Lemma 89.13.2. Let $\mathcal{F}$ be a predeformation category satisfying (S1) and (S2). Let $\xi$ be a formal object of $\mathcal{F}$ corresponding to $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F}$, see Remark 89.7.12. Then $\xi$ is versal if and only if the following two conditions hold:

1. the map $d\underline{\xi } : T\underline{R}|_{\mathcal{C}_\Lambda } \to T\mathcal{F}$ on tangent spaces is surjective, and

2. given a diagram in $\widehat{\mathcal{F}}$

$\vcenter { \xymatrix{ & y \ar[d] \\ \xi \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & B \ar[d]^{f} \\ R \ar[r] & A } }$

in $\widehat{\mathcal{C}}_\Lambda$ with $B \to A$ a small extension of Artinian rings, then there exists a ring map $R \to B$ such that

$\xymatrix{ & B \ar[d]^{f} \\ R \ar[ur] \ar[r] & A }$

commutes.

Proof. If $\xi$ is versal then (1) holds by Lemma 89.8.8 and (2) holds by Remark 89.8.10. Assume (1) and (2) hold. By Remark 89.8.10 we must show that given a diagram in $\widehat{\mathcal{F}}$ as in (2), there exists $\xi \to y$ such that

$\xymatrix{ & y \ar[d] \\ \xi \ar[ur] \ar[r] & x }$

commutes. Let $b : R \to B$ be the map guaranteed by (2). Denote $y' = b_*\xi$ and choose a factorization $\xi \to y' \to x$ lying over $R \to B \to A$ of the given morphism $\xi \to x$. By (S1) we obtain a commutative diagram

$\vcenter { \xymatrix{ z \ar[r] \ar[d] & y \ar[d] \\ y' \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B \times _ A B \ar[d] \ar[r] & B \ar[d]^{f} \\ B \ar[r]^{f} & A . } }$

Set $I = \mathop{\mathrm{Ker}}(f)$. Let $\overline{g} : B \times _ A B \to k[I]$ be the ring map $(u, v) \mapsto \overline{u} \oplus (v - u)$, cf. Lemma 89.10.8. By (1) there exists a morphism $\xi \to \overline{g}_*z$ which lies over a ring map $i : R \to k[\epsilon ]$. Choose an Artinian quotient $b_1 : R \to B_1$ such that both $b : R \to B$ and $i : R \to k[\epsilon ]$ factor through $R \to B_1$, i.e., giving $h : B_1 \to B$ and $i' : B_1 \to k[\epsilon ]$. Choose a pushforward $y_1 = b_{1, *}\xi$, a factorization $\xi \to y_1 \to y'$ lying over $R \to B_1 \to B$ of $\xi \to y'$, and a factorization $\xi \to y_1 \to \overline{g}_*z$ lying over $R \to B_1 \to k[\epsilon ]$ of $\xi \to \overline{g}_*z$. Applying (S1) once more we obtain

$\vcenter { \xymatrix{ z_1 \ar[r] \ar[d] & z \ar[r] \ar[d] & y \ar[d] \\ y_1 \ar[r] & y' \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B_1 \times _ A B \ar[d] \ar[r] & B \times _ A B \ar[r] \ar[d] & B \ar[d]^{f} \\ B_1 \ar[r] & B \ar[r] & A . } }$

Note that the map $g : B_1 \times _ A B \to k[I]$ of Lemma 89.10.8 (defined using $h$) is the composition of $B_1 \times _ A B \to B \times _ A B$ and the map $\overline{g}$ above. By construction there exists a morphism $y_1 \to g_*z_1 \cong \overline{g}_*z$! Hence Lemma 89.10.8 applies (to the outer rectangles in the diagrams above) to give a morphism $y_1 \to y$ and precomposing with $\xi \to y_1$ gives the desired morphism $\xi \to y$. $\square$

Comment #2639 by Xiaowen Hu on

Set $I=\ker(k)$ should be $I=\ker(f)$.I=Ker(k)

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