4 Lemmas
We now give some additional lemmas needed for the proof.
4.1 Essential morphisms
A surjective morphism \(p : B \to A\) in \(\mathbf{C}_\Lambda \) is essential if for any morphism \(q : C \to B\) such that \(q \circ p\) is surjective, \(q\) is surjective.
Let \(p : B \to A\) be essential and \(q : C \to B\) be a morphism in \(\hat{\mathbf{C}}_\Lambda \) such that \(q \circ p\) is surjective. Then \(q\) is surjective.
This is trivial considering that \(q\) factorizes through \(q_n : C/\mathfrak {m}_C^n \to B\) for some \(n\) which is a morphism in \(\mathbf{C}_\Lambda \) (lemma 14).
If \(p : B \to A\) is a small extension, then \(p\) is not essential if and only if \(p\) admits a section, that is there exists a morphism \(s : A \to B\) such that \(p \circ s = \mathbb {1}_A\).
If \(p\) has a section \(s\), then \(p\) is not essential since \(s\) is not surjective as we have \(\operatorname{length}_\Lambda (B) = \operatorname{length}_\Lambda (A) + 1\).
Now, suppose that \(p\) is not essential, and let \(q : C \to B\) be such that \(q \circ p\) is surjective but \(q\) is not. Let \(C'\) be the range of \(q\). We have \(\operatorname{length}_\Lambda (C') {\lt} \operatorname{length}_\Lambda (B)\) since \(q\) is not surjective so \(\operatorname{length}_\Lambda (C') \le \operatorname{length}_\Lambda (A)\), and \(\operatorname{length}_\Lambda (C') \ge \operatorname{length}_\Lambda (A)\) because \(p|_{C'}\) is surjective (because \(q \circ p\) is). Thus \(p|_{C'}\) is bijective which gives the section that we wanted.
4.2 Action of a small extension
Let \(p : B \to A\) be a small extension in \(\mathbf{C}_\Lambda \), and write \(I = \ker p\).
There is an isomorphism \(B \times k[I] \cong B \times _A B\) whose first component is the identity.
It is given by \((x, r + y) \mapsto (x, x + y)\). The inverse map is \((x, y) \mapsto (x, x_0 + y - x)\) where \(x_0\) is the \(k\) residue of \(x\). It is easy to check that this defines an isomorphism.
Using this isomorphism, we can get the following result as in [ :
Let \(F : \mathbf{C}_\Lambda \to Sets\) such that \(F(k)\) is a singleton and \(F\) satisfies \((H_2)\). Using the isomorphism from lemma 50 together with the comparison maps (eq. 1), we get a map
This defines an action of \(F(k[I])\) on \(F(B)\), with orbits contained in the fibers of \(p\).
This is simply a formal check. In practice we only need the fact that \(0\) acts trivially, so there is less to check.
The condition \((H_1)\) implies that the map in lemma 51 is surjective, i.e. the action is transitive on fibers of \(p\).
The condition \((H_4)\) implies that the map in lemma 51 is bijective, i.e. the fibers of \(p\) are principal homogeneous spaces under \(F(k[I])\).
The action defined in lemma 51 is functorial: if \(G\) is another functor with the necessary conditions, and \(\eta : F \to G\) is a natural transformation, then the following diagram commutes:
![\begin{tikzcd}
{G(B) \times G(k[I])} \arrow[rr] & & G(B) \times_{G(A)} G(B) \\
{F(B) \times F(k[I])} \arrow[rr] \arrow[u] & & F(B) \times_{F(A)} F(B) \arrow[u]
\end{tikzcd}](images/img-0001.svg)
4.3 Topology
We need the following lemma [ :
Let \(R\) be an object of \(\hat{\mathbf{C}}_\Lambda \), and let \((J_n)_{n\in \mathbb {N}}\) be a decreasing sequence of ideals of \(R\) such that \(\mathfrak {m}_R^n \le J_n\). Set \(J = \bigcap _{n\in \mathbb {N}} J_n\). Then the sequence \((J_n/J)_{n\in \mathbb {N}}\) defines the \(\mathfrak {m}_{R/J}\)-adic topology on \(R/J\).
We need to show that for any \(n \in \mathbb {N}\), there exists \(N\in \mathbb {N}\) such that \(J_N \le J + \mathfrak {m}_R^n\). For every \(k \in \mathbb {N}\), \(R / \mathfrak {m}_R^k\) is Artinian, so the decreasing sequence \((J_i + \mathfrak {m}_R^k)_{i \in \mathbb {N}}\) stabilizes. We write \(N_k\) for some index such that \(J_{N_k} + \mathfrak {m}_R^k\) is the minimum of the sequence. Note that we have \(J_{N_k} + \mathfrak {m}_R^k \le J_k + \mathfrak {m}_R^k\) so since we also have \(\mathfrak {m}_R^k \le J_k\) we get that \(J_{N_k} \le J_k\). We claim that \(J_{N_n} \le J + \mathfrak {m}_R^n\). Let \(x \in J_{N_n}\). We define a Cauchy sequence \((x_i)\) in \(R\) with \(x_i \in J_{N_i}\) for \(i \ge n\) by induction, by first setting \(x_i = x\) if \(i \le n\). If \(i {\gt} n\), then there exists \(x_{i+1} \in J_{N_{i+1}}\) such that \(x_i - x_{i+1} \in \mathfrak {m}_R ^i\): indeed if \(N_{i+1} \le N_i\) we can take \(x_{i+1}= x_i\), and otherwise we have \(J_{N_{i+1}} + \mathfrak {m}_R^i = J_{N_i} + \mathfrak {m}_R^i\) (note that have \(x_i \in J_{N_i}\)). Since \(R\) is complete, this sequence admits a limit \(y\). We have \(y \in J_{N_i} + \mathfrak {m}_R^i\) for all \(i\) so \(y \in J\) and \(y - x = y - x_n \in \mathfrak {m}_R ^n\), so \(x \in J + \mathfrak {m}_R^n\).
In particular, this means that if \(J \ne R\), we have \(\hat{F}(R/J) \cong \varprojlim F(R / J_n)\).