Lemma 97.24.4. In Situation 97.24.2 assume that (iv) of Lemma 97.24.3 holds and that $K^\bullet$ is a perfect object of $D(A)$. In this case, if $x$ is versal at a closed point $u_0 \in U$ then there exists an open neighbourhood $u_0 \in U' \subset U$ such that $x$ is versal at every finite type point of $U'$.

Proof. We may assume that $K^\bullet$ is a finite complex of finite projective $A$-modules. Thus the derived tensor product with $K^\bullet$ is the same as simply tensoring with $K^\bullet$. Let $E^\bullet$ be the dual perfect complex to $K^\bullet$, see More on Algebra, Lemma 15.74.15. (So $E^ n = \mathop{\mathrm{Hom}}\nolimits _ A(K^{-n}, A)$ with differentials the transpose of the differentials of $K^\bullet$.) Let $E \in D^{-}(A)$ denote the object represented by the complex $E^\bullet [-1]$. Let $\xi \in H^1(\text{Tot}(K^\bullet \otimes _ A \mathop{N\! L}\nolimits _{A/\Lambda }))$ be the element constructed in Lemma 97.24.3 and denote $\xi : E = E^\bullet [-1] \to \mathop{N\! L}\nolimits _{A/\Lambda }$ the corresponding map (loc.cit.). We claim that the pair $(E, \xi )$ satisfies all the assumptions of Lemma 97.23.4 which finishes the proof.

Namely, assumption (i) of Lemma 97.23.4 follows from conclusion (1) of Lemma 97.24.3 and the fact that $H^2(K^\bullet \otimes _ A^\mathbf {L} -) = \mathop{\mathrm{Ext}}\nolimits ^1(E, -)$ by loc.cit. Assumption (ii) of Lemma 97.23.4 follows from conclusion (2) of Lemma 97.24.3 and the fact that $H^1(K^\bullet \otimes _ A^\mathbf {L} -) = \mathop{\mathrm{Ext}}\nolimits ^0(E, -)$ by loc.cit. Assumption (iii) of Lemma 97.23.4 is clear. $\square$

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).