Lemma 96.19.2. Let $S$ be a locally Noetherian scheme. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume

$\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces,

$\mathcal{X}$ has (RS*),

$\mathcal{X}$ is limit preserving.

Let $x$ be an object of $\mathcal{X}$ over a scheme $U$ of finite type over $S$. Let $u \leadsto u_0$ be a specialization of finite type points of $U$ such that $x$ is versal at $u_0$. Then $x$ is versal at $u$.

**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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