Suppose that \(F\) satisfies \((H_1)\), \((H_2)\) and \((H_3)\). Let \(r = \dim _k(t_F)\), \(S = \Lambda [[X_1, \ldots , X_r]]\) and \(\mathfrak {n}\) the maximal ideal of \(S\).

We first construct \(R\) as a projective limit of quotients of \(S\). More specifically we define a sequence \((J_q, \xi _q)_{q\ge 2}\) by induction where \((J_q)\) is a decreasing sequence of ideals of \(S\) such that \(R_q = S / J_q \in \mathbf{C}_\Lambda \) and \(\xi _q \in F(R_q)\) is such that \((\xi _q) \in \varprojlim F(R_q)\).

First we set \(J_2 = \mathfrak {n}^2 + \mathfrak {m}_\Lambda S\) and \(R_2 = S/J_2\).

There exists \(\xi _2 \in R_2\) inducing an isomorphism \(\operatorname{Hom}(R_2, k[\varepsilon ]) \cong t_F\). To see this, first note that \(R_2 \cong k[k^r] \cong k[\varepsilon ]^r\). By applying \((H_2)\), this yields \(F(R_2) \cong t_F^r\), so choosing a basis \((e_1, \ldots , e_r)\) of \(t_F\) gives an element \(\xi _2 \in F(R_2)\). We then check that \(u \mapsto F(u)(\xi _2)\) is an isomorphism. In fact, this map is linear and sends the morphism \(R_2 \to k[\varepsilon ]\) corresponding to the \(i\)-th projection \(k^r \to k\) to \(e_i\).

For the induction step, suppose that we have an ideal \(J_q\) such that \(R_q = S/J_q \in \mathbf{C}_\Lambda \) and \(\xi _q \in F(R_q)\). Consider the set \(\mathscr {S}\) of ideals \(J\) of \(S\) such that \(\mathfrak {n}J_q \le J \le J_q\) and \(\xi _q\) lifts to \(F(S/J)\). We will set \(J_{q+1}\) to be the minimum of \(\mathscr {S}\), with \(\xi _{q+1}\) some lift of \(\xi _q\), after proving that it exists. First note that \(\mathscr {S}\) is not empty as it contains \(J_q\), and since \(S / \mathfrak {n}J_q\) is Artinian, we know that \(\mathscr {S}\) admits a minimal element, so we only need to show that \(\mathscr {S}\) is stable by pairwise intersection to get a minimum. Let \(J\) and \(K\) be two elements of \(\mathscr {S}\). The elements of \(\mathscr {S}\) are in correspondence with \(k\)-vector subspaces of \(J/\mathfrak {n}J_q\), so we can extend \(J\) into some ideal \(J' \le J_q\) such that \(J' \cap K = J \cap K\) and \(J' + K = J_q\), and we have \(J' \in \mathscr {S}\) because \(J \le J'\). We now use the isomorphism

together with lemma 41 to conclude that \(\xi _q\) lifts to \(S/(J \cap K)\), i.e. that \(J \cap K \in \mathscr {S}\), which concludes the induction step.

We set \(J = \bigcap _{q \ge 2} J_q\) and \(R = S / J\). By definition of the \(J_q\)’s, we have \(\mathfrak {n}^q \le J_q\), so lemma 55 applies and \(\hat{F}(R) = \varprojlim F(S/J_q)\), so we can set \(\xi = \varprojlim \xi _q \in \hat{F}(R)\).

We have \(\operatorname{Hom}(R, k[\varepsilon ]) \cong \operatorname{Hom}(R_2, k[\varepsilon ])\) because any map \(S \to k[\varepsilon ]\) factorizes through \(S/J_2\), and since \(\xi \) projects to \(\xi _2\) we have \(t_R \cong t_F\) by definition of \(\xi _2\).

It remains to show that \(\nu (\xi ) : h_R \to F\) is smooth.

Using lemma 31, let \(p : A' \to A\) be a small extension, \(u : R \to A\), \(\eta ' \in F(A')\) and \(\eta \in F(A)\) such that \(\nu (\xi ) (u) = F(p)(\eta ') = \eta \). We need to find \(u' : R \to A'\) such that \(\nu (\xi ) (u') = \eta '\) and \(p \circ u' = u\).

First we reduce this to finding \(u'' : R \to A'\) such that \(p \circ u'' = u\). Indeed, if we have such a \(u''\), then \(\nu (\xi ) (u'')\) and \(\eta '\) both lie over \(\eta \), so by lemma 52, there exists \(\sigma \in F(k[I])\) sending \(\nu (\xi ) (u'')\) to \(\eta '\), where \(I = \ker p\). By using lemma 54 and the fact that \(h_R (k[I]) \cong F(k[I])\) (using lemma 21), we can also apply \(\sigma \) to \(u''\) giving us the \(u' : R \to A'\) that we wanted.

Since \(u\) factorizes through \(R/\mathfrak {m}_R^n \to A\) for some \(n\), and the topology is induced by the \(J_q\)’s (lemma 55), \(u\) factorizes through \(R_q \to A\) for some \(q\). It suffices to find \(v\) completing the following diagram:

where \(w\) is defined using the universal property of \(S = \Lambda [[X_1, \ldots , X_r]]\) to make the diagram commute. More specifically, to define \(w\) we use that \(pr_1\) is surjective so that we can first define \(w\) on the \(X_i\)’s, and since all maps are local the images of the \(X_i\)’s are in the maximal ideal so we can extend the map to \(S\). In this context we can easily show that \(pr_1\) is a small extension, so we can apply lemma 49. If \(pr_1\) admits a section then we are done. Otherwise, \(pr_1\) is essential, and \(pr_1 \circ w\) is the quotient map \(S \to R_q\), so \(w\) is surjective by lemma 48. By \((H_1)\), \(\xi _q\) lifts to \(R_q \times _A A'\), so \(\ker w \in \mathscr {S}\) and by minimality we have \(J_{q+1} \le \ker w\). Thus \(w\) factors through \(R_{q+1}\) giving the \(v\) that we wanted.

Suppose that \(F\) admits a hull \((R, \xi )\).

First since \(t_R \cong t_F\) (linearly by lemma 20), and \(h_R\) satisfies \((H_3)\), \(F\) satisfies \((H_3)\).

To check \((H_1)\), let \(f : X \to Z\) and \(g : Y \to Z\) be morphisms in \(\mathbf{C}_\Lambda \) with \(g\) surjective, and \(x \in F(X)\) and \(y \in F(Y)\) both lying over \(z \in F(Z)\). First, since \(h_R \to F\) is smooth, by lemma 32 there exists \(u_x : R \to X\) such that \(\nu (\xi ) (u_x) = x\). Then by smoothness applied to \(g\), there exists \(u_y : R \to Y\) such that \(\nu (\xi ) (u_y) = y\) and \(g \circ u_y = f \circ u_x\). Then \(\nu (\xi )(u_x \times u_y) \in F(X \times _Z Y)\) projects to \(x\) and \(y\), so \(p_{f, g}\) is surjective.

To check \((H_2)\), we now only need to check injectivity, since we have already shown \((H_1)\). Let \(A \in \mathbf{C}_\Lambda \) and \(\zeta _1\) and \(\zeta _2\) in \(F(A \times k[\varepsilon ])\) having the same projections \(a \in F(A)\) and \(e \in k[\varepsilon ]\). We proceed similarly as in the proof of \((H_1)\). First there is \(u' : R \to A\) such that \(\nu (\xi ) (u') = a\). Then by smoothness applied to the projection \(pr_1 : A \times k[\varepsilon ] \to A\), we get maps \(u_i : R \to A \times k[\varepsilon ]\) for \(i = 1, 2\) such that \(\nu (\xi ) (u_i) = \zeta _i\) and \(pr_1 \circ u_i = u'\). Since \(\nu (\xi ) (pr_2 \circ u_i) = e\) in both cases, and \(\nu (\xi )_{k[\varepsilon ]}\) is an isomorphism, we have \(u_1 = u_2\) so \(\zeta _1 = \zeta _2\).

Suppose that \(F\) satisfies \((H_1)\), \((H_2)\), \((H_3)\) and \((H_4)\). By the first point of the theorem, we know that \(F\) admits a hull \((R, \xi )\). Furthermore, since \(\nu (\xi )\) is smooth, we already know that \(\nu (\xi )_A\) is surjective for all \(A \in \mathbf{C}_\Lambda \) (lemma 32). We will show by induction on \(\operatorname{length}_\Lambda (A)\) that it is injective, using ??. If \(A = k\), then this is true by definition of a hull. Otherwise, let \(p : B \to A\) be a small extension, with \(I = \ker p\), and suppose that \(\nu (\xi )_A\) is an isomorphism. By applying lemma 53 together with lemma 54, it is easy check that \(\nu (\xi )_B\) is injective. Namely we take \(u_1\) and \(u_2\) in \(h_R(B)\) such that \(\nu (\xi ) (u_1) = \nu (\xi ) (u_2)\), and we need to show that the action of \(h_R(k[I])\) by \(0\) sends \(u_1\) to \(u_2\), which is clear from the commutative diagram in lemma 54.