Proof.
We prove that (1) is equivalent to (2) and (2) is equivalent to (3). The equivalence of (3) and (4) follows from Lemma 90.3.3.
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 90.16.5 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 90.16.6 and 90.10.5. 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.
$\square$
Comments (0)