Lemma 89.16.7. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda$ satisfying (RS). The following conditions are equivalent:

1. $\overline{\mathcal{F}}$ satisfies (RS).

2. Let $f_1: A_1 \to A$ and $f_2: A_2 \to A$ be ring maps in $\mathcal{C}_\Lambda$ with $f_2$ surjective. The induced map of sets of isomorphism classes

$\overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)} \to \overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2)$

is injective.

3. For every morphism $x' \to x$ in $\mathcal{F}$ lying over a surjective ring map $A' \to A$, the map $\text{Aut}_{A'}(x') \to \text{Aut}_ A(x)$ is surjective.

4. For every morphism $x' \to x$ in $\mathcal{F}$ lying over a small extension $A' \to A$, the map $\text{Aut}_{A'}(x') \to \text{Aut}_ A(x)$ is surjective.

Proof. We prove that (1) is equivalent to (2) and (2) is equivalent to (3). The equivalence of (3) and (4) follows from Lemma 89.3.3.

Let $f_1: A_1 \to A$ and $f_2: A_2 \to A$ be ring maps in $\mathcal{C}_\Lambda$ with $f_2$ surjective. By Remark 89.16.5 we see $\overline{\mathcal{F}}$ satisfies (RS) if and only if the map

$\overline{\mathcal{F}}(A_1 \times _ A A_2) \to \overline{\mathcal F}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2)$

is bijective for any such $f_1, f_2$. This map is at least surjective since that is the condition of (S1) and $\overline{\mathcal{F}}$ satisfies (S1) by Lemmas 89.16.6 and 89.10.5. Moreover, this map factors as

$\overline{\mathcal{F}}(A_1 \times _ A A_2) \longrightarrow \overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)} \longrightarrow \overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2),$

where the first map is a bijection since

$\mathcal{F}(A_1 \times _ A A_2) \longrightarrow \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$

is an equivalence by (RS) for $\mathcal{F}$. Hence (1) is equivalent to (2).

Assume (2) holds. Let $x' \to x$ be a morphism in $\mathcal{F}$ lying over a surjective ring map $f: A' \to A$. Let $a \in \text{Aut}_ A(x)$. The objects

$(x', x', a : x \to x), \ (x', x', \text{id} : x \to x)$

of $\mathcal{F}(A') \times _{\mathcal{F}(A)} \mathcal{F}(A')$ have the same image in $\overline{\mathcal{F}}(A') \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A')$. By (2) there exists maps $b_1, b_2 : x' \to x'$ such that

$\xymatrix{ x \ar[r]_ a \ar[d]_{f_*b_1} & x \ar[d]^{f_*b_2} \\ x \ar[r]^{\text{id}} & x }$

commutes. Hence $b_2^{-1} \circ b_1 \in \text{Aut}_{A'}(x')$ has image $a \in \text{Aut}_ A(x)$. Hence (3) holds.

Assume (3) holds. Suppose

$(x_1, x_2, a : (f_1)_*x_1 \to (f_2)_*x_2), \ (x'_1, x'_2, a' : (f_1)_*x'_1 \to (f_2)_*x'_2)$

are objects of $\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ with the same image in $\overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2)$. Then there are morphisms $b_1: x_1 \to x'_1$ in $\mathcal{F}(A_1)$ and $b_2: x_2 \to x'_2$ in $\mathcal F(A_2)$. By (3) we can modify $b_2$ by an automorphism of $x_2$ over $A_2$ so that the diagram

$\xymatrix{ (f_1)_*x_1 \ar[r]_ a \ar[d]_{(f_1)_*b_1} & (f_2)_*x_2 \ar[d]^{(f_2)_*b_2} \\ (f_1)_*x'_1 \ar[r]^{a'} & (f_2)_*x'_2. }$

commutes. This proves $(x_1, x_2, a) \cong (x'_1, x'_2, a')$ in $\overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)}$. Hence (2) holds. $\square$

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