Lemma 90.28.1. Being formally homotopic is an equivalence relation on sets of morphisms in $\widehat{\mathcal{C}}_\Lambda $.
90.28 Uniqueness of versal rings
Given $R, S$ in $\widehat{\mathcal{C}}_\Lambda $ we say maps $f, g : R \to S$ are formally homotopic if there exists an $r \geq 0$ and maps $h : R \to R[[t_1, \ldots , t_ r]]$ and $k : R[[t_1, \ldots , t_ r]] \to S$ in $\widehat{\mathcal{C}}_\Lambda $ such that for all $a \in R$ we have
$h(a) \bmod (t_1, \ldots , t_ r) = a$,
$f(a) = k(a)$,
$g(a) = k(h(a))$.
We will say $(r, h, k)$ is a formal homotopy between $f$ and $g$.
Proof. Suppose we have any $r \geq 1$ and two maps $h_1, h_2 : R \to R[[t_1, \ldots , t_ r]]$ such that $h_1(a) \bmod (t_1, \ldots , t_ r) = h_2(a) \bmod (t_1, \ldots , t_ r) = a$ for all $a \in R$ and a map $k : R[[t_1, \ldots , t_ r]] \to S$. Then we claim $k \circ h_1$ is formally homotopic to $k \circ h_2$. The symmetric inherent in this claim will show that our notion of formally homotopic is symmetric. Namely, the map
\[ \Psi : R[[t_1, \ldots , t_ r]] \longrightarrow R[[t_1, \ldots , t_ r]],\quad \sum a_ I t^ I \longmapsto \sum h_1(a_ I)t^ I \]
is an isomorphism. Set $h(a) = \Psi ^{-1}(h_2(a))$ for $a \in R$ and $k' = k \circ \Psi $, then we see that $(r, h, k')$ is a formal homotopy between $k \circ h_1$ and $k \circ h_2$, proving the claim
Say we have three maps $f_1, f_2, f_3 : R \to S$ as above and a formal homotopy $(r_1, h_1, k_1)$ between $f_1$ and $f_2$ and a formal homotopy $(r_2, h_2, k_2)$ between $f_3$ and $f_2$ (!). After relabeling the coordinates we may assume $h_2 : R \to R[[t_{r_1 + 1}, \ldots , t_{r_1 + r_2}]]$ and $k_2 : R[[t_{r_1 + 1}, \ldots , t_{r_1 + r_2}]] \to S$. By choosing a suitable isomorphism
\[ R[[t_1, \ldots , t_{r_1 + r_2}]] \longrightarrow R[[t_{r_1 + 1}, \ldots , t_{r_1 + r_2}]] \widehat{\otimes }_{h_2, R, h_1} R[[t_1, \ldots , t_{r_1}]] \]
we may fit these maps into a commutative diagram
\[ \xymatrix{ R \ar[r]_{h_1} \ar[d]^{h_2} & R[[t_1, \ldots , t_{r_1}]] \ar[d]^{h_2'} \\ R[[t_{r_1 + 1}, \ldots , t_{r_1 + r_2}]] \ar[r]^{h_1'} & R[[t_1, \ldots , t_{r_1 + r_2}]] } \]
with $h_2'(t_ i) = t_ i$ for $1 \leq i \leq r_1$ and $h_1'(t_ i) = t_ i$ for $r_1 + 1 \leq i \leq r_2$. Some details omitted. Since this diagram is a pushout in the category $\widehat{\mathcal{C}}_\Lambda $ (see proof of Lemma 90.4.3) and since $k_1 \circ h_1 = f_2 = k_2 \circ h_2$ we conclude there exists a map
\[ k : R[[t_1, \ldots , t_{r_1 + r_2}]] \to S \]
with $k_1 = k \circ h_2'$ and $k_2 = k \circ h_1'$. Denote $h = h_1' \circ h_2 = h_2' \circ h_1$. Then we have
$k(h_1'(a)) = k_2(a) = f_3(a)$, and
$k(h_2'(a)) = k_1(a) = f_1(a)$.
By the claim in the first paragraph of the proof this shows that $f_1$ and $f_3$ are formally homotopic. $\square$
Lemma 90.28.2. In the category $\widehat{\mathcal{C}}_\Lambda $, if $f_1, f_2 : R \to S$ are formally homotopic and $g : S \to S'$ is a morphism, then $g \circ f_1$ and $g \circ f_2$ are formally homotopic.
Proof. Namely, if $(r, h, k)$ is a formal homotopy between $f_1$ and $f_2$, then $(r, h, g \circ k)$ is a formal homotopy between $g \circ f_1$ and $g \circ f_2$. $\square$
Lemma 90.28.3. Let $\mathcal{F}$ be a deformation category over $\mathcal{C}_\Lambda $ with $\dim _ k T\mathcal{F} < \infty $ and $\dim _ k \text{Inf}(\mathcal{F}) < \infty $. Let $\xi $ be a versal formal object lying over $R$. Let $\eta $ be a formal object lying over $S$. Then any two maps such that $f_*\xi \cong \eta \cong g_*\xi $ are formally homotopic.
\[ f, g : R \to S \]
Proof. By Theorem 90.26.4 and its proof, $\mathcal{F}$ has a presentation by a smooth prorepresentable groupoid
\[ (\underline{R}, \underline{R_1}, s, t, c, e, i)|_{\mathcal{C}_\Lambda } \]
in functors on $\mathcal{C}_\lambda $ such that $\mathcal{F}$. Then the maps $s : R \to R_1$ and $t : R \to R_1$ are formally smooth ring maps and $e : R_1 \to R$ is a section. In particular, we can choose an isomorphism $R_1 = R[[t_1, \ldots , t_ r]]$ for some $r \geq 0$ such that $s$ is the embedding $R \subset R[[t_1, \ldots , t_ r]]$ and $t$ corresponds to a map $h : R \to R[[t_1, \ldots , t_ r]]$ with $h(a) \bmod (t_1, \ldots , t_ r) = a$ for all $a \in R$. The existence of the isomorphism $\alpha : f_*\xi \to g_*\xi $ means exactly that there is a map $k : R_1 \to S$ such that $f = k \circ s$ and $g = k \circ t$. This exactly means that $(r, h, k)$ is a formal homotopy between $f$ and $g$. $\square$
Lemma 90.28.4. In the category $\widehat{\mathcal{C}}_\Lambda $, if $f_1, f_2 : R \to S$ are formally homotopic and $\mathfrak p \subset R$ is a minimal prime ideal, then $f_1(\mathfrak p)S = f_2(\mathfrak p)S$ as ideals.
Proof. Suppose $(r, h, k)$ is a formal homotopy between $f_1$ and $f_2$. We claim that $\mathfrak pR[[t_1, \ldots , t_ r]] = h(\mathfrak p)R[[t_1, \ldots , t_ r]]$. The claim implies the lemma by further composing with $k$. To prove the claim, observe that the map $\mathfrak p \mapsto \mathfrak pR[[t_1, \ldots , t_ r]]$ is a bijection between the minimal prime ideals of $R$ and the minimal prime ideals of $R[[t_1, \ldots , t_ r]]$. Finally, $h(\mathfrak p)R[[t_1, \ldots , t_ r]]$ is a minimal prime as $h$ is flat, and hence of the form $\mathfrak q R[[t_1, \ldots , t_ r]]$ for some minimal prime $\mathfrak q \subset R$ by what we just said. But since $h \bmod (t_1, \ldots , t_ r) = \text{id}_ R$ by definition of a formal homotopy, we conclude that $\mathfrak q = \mathfrak p$ as desired. $\square$
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