Lemma 99.5.2. In Situation 99.5.1 the functor $p : \mathcal{C}\! \mathit{oh}_{X/B} \longrightarrow (\mathit{Sch}/S)_{fppf}$ is fibred in groupoids.
Proof. We show that $p$ is fibred in groupoids by checking conditions (1) and (2) of Categories, Definition 4.35.1. Given an object $(T', g', \mathcal{F}')$ of $\mathcal{C}\! \mathit{oh}_{X/B}$ and a morphism $h : T \to T'$ of schemes over $S$ we can set $g = h \circ g'$ and $\mathcal{F} = (h')^*\mathcal{F}'$ where $h' : X_ T \to X_{T'}$ is the base change of $h$. Then it is clear that we obtain a morphism $(T, g, \mathcal{F}) \to (T', g', \mathcal{F}')$ of $\mathcal{C}\! \mathit{oh}_{X/B}$ lying over $h$. This proves (1). For (2) suppose we are given morphisms
\[ (h_1, \varphi _1) : (T_1, g_1, \mathcal{F}_1) \to (T, g, \mathcal{F}) \quad \text{and}\quad (h_2, \varphi _2) : (T_2, g_2, \mathcal{F}_2) \to (T, g, \mathcal{F}) \]
of $\mathcal{C}\! \mathit{oh}_{X/B}$ and a morphism $h : T_1 \to T_2$ such that $h_2 \circ h = h_1$. Then we can let $\varphi $ be the composition
\[ (h')^*\mathcal{F}_2 \xrightarrow {(h')^*\varphi _2^{-1}} (h')^*(h_2)^*\mathcal{F} = (h_1)^*\mathcal{F} \xrightarrow {\varphi _1} \mathcal{F}_1 \]
to obtain the morphism $(h, \varphi ) : (T_1, g_1, \mathcal{F}_1) \to (T_2, g_2, \mathcal{F}_2)$ that witnesses the truth of condition (2). $\square$
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