Lemma 57.16.5. Let $k$ be an algebraically closed field. Let $X$ be a smooth proper scheme over $k$. Let $f : Y \to S$ be a smooth proper morphism with $S$ of finite type over $k$. Let $K$ be the Fourier-Mukai kernel of a relative equivalence from $X \times S$ to $Y$ over $S$. Then $S$ can be covered by open subschemes $U$ such that there is a $U$-isomorphism $f^{-1}(U) \cong Y_0 \times U$ for some $Y_0$ proper and smooth over $k$.

Proof. Choose a closed point $s \in S$. Since $k$ is algebraically closed this is a $k$-rational point. Set $Y_0 = Y_ s$. The restriction $K_0$ of $K$ to $X \times Y_0$ is the Fourier-Mukai kernel of a relative equivalence from $X$ to $Y_0$ over $\mathop{\mathrm{Spec}}(k)$ by Lemma 57.15.3. Let $K'_0$ in $D_{perf}(\mathcal{O}_{Y_0 \times X})$ be the object assumed to exist in Definition 57.15.1. Then $K'_0$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0$ to $X$ over $\mathop{\mathrm{Spec}}(k)$ by the symmetry inherent in Definition 57.15.1. Hence by Lemma 57.15.3 we see that the pullback

$M = (Y_0 \times X \times S \to Y_0 \times X)^*K'_0$

on $(Y_0 \times S) \times _ S (X \times S) = Y_0 \times X \times S$ is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $X \times S$ over $S$. Now consider the kernel

$K_{new} = R\text{pr}_{13, *}(L\text{pr}_{12}^*M \otimes _{\mathcal{O}_{(Y_0 \times S) \times _ S (X \times S) \times _ S Y}}^\mathbf {L} L\text{pr}_{23}^*K)$

on $(Y_0 \times S) \times _ S Y$. This is the Fourier-Mukai kernel of a relative equivalence from $Y_0 \times S$ to $Y$ over $S$ since it is the composition of two invertible arrows in the category constructed in Section 57.14. Moreover, this composition passes through base change (Lemma 57.14.1). Hence we see that the pullback of $K_{new}$ to $((Y_0 \times S) \times _ S Y)_ s = Y_0 \times Y_0$ is equal to the composition of $K_0$ and $K'_0$ and hence equal to the identity in this category. In other words, we have

$L(Y_0 \times Y_0 \to (Y_0 \times S) \times _ S Y)^*K_{new} \cong \Delta _{Y_0/k, *}\mathcal{O}_{Y_0}$

Thus by Lemma 57.16.4 we conclude that $Y \to S$ is isomorphic to $Y_0 \times S$ in an open neighbourhood of $s$. This finishes the proof. $\square$

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