We prove that versality is preserved under generalizations for stacks which have (RS*) and are limit preserving. We suggest skipping this section on a first reading.

**Proof.**
Let $R = \mathcal{O}_{U, u}^\wedge $. Let $k = \kappa (u)$ be the residue field of $R$. Let $\xi $ be the formal object of $\mathcal{X}$ over $R$ associated to $x$. Since $x$ is not versal at $u$, we see that $\xi $ is not versal, see Lemma 96.12.3. By the discussion following Definition 96.12.1 this means we can find morphisms $\xi _1 \to x_ A \to x_ B$ of $\mathcal{X}$ lying over closed immersions $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$ where $A, B$ are Artinian local rings with residue field $k$, an $n \geq 1$ and a commutative diagram

\[ \vcenter { \xymatrix{ & x_ A \ar[ld] \\ \xi _ n & \xi _1 \ar[u] \ar[l] } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & \mathop{\mathrm{Spec}}(A) \ar[ld] \\ \mathop{\mathrm{Spec}}(R/\mathfrak m^ n) & \mathop{\mathrm{Spec}}(k) \ar[u] \ar[l] } } \]

such that there does **not** exist an $m \geq n$ and a commutative diagram

\[ \vcenter { \xymatrix{ & & x_ B \ar[lldd] \\ & & x_ A \ar[ld] \ar[u] \\ \xi _ m & \xi _ n \ar[l] & \xi _1 \ar[u] \ar[l] } } \quad \text{lying over} \vcenter { \xymatrix{ & & \mathop{\mathrm{Spec}}(B) \ar[lldd] \\ & & \mathop{\mathrm{Spec}}(A) \ar[ld] \ar[u] \\ \mathop{\mathrm{Spec}}(R/\mathfrak m^ m) & \mathop{\mathrm{Spec}}(R/\mathfrak m^ n) \ar[l] & \mathop{\mathrm{Spec}}(k) \ar[u] \ar[l] } } \]

We may moreover assume that $B \to A$ is a small extension, i.e., that the kernel $I$ of the surjection $B \to A$ is isomorphic to $k$ as an $A$-module. This follows from Formal Deformation Theory, Remark 88.8.10. Then we simply define

\[ T = U \amalg _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(B) \]

By property (RS*) we find $y$ over $T$ whose restriction to $\mathop{\mathrm{Spec}}(B)$ is $x_ B$ and whose restriction to $U$ is $x$ (this gives the arrow $x \to y$ lying over $U \to T$). To finish the proof we verify conditions (1), (2), and (3).

By the construction of the pushout we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & I \ar[r] & B \ar[r] & A \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & \Gamma (T, \mathcal{O}_ T) \ar[r] \ar[u] & \Gamma (U, \mathcal{O}_ U) \ar[r] \ar[u] & 0 } \]

with exact rows. This immediately proves (1) and (2). To finish the proof we will argue by contradiction. Assume we have a pair $(W, \beta )$ as in (3). Since $\mathop{\mathrm{Spec}}(B) \to T$ factors through $W$ we get the morphism

\[ x_ B \to y|_ W \xrightarrow {\beta } x \]

Since $B$ is Artinian local with residue field $k = \kappa (u)$ we see that $x_ B \to x$ lies over a morphism $\mathop{\mathrm{Spec}}(B) \to U$ which factors through $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}/\mathfrak m_ u^ m)$ for some $m \geq n$. In other words, $x_ B \to x$ factors through $\xi _ m$ giving a map $x_ B \to \xi _ m$. The compatibility condition on the morphism $\alpha $ in condition (3) translates into the condition that

\[ \xymatrix{ x_ B \ar[d] & x_ A \ar[d] \ar[l] \\ \xi _ m & \xi _ n \ar[l] } \]

is commutative. This gives the contradiction we were looking for.
$\square$

**Proof.**
After shrinking $U$ we may assume $U$ is affine and $U$ maps into an affine open $\mathop{\mathrm{Spec}}(\Lambda )$ of $S$. If $x$ is not versal at $u$ then we may pick $x \to y$ lying over $U \to T$ as in Lemma 96.19.1. Write $U = \mathop{\mathrm{Spec}}(R_0)$ and $T = \mathop{\mathrm{Spec}}(R)$. The morphism $U \to T$ corresponds to a surjective ring map $R \to R_0$ whose kernel is an ideal of square zero. By assumption (3) we get that $y$ comes from an object $x'$ over $U' = \mathop{\mathrm{Spec}}(R')$ for some finite type $\Lambda $-subalgebra $R' \subset R$. After increasing $R'$ we may and do assume that $R' \to R_0$ is surjective, so that $U \subset U'$ is a first order thickening. Thus we now have

\[ x \to y \to x' \text{ lying over } U \to T \to U' \]

By assumption (1) there is an algebraic space $Z$ over $S$ representing

\[ (\mathit{Sch}/U)_{fppf} \times _{x, \mathcal{X}, x'} (\mathit{Sch}/U')_{fppf} \]

see Algebraic Stacks, Lemma 92.10.11. By construction of $2$-fibre products, a $V$-valued point of $Z$ corresponds to a triple $(a, a', \alpha )$ consisting of morphisms $a : V \to U$, $a' : V \to U'$ and a morphism $\alpha : a^*x \to (a')^*x'$. We obtain a commutative diagram

\[ \xymatrix{ U \ar[rd] \ar[rdd] \ar[rrd] \\ & Z \ar[r]_{p'} \ar[d]^ p & U' \ar[d] \\ & U \ar[r] & S } \]

The morphism $i : U \to Z$ comes the isomorphism $x \to x'|_ U$. Let $z_0 = i(u_0) \in Z$. By Lemma 96.12.6 we see that $Z \to U'$ is smooth at $z_0$. After replacing $U$ by an affine open neighbourhood of $u_0$, replacing $U'$ by the corresponding open, and replacing $Z$ by the intersection of the inverse images of these opens by $p$ and $p'$, we reach the situation where $Z \to U'$ is smooth along $i(U)$. Since $u \leadsto u_0$ the point $u$ is in this open. Condition (3) of Lemma 96.19.1 is clearly preserved by shrinking $U$ (all of the schemes $U$, $T$, $U'$ have the same underlying topological space). Since $U \to U'$ is a first order thickening of affine schemes, we can choose a morphism $i' : U' \to Z$ such that $p' \circ i' = \text{id}_{U'}$ and whose restriction to $U$ is $i$ (More on Morphisms of Spaces, Lemma 74.19.6). Pulling back the universal morphism $p^*x \to (p')^*x'$ by $i'$ we obtain a morphism

\[ x' \to x \]

lying over $p \circ i' : U' \to U$ such that the composition

\[ x \to x' \to x \]

is the identity. Recall that we have $y \to x'$ lying over the morphism $T \to U'$. Composing we get a morphism $y \to x$ whose existence contradicts condition (3) of Lemma 96.19.1. This contradiction finishes the proof.
$\square$

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