. Let me get back to it over the weekend. Commented Nov 15, 2021 at 22:43
• I have a question. For each $F_i$, you mean $F_i : (Sch/X)^{opp} \to Sets, (f:T \to X) \mapsto \operatorname{Hom}_{(\mathcal{O}_{U_i}-Alg)}(\mathcal{R}|_{U_i}, f_{*}\mathcal{O}_{T}|_{U_i})$ ? But why these functor are representable? Commented Dec 8, 2021 at 2:09
• @Plantation Observation $1$ in the edited answer above answer your question. You should also look at Observations $4,5$ and $6$. Commented Dec 18, 2021 at 22:07