The Stacks project

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

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

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

@#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.

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 $$\begin{align*} G:\text{Sch}^\text{opp}/S&\to\text{Sets}\\ (f:T\to S)&\mapsto \operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T), \end{align*}$$ 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 $$\begin{matrix} G(f)&\longrightarrow&\text{Hom}_S(T,\operatorname{Spec}A)&\longrightarrow& G(f)\\ &\searrow&&\swarrow\\ &&\text{Hom}_{R\text{-Alg}}(A,\Gamma(T,\mathcal{O}_T)) \end{matrix}$$ commutes. But this is easy.