Lemma 90.12.6. Let $\mathcal{F}$ be a predeformation category over $\mathcal{C}_\Lambda$. If $\overline{\mathcal{F}}$ has (S2) then the maps $\gamma _ V$ are $k$-linear and we have $a_ V(D, x) = x + \gamma _ V(D)$.

Proof. In the proof of Lemma 90.12.2 we have seen that the functor $V \mapsto \overline{\mathcal{F}}(k[V])$ transforms $0$ to a singleton and products to products. The same is true of the functor $V \mapsto \text{Der}_\Lambda (k, V)$. Hence $\gamma _ V$ is linear by Lemma 90.11.5. Let $D : k \to V$ be a $\Lambda$-derivation. Set $D_1 : k \to V^{\oplus 2}$ equal to $a \mapsto (D(a), 0)$. Then

$\xymatrix{ k[V \times V] \ar[r]_{+} \ar[d]^{f_{1, D_1}} & k[V] \ar[d]^{f_{1, D}} \\ k[V \times V] \ar[r]^{+} & k[V] }$

commutes. Unwinding the definitions and using that $\overline{F}(V \times V) = \overline{F}(V) \times \overline{F}(V)$ this means that $a_ D(x_1) + x_2 = a_ D(x_1 + x_2)$ for all $x_1, x_2 \in \overline{F}(V)$. Thus it suffices to show that $a_ V(D, 0) = 0 + \gamma _ V(D)$ where $0 \in \overline{F}(V)$ is the zero vector. By definition this is the element $f_{0, *}(x_0)$. Since $f_ D = f_{1, D} \circ f_0$ the desired result follows. $\square$

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