Lemma 88.29.2. With notation and assumptions as in Situation 88.29.1.
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 88.6.2.
Part (3) follows as the functor (88.29.1.1) commutes with fibre products and transforms surjective maps into surjective maps, see Definition 88.10.1.
Part (4). To see this consider a diagram
in $\mathcal{C}_{\Lambda , l}$ as in Definition 88.10.1. Applying the functor (88.29.1.1) we obtain
where $l\epsilon $ denotes the finite dimensional $k$-vector space $l\epsilon \subset l[\epsilon ]$. According to Lemma 88.10.4 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 88.16.4 part (2) and the fact that (88.29.1.1) commutes with fibre products. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)