Lemma 27.4.2. In Situation 27.3.1. Let $F$ be the functor associated to $(S, \mathcal{A})$ above. If $S$ is affine, then $F$ is representable by the affine scheme $\mathop{\mathrm{Spec}}(\Gamma (S, \mathcal{A}))$.

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and $A = \Gamma (S, \mathcal{A})$. Then $A$ is an $R$-algebra and $\mathcal{A} = \widetilde A$. The ring map $R \to A$ gives rise to a canonical map

$f_{univ} : \mathop{\mathrm{Spec}}(A) \longrightarrow S = \mathop{\mathrm{Spec}}(R).$

We have $f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A}$ by Schemes, Lemma 26.7.3. Hence there is a canonical map

$\varphi _{univ} : f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A} \longrightarrow \widetilde A = \mathcal{O}_{\mathop{\mathrm{Spec}}(A)}$

coming from the $A$-module map $A \otimes _ R A \to A$, $a \otimes a' \mapsto aa'$. We claim that the pair $(f_{univ}, \varphi _{univ})$ represents $F$ in this case. In other words we claim that for any scheme $T$ the map

$\mathop{\mathrm{Mor}}\nolimits (T, \mathop{\mathrm{Spec}}(A)) \longrightarrow \{ \text{pairs } (f, \varphi )\} ,\quad a \longmapsto (f_{univ} \circ a, a^*\varphi _{univ})$

is bijective.

Let us construct the inverse map. For any pair $(f : T \to S, \varphi )$ we get the induced ring map

$\xymatrix{ A = \Gamma (S, \mathcal{A}) \ar[r]^{f^*} & \Gamma (T, f^*\mathcal{A}) \ar[r]^{\varphi } & \Gamma (T, \mathcal{O}_ T) }$

This induces a morphism of schemes $T \to \mathop{\mathrm{Spec}}(A)$ by Schemes, Lemma 26.6.4.

The verification that this map is inverse to the map displayed above is omitted. $\square$

Comment #7016 by on

Just before the construction of the inverse map : is that $\varphi_{\mathrm{univ}}$ instead of $\varphi$ ?

Comment #7121 by Elías Guisado on

Who is exactly the map $A = \Gamma (S, \mathcal{A}) \xrightarrow{f^*} \Gamma (T, f^*\mathcal{A})$ in the description of the inverse?

Comment #7122 by Elías Guisado on

Is it $\mathcal{A}\to f_*f^*\mathcal{A}$, the unit of the pushforward-pullback adjunction of sheaves of modules?

Comment #7236 by on

@#7016. Thanks and fixed here.

@#7121 and #7122. Yes. Discussion. Say we have a continuous map $f : X \to Y$ of topological spaces and we have a sheaf $\mathcal{G}$ on $Y$, then we have a map $f^{-1} : \Gamma(Y, \mathcal{G}) \to \Gamma(X, f^{-1}\mathcal{G})$ often called a pullback map. Now it is indeed true that you can write $\Gamma(X, f^{-1}\mathcal{G}) = \Gamma(Y, f_*f^{-1}\mathcal{G})$ and then you can get the map by appying the adjunction map $\mathcal{G} \to f_*f^{-1}\mathcal{G}$. Another method, is to say that $\Gamma(Y, \mathcal{G}) = \text{Mor}(*, \mathcal{G})$, use the fact that $f^{-1}$ is a functor, and use that $f^{-1}* = *$. Here $*$ is the singleton sheaf AKA the final object of the category of sheaves of sets. These constructions give the same thing.

Of course, if we have a map of ringed spaces and $\mathcal{G}$ is a sheaf of modules, then we also have a pullback map $\Gamma(Y, \mathcal{G}) \to \Gamma(X, f^*\mathcal{G})$. It can be constructed by either method discussed above or its existence can be deduced from the construction of the map for sheaves of sets.

Comment #8436 by on

I would propose writing the whole proof in this way: I'm going to adopt the POV of representability of $G$ instead of $F$, see https://stacks.math.columbia.edu/tag/01LQ#comment-8435 . We have $\mathcal{A}=\widetilde{A_R}$ (where $A_R$ just means $A$ regarded as an $R$-module, notation introduced in the paragraph before 10.14.3). To show that $\pi:\operatorname{Spec} A\to\operatorname{Spec} R$ represents we check that the identity map $\varphi_\text{univ}:\widetilde{A_R}\to\pi_*\widetilde{A}=\widetilde{A_R}$ is a universal element for $G$. Fix $f:T\to S$. We have a map $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$, induced by $\varphi_\text{univ}$. To construct the inverse, for an $\mathcal{O}_S$-linear map $\varphi:\widetilde{A_R}\to f_*\mathcal{O}_T$ (an element of $G(f)$), we have a map on global sections $A\to\Gamma(T,\mathcal{O}_T)$, which is $R$-linear. Therefore, it induces a morphism $T\to\operatorname{Spec}A$ over $\operatorname{Spec} R$, by Schemes, 26.6.4. On the one hand, it is easy to see that the composite $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)\to\text{Hom}_S(T,\operatorname{Spec}A)$ is the identity (use uniqueness in Schemes, 26.6.4). On the other hand, to see that $G(f)\to \text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$ is the identity, we note that all the involved maps are compatible with restrictions to distinguished open subsets of $S$ (i.e., given $U\subset S$ distinguished open, one restricts to $f^{-1}(U)\to U$ and to $\pi^{-1}(U)\to U$). Hence, it suffices to see that commutes. But this is easy.

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