The Stacks project

Proof. Let $B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda $. Let $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{G}(B))$, $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$, and $y \to \varphi (x)$ be a morphism lying over $B \to A$. By Lemma 90.3.3 we can factor $B \to A$ into small extensions $B = B_ n \to B_{n-1} \to \ldots \to B_0 = A$. We argue by induction on $n$. If $n = 1$ the result is true by assumption. If $n > 1$, then denote $f : B = B_ n \to B_{n - 1}$ and denote $g : B_{n - 1} \to B_0 = A$. Choose a pushforward $y \to f_* y$ of $y$ along $f$, so that the morphism $y \to \varphi (x)$ factors as $y \to f_* y \to \varphi (x)$. By the induction hypothesis we can find $x_{n - 1} \to x$ lying over $g : B_{n - 1} \to A$ and $a : \varphi (x_{n - 1}) \to f_*y$ lying over $\text{id} : B_{n - 1} \to B_{n - 1}$ such that

commutes. We can apply the assumption to the composition $y \to \varphi (x_{n - 1})$ of $y \to f_*y$ with $a^{-1} : f_*y \to \varphi (x_{n - 1})$. We obtain $x_ n \to x_{n - 1}$ lying over $B_ n \to B_{n - 1}$ and $\varphi (x_ n) \to y$ lying over $\text{id} : B_ n \to B_ n$ so that the diagram

commutes. Then the composition $x_ n \to x_{n - 1} \to x$ and $\varphi (x_ n) \to y$ are the morphisms required by the definition of smoothness. $\square$

Proof. Assume $S$ is a power series ring over $R$. Say $S = R[[x_1, \ldots , x_ n]]$. Smoothness of $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ means the following (see Remark 90.8.4): Given a surjective ring map $B \to A$ in $\mathcal{C}_\Lambda $, a ring map $R \to B$, a ring map $S \to A$ such that the solid diagram

is commutative then a dotted arrow exists making the diagram commute. (Note the similarity with Algebra, Definition 10.138.1.) To construct the dotted arrow choose elements $b_ i \in B$ whose images in $A$ are equal to the images of $x_ i$ in $A$. Note that $b_ i \in \mathfrak m_ B$ as $x_ i$ maps to an element of $\mathfrak m_ A$. Hence there is a unique $R$-algebra map $R[[x_1, \ldots , x_ n]] \to B$ which maps $x_ i$ to $b_ i$ and which can serve as our dotted arrow.

Conversely, assume $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. Let $x_1, \ldots , x_ n \in S$ be elements whose images form a basis in the relative cotangent space $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ of $S$ over $R$. Set $T = R[[X_1, \ldots , X_ n]]$. Note that both

and

Let $S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the local $R$-algebra isomorphism given by mapping the class of $x_ i$ to the class of $X_ i$. Let $f_1 : S \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the composition $S \to S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$. The assumption that $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth means we can lift $f_1$ to a map $f_2 : S \to T/\mathfrak {m}_ T^2$, then to a map $f_3 : S \to T/\mathfrak {m}_ T^3$, and so on, for all $n \geq 1$. Thus we get an induced map $f : S \to T = \mathop{\mathrm{lim}}\nolimits T/\mathfrak m_ T^ n$ of local $R$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to \mathfrak m_ T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ of relative cotangent spaces. Hence $f$ is surjective by Lemma 90.4.2 (where we think of $f$ as a map in $\widehat{\mathcal{C}}_ R$). Choose preimages $y_ i \in S$ of $X_ i \in T$ under $f$. As $T$ is a power series ring over $R$ there exists a local $R$-algebra homomorphism $s : T \to S$ mapping $X_ i$ to $y_ i$. By construction $f \circ s = \text{id}$. Then $s$ is injective. But $s$ induces an isomorphism on relative cotangent spaces since $f$ does, so it is also surjective by Lemma 90.4.2 again. Hence $s$ and $f$ are isomorphisms. $\square$

Proof. Statements (1) and (2) follow immediately from the definitions. Proof of (3) omitted. Hints: use the formulation of smoothness given in Remark 90.8.3 and use that $\mathcal{F} \times _\mathcal {G} \mathcal{G}'$ is the $2$-fibre product, see Remarks 90.5.2 (13). $\square$

Proof. Let $y$ be an object of $\mathcal{G}$ lying over $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$. Let $y \to y_0$ be a pushforward of $y$ along $A \to k$. By the assumption on essential surjectivity of $\varphi : \mathcal{F}(k) \to \mathcal{G}(k)$ there exist an object $x_0$ of $\mathcal{F}$ lying over $k$ and an isomorphism $y_0 \to \varphi (x_0)$. Smoothness of $\varphi $ implies there exists an object $x$ of $\mathcal{F}$ over $A$ whose image $\varphi (x)$ is isomorphic to $y$. Thus $\varphi : \mathcal{F} \to \mathcal{G}$ is essentially surjective.

Let $\eta = (R, \eta _ n, g_ n)$ be an object of $\widehat{\mathcal{G}}$. We construct an object $\xi $ of $\widehat{\mathcal{F}}$ with an isomorphism $\eta \to \varphi (\xi )$. By the assumption on essential surjectivity of $\varphi : \mathcal{F}(k) \to \mathcal{G}(k)$, there exists a morphism $\eta _1 \to \varphi (\xi _1)$ in $\mathcal{G}(k)$ for some $\xi _1 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. The morphism $\eta _2 \xrightarrow {g_1} \eta _1 \to \varphi (\xi _1)$ lies over the surjective ring map $R/\mathfrak m_ R^2 \to k$, hence by smoothness of $\varphi $ there exists $\xi _2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(R/\mathfrak m_ R^2))$, a morphism $f_1: \xi _2 \to \xi _1$ lying over $R/\mathfrak m_ R^2 \to k$, and a morphism $\eta _2 \to \varphi (\xi _2)$ such that

commutes. Continuing in this way we construct an object $\xi = (R, \xi _ n, f_ n)$ of $\widehat{\mathcal{F}}$ and a morphism $\eta \to \varphi (\xi ) = (R, \varphi (\xi _ n), \varphi (f_ n))$ in $\widehat{\mathcal{G}}(R)$. $\square$

Proof. By assumption the morphism $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F}$ is smooth. Then $\iota (\xi ) : \underline{R} \to \widehat{\mathcal{F}}$ is the completion of $\underline{\xi }$, see Remark 90.7.12. By Lemma 90.8.8 there exists an object $f$ of $\underline{R}$ such that $\iota (\xi )(f) = \eta $. Then $f$ is a ring map $f : R \to S$ in $\widehat{\mathcal{C}}_\Lambda $. And $\iota (\xi )(f) = \eta $ means that $f_*\xi \cong \eta $ which means exactly that there is a morphism $\xi \to \eta $ lying over $f$. $\square$