Lemma 96.3.1. The functor $p : \mathcal{F} \to \mathcal{C}_\Lambda$ defined above is a predeformation category.

Proof. We have to show that $\mathcal{F}$ is (a) cofibred in groupoids over $\mathcal{C}_\Lambda$ and (b) that $\mathcal{F}(k)$ is a category equivalent to a category with a single object and a single morphism.

Proof of (a). The fibre categories of $\mathcal{F}$ over $\mathcal{C}_\Lambda$ are groupoids as the fibre categories of $\mathcal{X}$ are groupoids. Let $A \to A'$ be a morphism of $\mathcal{C}_\Lambda$ and let $x_0 \to x$ be an object of $\mathcal{F}(A)$. Because $\mathcal{X}$ is fibred in groupoids, we can find a morphism $x' \to x$ lying over $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$. Since the composition $A \to A' \to k$ is equal the given map $A \to k$ we see (by uniqueness of pullbacks up to isomorphism) that the pullback via $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(A')$ of $x'$ is $x_0$, i.e., that there exists a morphism $x_0 \to x'$ lying over $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(A')$ compatible with $x_0 \to x$ and $x' \to x$. This proves that $\mathcal{F}$ has pushforwards. We conclude by (the dual of) Categories, Lemma 4.35.2.

Proof of (b). If $A = k$, then $\mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(A)$ and since $\mathcal{X}$ is fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ we see that given any object $x_0 \to x$ in $\mathcal{F}(k)$ the morphism $x_0 \to x$ is an isomorphism. Hence every object of $\mathcal{F}(k)$ is isomorphic to $x_0 \to x_0$. Clearly the only self morphism of $x_0 \to x_0$ in $\mathcal{F}$ is the identity. $\square$

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