Lemma 85.2.1. Choose a category of schemes $\mathit{Sch}_\alpha$ as in Sets, Lemma 3.9.2. Given a formal scheme $\mathfrak X$ let

$h_\mathfrak X : (\mathit{Sch}_\alpha )^{opp} \longrightarrow \textit{Sets},\quad h_\mathfrak X(S) = \mathop{Mor}\nolimits _{\textit{Formal Schemes}}(S, \mathfrak X)$

be its functor of points. Then we have

$\mathop{Mor}\nolimits _{\textit{Formal Schemes}}(\mathfrak X, \mathfrak Y) = \mathop{Mor}\nolimits _{\textit{PSh}(\mathit{Sch}_\alpha )}(h_\mathfrak X, h_\mathfrak Y)$

provided the size of $\mathfrak X$ is not too large.

Proof. First we observe that $h_\mathfrak X$ satisfies the sheaf property for the Zariski topology for any formal scheme $\mathfrak X$ (see Schemes, Definition 26.15.3). This follows from the local nature of morphisms in the category of formal schemes. Also, for an open immersion $\mathfrak V \to \mathfrak W$ of formal schemes, the corresponding transformation of functors $h_\mathfrak V \to h_\mathfrak W$ is injective and representable by open immersions (see Schemes, Definition 26.15.3). Choose an open covering $\mathfrak X = \bigcup \mathfrak U_ i$ of a formal scheme by affine formal schemes $\mathfrak U_ i$. Then the collection of functors $h_{\mathfrak U_ i}$ covers $h_\mathfrak X$ (see Schemes, Definition 26.15.3). Finally, note that

$h_{\mathfrak U_ i} \times _{h_\mathfrak X} h_{\mathfrak U_ j} = h_{\mathfrak U_ i \cap \mathfrak U_ j}$

Hence in order to give a map $h_\mathfrak X \to h_\mathfrak Y$ is equivalent to giving a family of maps $h_{\mathfrak U_ i} \to h_\mathfrak Y$ which agree on overlaps. Thus we can reduce the bijectivity (resp. injectivity) of the map of the lemma to bijectivity (resp. injectivity) for the pairs $(\mathfrak U_ i, \mathfrak Y)$ and injectivity (resp. nothing) for $(\mathfrak U_ i \cap \mathfrak U_ j, \mathfrak Y)$. In this way we reduce to the case where $\mathfrak X$ is an affine formal scheme. Say $\mathfrak X = \text{Spf}(A)$ for some admissible topological ring $A$. Also, choose a fundamental system of ideals of definition $I_\lambda \subset A$.

We can also localize on $\mathfrak Y$. Namely, suppose that $\mathfrak V \subset \mathfrak Y$ is an open formal subscheme and $\varphi : h_\mathfrak X \to h_\mathfrak Y$. Then

$h_\mathfrak V \times _{h_\mathfrak Y, \varphi } h_\mathfrak X \to h_\mathfrak X$

is representable by open immersions. Pulling back to $\mathop{\mathrm{Spec}}(A/I_\lambda )$ for all $\lambda$ we find an open subscheme $U_\lambda \subset \mathop{\mathrm{Spec}}(A/I_\lambda )$. However, for $I_\lambda \subset I_\mu$ the morphism $\mathop{\mathrm{Spec}}(A/I_\lambda ) \to \mathop{\mathrm{Spec}}(A/I_\mu )$ pulls back $U_\mu$ to $U_\lambda$. Thus these glue to give an open formal subscheme $\mathfrak U \subset \mathfrak X$. A straightforward argument (omitted) shows that

$h_\mathfrak U = h_\mathfrak V \times _{h_\mathfrak Y} h_\mathfrak X$

In this way we see that given an open covering $\mathfrak Y = \bigcup \mathfrak V_ j$ and a transformation of functors $\varphi : h_\mathfrak X \to h_\mathfrak Y$ we obtain a corresponding open covering of $\mathfrak X$. Since $\mathfrak X$ is affine, we can refine this covering by a finite open covering $\mathfrak X = \mathfrak U_1 \cup \ldots \cup \mathfrak U_ n$ by affine formal subschemes. In other words, for each $i$ there is a $j$ and a map $\varphi _ i : h_{\mathfrak U_ i} \to h_{\mathfrak V_ j}$ such that

$\xymatrix{ h_{\mathfrak U_ i} \ar[r]_{\varphi _ i} \ar[d] & h_{\mathfrak V_ j} \ar[d] \\ h_{\mathfrak X} \ar[r]^\varphi & h_\mathfrak Y }$

commutes. With a few additional arguments (which we omit) this implies that it suffices to prove the bijectivity of the lemma in case both $\mathfrak X$ and $\mathfrak Y$ are affine formal schemes.

Assume $\mathfrak X$ and $\mathfrak Y$ are affine formal schemes. Say $\mathfrak X = \text{Spf}(A)$ and $\mathfrak Y = \text{Spf}(B)$. Let $\varphi : h_\mathfrak X \to h_\mathfrak Y$ be a transformation of functors. Let $I_\lambda \subset A$ be a fundamental system of ideals of definition. The canonical inclusion morphism $i_\lambda : \mathop{\mathrm{Spec}}(A/I_\lambda ) \to \mathfrak X$ maps to a morphism $\varphi (i_\lambda ) : \mathop{\mathrm{Spec}}(A/I_\lambda ) \to \mathfrak Y$. By (85.2.0.1) this corresponds to a continuous map $\chi _\lambda : B \to A/I_\lambda$. Since $\varphi$ is a transformation of functors it follows that for $I_\lambda \subset I_\mu$ the composition $B \to A/I_\lambda \to A/I_\mu$ is equal to $\chi _\mu$. In other words we obtain a ring map

$\chi = \mathop{\mathrm{lim}}\nolimits \chi _\lambda : B \longrightarrow \mathop{\mathrm{lim}}\nolimits A/I_\lambda = A$

This is a continuous homomorphism because the inverse image of $I_\lambda$ is open for all $\lambda$ (as $A/I_\lambda$ has the discrete topology and $\chi _\lambda$ is continuous). Thus we obtain a morphism $\text{Spf}(\chi ) : \mathfrak X \to \mathfrak Y$ by (85.2.0.1). We omit the verification that this construction is the inverse to the map of the lemma in this case.

Set theoretic remarks. To make this work on the given category of schemes $\mathit{Sch}_\alpha$ we just have to make sure all the schemes used in the proof above are isomorphic to objects of $\mathit{Sch}_\alpha$. In fact, a careful analysis shows that it suffices if the schemes $\mathop{\mathrm{Spec}}(A/I_\lambda )$ occurring above are isomorphic to objects of $\mathit{Sch}_\alpha$. For this it certainly suffices to assume the size of $\mathfrak X$ is at most the size of a scheme contained in $\mathit{Sch}_\alpha$. $\square$

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