**Proof.**
Assume (1). By Lemma 89.16.2 we see that every object of $\mathcal{F}(A_1 \times _ A A_2)$ is of the form $x_1 \times _ x x_2$. Moreover

\[ \mathop{\mathrm{Mor}}\nolimits _{A_1 \times _ A A_2}(x_1 \times _ x x_2, y_1 \times _ y y_2) = \mathop{\mathrm{Mor}}\nolimits _{A_1}(x_1, y_1) \times _{\mathop{\mathrm{Mor}}\nolimits _ A(x, y)} \mathop{\mathrm{Mor}}\nolimits _{A_2}(x_2, y_2). \]

Hence we see that $\mathcal{F}(A_1 \times _ A A_2)$ is a $2$-fibre product of $\mathcal{F}(A_1)$ with $\mathcal{F}(A_2)$ over $\mathcal{F}(A)$ by Categories, Remark 4.31.5. In other words, we see that (2) holds.

The implication (2) $\Rightarrow $ (3) is immediate.

Assume (3). Let $q_1 : A_1 \to A$ and $q_2 : A_2 \to A$ be given with $q_2$ a small extension. We will use the description of the $2$-fibre product $\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ from Categories, Remark 4.31.5. Hence let $y \in \mathcal{F}(A_1 \times _ A A_2)$ correspond to $(x_1, x_2, x, a_1 : x_1 \to x, a_2 : x_2 \to x)$. Let $z$ be an object of $\mathcal{F}$ lying over $C$. Then

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, y) & = \{ (f, \alpha ) \mid f : C \to A_1 \times _ A A_2, \alpha : f_*z \to y\} \\ & = \{ (f_1, f_2, \alpha _1, \alpha _2) \mid f_ i : C \to A_ i, \ \alpha _ i : f_{i, *}z \to x_ i, \\ & \quad \quad \quad \quad q_1 \circ f_1 = q_2 \circ f_2, \ q_{1, *} \alpha _1 = q_{2, *}\alpha _2\} \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x_1) \times _{\mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x)} \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x_2) \end{align*}

whence $y$ is a fibre product of $x_1$ and $x_2$ over $x$. Thus we see that $\mathcal{F}$ satisfies (RS) in case $A_2 \to A$ is a small extension. Hence (RS) holds by Lemma 89.16.3.
$\square$

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