Lemma 97.23.6. Let $S$ and $\mathcal{X}$ be as in Definition 97.23.5 and let $\mathcal{X}$ be endowed with a naive obstruction theory. Let $A \to B$ and $y \to x$ be as in (3). Let $k$ be a $B$-algebra which is a field. Then the functoriality map $E_ x \to E_ y$ induces bijections

$H^ i(E_ x \otimes _ A^{\mathbf{L}} k) \to H^ i(E_ y \otimes _ B^{\mathbf{L}} k)$

for $i = 0, 1$.

Proof. Let $z = x|_{\mathop{\mathrm{Spec}}(k)}$. Then (RS*) implies that

$\textit{Lift}(x, A[k]) = \textit{Lift}(z, k[k]) \quad \text{and}\quad \textit{Lift}(y, B[k]) = \textit{Lift}(z, k[k])$

because $A[k] = A \times _ k k[k]$ and $B[k] = B \times _ k k[k]$. Hence the properties of a naive obstruction theory imply that the functoriality map $E_ x \to E_ y$ induces bijections $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(E_ x, k) \to \text{Ext}^ i_ B(E_ y, k)$ for $i = -1, 0$. By Lemma 97.23.1 our maps $H^ i(E_ x \otimes _ A^{\mathbf{L}} k) \to H^ i(E_ y \otimes _ B^{\mathbf{L}} k)$, $i = 0, 1$ induce isomorphisms on dual vector spaces hence are isomorphisms. $\square$

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