Lemma 90.29.2. With notation and assumptions as in Situation 90.29.1.
We have $\overline{\mathcal{F}_{l/k}} = (\overline{\mathcal{F}})_{l/k}$.
If $\mathcal{F}$ is a predeformation category, then $\mathcal{F}_{l/k}$ is a predeformation category.
If $\mathcal{F}$ satisfies (S1), then $\mathcal{F}_{l/k}$ satisfies (S1).
If $\mathcal{F}$ satisfies (S2), then $\mathcal{F}_{l/k}$ satisfies (S2).
If $\mathcal{F}$ satisfies (RS), then $\mathcal{F}_{l/k}$ satisfies (RS).
Proof. Part (1) is immediate from the definitions.
Since $\mathcal{F}_{l/k}(l) = \mathcal{F}(k)$ part (2) follows from the definition, see Definition 90.6.2.
Part (3) follows as the functor (90.29.1.1) commutes with fibre products and transforms surjective maps into surjective maps, see Definition 90.10.1.
Part (4). To see this consider a diagram
in $\mathcal{C}_{\Lambda , l}$ as in Definition 90.10.1. Applying the functor (90.29.1.1) we obtain
where $l\epsilon $ denotes the finite dimensional $k$-vector space $l\epsilon \subset l[\epsilon ]$. According to Lemma 90.10.4 the condition of (S2) for $\mathcal{F}$ also holds for this diagram. Hence (S2) holds for $\mathcal{F}_{l/k}$.
Part (5) follows from the characterization of (RS) in Lemma 90.16.4 part (2) and the fact that (90.29.1.1) commutes with fibre products. $\square$
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