Lemma 89.29.2. With notation and assumptions as in Situation 89.29.1.
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 89.6.2.
Part (3) follows as the functor (89.29.1.1) commutes with fibre products and transforms surjective maps into surjective maps, see Definition 89.10.1.
Part (4). To see this consider a diagram
in $\mathcal{C}_{\Lambda , l}$ as in Definition 89.10.1. Applying the functor (89.29.1.1) we obtain
where $l\epsilon $ denotes the finite dimensional $k$-vector space $l\epsilon \subset l[\epsilon ]$. According to Lemma 89.10.4 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 89.16.4 part (2) and the fact that (89.29.1.1) commutes with fibre products. $\square$
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