Lemma 90.10.5. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$.

1. If $\mathcal{F}$ satisfies (S1), then so does $\overline{\mathcal{F}}$.

2. If $\mathcal{F}$ satisfies (S2), then so does $\overline{\mathcal{F}}$ provided at least one of the following conditions is satisfied

1. $\mathcal{F}$ is a predeformation category,

2. the category $\mathcal{F}(k)$ is a set or a setoid, or

3. for any morphism $x_\epsilon \to x_0$ of $\mathcal{F}$ lying over $k[\epsilon ] \to k$ the pushforward map $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective.

Proof. Assume $\mathcal{F}$ has (S1). Suppose we have ring maps $f_ i : A_ i \to A$ in $\mathcal{C}_\Lambda$ with $f_2$ surjective. Let $x_ i \in \mathcal{F}(A_ i)$ such that the pushforwards $f_{1, *}(x_1)$ and $f_{2, *}(x_2)$ are isomorphic. Then we can denote $x$ an object of $\mathcal{F}$ over $A$ isomorphic to both of these and we obtain a diagram as in (S1). Hence we find an object $y$ of $\mathcal{F}$ over $A_1 \times _ A A_2$ whose pushforward to $A_1$, resp. $A_2$ is isomorphic to $x_1$, resp. $x_2$. In this way we see that (S1) holds for $\overline{\mathcal{F}}$.

Assume $\mathcal{F}$ has (S2). The first part of (S2) for $\overline{\mathcal{F}}$ follows as in the argument above. The second part of (S2) for $\overline{\mathcal{F}}$ signifies that the map

$\overline{\mathcal{F}}(A \times _ k k[\epsilon ]) \to \overline{\mathcal{F}}(A) \times _{\overline{\mathcal{F}}(k)} \overline{\mathcal{F}}(k[\epsilon ])$

is injective for any ring $A$ in $\mathcal{C}_\Lambda$. Suppose that $y, y' \in \mathcal{F}(A \times _ k k[\epsilon ])$. Using the axioms of cofibred categories we can choose commutative diagrams

$\vcenter { \xymatrix{ y \ar[r]_ c \ar[d]_ a & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x'_\epsilon \ar[d]^{e'} \\ x' \ar[r]^{d'} & x'_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } }$

Assume that there exist isomorphisms $\alpha : x \to x'$ in $\mathcal{F}(A)$ and $\beta : x_\epsilon \to x'_\epsilon$ in $\mathcal{F}(k[\epsilon ])$. This also means there exists an isomorphism $\gamma : x_0 \to x'_0$ compatible with $\alpha$. To prove (S2) for $\overline{\mathcal{F}}$ we have to show that there exists an isomorphism $y \to y'$ in $\mathcal{F}(A \times _ k k[\epsilon ])$. By (S2) for $\mathcal{F}$ such a morphism will exist if we can choose the isomorphisms $\alpha$ and $\beta$ and $\gamma$ such that

$\xymatrix{ x \ar[d]^\alpha \ar[r] & x_0 \ar[d]^\gamma & x_\epsilon \ar[d]^\beta \ar[l]^ e \\ x' \ar[r] & x'_0 & x'_\epsilon \ar[l]_{e'} }$

is commutative (because then we can replace $x$ by $x'$ and $x_\epsilon$ by $x'_\epsilon$ in the previous displayed diagram). The left hand square commutes by our choice of $\gamma$. We can factor $e' \circ \beta$ as $\gamma ' \circ e$ for some second map $\gamma ' : x_0 \to x'_0$. Now the question is whether we can arrange it so that $\gamma = \gamma '$? This is clear if $\mathcal{F}(k)$ is a set, or a setoid. Moreover, if $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective, then we can adjust the choice of $\beta$ by precomposing with an automorphism of $x_\epsilon$ whose image is $\gamma ^{-1} \circ \gamma '$ to make things work. $\square$

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