Lemma 86.9.4. Let $S$ be a scheme. Let $X \to Y$ be a proper, flat morphism of algebraic spaces which is of finite presentation. Let $(\omega ^\bullet , \tau )$ be a pair consisting of a $Y$-perfect object of $D(\mathcal{O}_ X)$ and a map $\tau : Rf_*\omega ^\bullet \to \mathcal{O}_ Y$. Assume we have cartesian diagrams
\[ \xymatrix{ X_ i \ar[r]_{g_ i'} \ar[d]_{f_ i} & X \ar[d]^ f \\ Y_ i \ar[r]^{g_ i} & Y } \]
with $Y_ i$ affine such that $\{ g_ i : Y_ i \to Y\} $ is an étale covering and isomorphisms of pairs $(\omega ^\bullet |_{X_ i}, \tau |_{Y_ i}) \to (a_ i(\mathcal{O}_{Y_ i}), \text{Tr}_{f_ i, \mathcal{O}_{Y_ i}})$ as in Definition 86.9.1. Then $(\omega ^\bullet , \tau )$ is a relative dualizing complex for $X$ over $Y$.
Proof.
Let $g : Y' \to Y$ and $X', f', g', a'$ be as in Definition 86.9.1. Set $((\omega ')^\bullet , \tau ') = (L(g')^*\omega ^\bullet , Lg^*\tau )$. We can find a finite étale covering $\{ Y'_ j \to Y'\} $ by affines which refines $\{ Y_ i \times _ Y Y' \to Y'\} $ (Topologies, Lemma 34.4.4). Thus for each $j$ there is an $i_ j$ and a morphism $k_ j : Y'_ j \to Y_{i_ j}$ over $Y$. Consider the fibre products
\[ \xymatrix{ X'_ j \ar[r]_{h_ j'} \ar[d]_{f'_ j} & X' \ar[d]^{f'} \\ Y'_ j \ar[r]^{h_ j} & Y' } \]
Denote $k'_ j : X'_ j \to X_{i_ j}$ the induced morphism (base change of $k_ j$ by $f_{i_ j}$). Restricting the given isomorphisms to $Y'_ j$ via the morphism $k'_ j$ we get isomorphisms of pairs $((\omega ')^\bullet |_{X'_ j}, \tau '|_{Y'_ j}) \to (a_ j(\mathcal{O}_{Y'_ j}), \text{Tr}_{f'_ j, \mathcal{O}_{Y'_ j}})$. After replacing $f : X \to Y$ by $f' : X' \to Y'$ we reduce to the problem solved in the next paragraph.
Assume $Y$ is affine. Problem: show $(\omega ^\bullet , \tau )$ is isomorphic to $(\omega _{X/Y}^\bullet , \text{Tr}) = (a(\mathcal{O}_ Y), \text{Tr}_{f, \mathcal{O}_ Y})$. We may assume our covering $\{ Y_ i \to Y\} $ is given by a single surjective étale morphism $\{ g : Y' \to Y\} $ of affines. Namely, we can first replace $\{ g_ i: Y_ i \to Y\} $ by a finite subcovering, and then we can set $g = \coprod g_ i : Y' = \coprod Y_ i \to Y$; some details omitted. Set $X' = Y' \times _ Y X$ with maps $f', g'$ as in Definition 86.9.1. Then all we're given is that we have an isomorphism
\[ (\omega ^\bullet |_{X'}, \tau |_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) \]
Since $(\omega _{X/Y}^\bullet , \text{Tr})$ is a relative dualizing complex (see discussion following Definition 86.9.1) there is a unique isomorphism
\[ (\omega _{X/Y}^\bullet |_{X'}, \text{Tr}|_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) \]
Uniqueness by Lemma 86.9.3 for example. Combining the displayed isomorphisms we find an isomorphism
\[ \alpha : (\omega ^\bullet |_{X'}, \tau |_{Y'}) \to (\omega _{X/Y}^\bullet |_{X'}, \text{Tr}|_{Y'}) \]
Set $Y'' = Y' \times _ Y Y'$ and $X'' = Y'' \times _ Y X$ the two pullbacks of $\alpha $ to $X''$ have to be the same by uniqueness again. Since we have vanishing negative self exts for $\omega _{X'/Y'}^\bullet $ over $X'$ (Lemma 86.9.2) and since this remains true after pulling back by any projection $Y' \times _ Y \ldots \times _ Y Y' \to Y'$ (small detail omitted – compare with the proof of Lemma 86.9.3), we find that $\alpha $ descends to an isomorphism $\omega ^\bullet \to \omega _{X/Y}^\bullet $ over $X$ by Simplicial Spaces, Lemma 85.35.1.
$\square$
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