Theorem 89.18.2. Let $F: \mathcal{C}_\Lambda \to \textit{Sets}$ be a functor. Then $F$ is prorepresentable if and only if (a) $F$ is a deformation functor, (b) $\dim _ k TF$ is finite, and (c) $\gamma : \text{Der}_\Lambda (k, k) \to TF$ is injective.

Proof. Assume $F$ is prorepresentable by $R \in \widehat{\mathcal{C}}_\Lambda$. We see $F$ is a deformation functor by Example 89.16.10. We see $\dim _ k TF$ is finite by Example 89.11.11. Finally, $\text{Der}_\Lambda (k, k) \to TF$ is identified with $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$ by Example 89.11.14 which is injective because $R \to k$ is surjective.

Conversely, assume (a), (b), and (c) hold. By Lemma 89.16.6 we see that (S1) and (S2) hold. Hence by Theorem 89.15.5 there exists a minimal versal formal object $\xi$ of $F$ such that (89.15.0.2) holds. Say $\xi$ lies over $R$. The map

$d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F}$

is bijective on $\text{Der}_\Lambda (k, k)$-orbits. Since the action of $\text{Der}_\Lambda (k, k)$ on the left hand side is free by (c) and Lemma 89.12.6 we see that the map is bijective. Thus we see that $\underline{\xi }$ is an isomorphism by Lemma 89.18.1. $\square$

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).