Proof.
Consider a factorization
\[ \mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(B) \to S \]
as in (98.3.0.1). By definition we have
\[ (\mathcal{F}_{\mathcal{X}, k , x_0})_{l/k}(B) = \mathcal{F}_{\mathcal{X}, k, x_0}(B \times _ l k) \]
see Formal Deformation Theory, Situation 90.29.1. Thus an object of this is a morphism $x_0 \to x$ of $\mathcal{X}$ lying over the morphism $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(B \times _ l k)$. Choosing pullback functor for $\mathcal{X}$ we can associate to $x_0 \to x$ the morphism $x_{l, 0} \to x_ B$ where $x_ B$ is the restriction of $x$ to $\mathop{\mathrm{Spec}}(B)$ (via the morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B \times _ l k)$ coming from $B \times _ l k \subset B$). This construction is functorial in $B$ and compatible with morphisms.
Next, assume $\mathcal{X}$ satisfies (RS). Consider the diagrams
\[ \vcenter { \xymatrix{ l & B \ar[l] \\ k \ar[u] & B \times _ l k \ar[l] \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathop{\mathrm{Spec}}(l) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(B) \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & \mathop{\mathrm{Spec}}(B \times _ l k) } } \]
The diagram on the left is a fibre product of rings. The diagram on the right is a pushout in the category of schemes, see More on Morphisms, Lemma 37.14.3. These schemes are all of finite type over $S$ (see remarks following Definition 98.5.1). Hence (RS) kicks in to give an equivalence of fibre categories
\[ \mathcal{X}_{\mathop{\mathrm{Spec}}(B \times _ l k)} \longrightarrow \mathcal{X}_{\mathop{\mathrm{Spec}}(k)} \times _{\mathcal{X}_{\mathop{\mathrm{Spec}}(l)}} \mathcal{X}_{\mathop{\mathrm{Spec}}(B)} \]
This implies that the functor defined above gives an equivalence of fibre categories. Hence the functor is an equivalence on categories cofibred in groupoids by (the dual of) Categories, Lemma 4.35.9.
$\square$
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