Lemma 48.28.6. Consider a cartesian square
\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }
of schemes. Assume X \to S is flat and locally of finite presentation. Let (K, \xi ) be a relative dualizing complex for f. Set K' = L(g')^*K. Let \xi ' be the derived base change of \xi (see proof). Then (K', \xi ') is a relative dualizing complex for f'.
Proof.
Consider the cartesian square
\xymatrix{ X' \ar[d]_{\Delta _{X'/S'}} \ar[r] & X \ar[d]^{\Delta _{X/S}} \\ X' \times _{S'} X' \ar[r]^{g' \times g'} & X \times _ S X }
Choose W \subset X \times _ S X open such that \Delta _{X/S} factors through a closed immersion \Delta : X \to W. Choose W' \subset X' \times _{S'} X' open such that \Delta _{X'/S'} factors through a closed immersion \Delta ' : X \to W' and such that (g' \times g')(W') \subset W. Let us still denote g' \times g' : W' \to W the induced morphism. We have
L(g' \times g')^*\Delta _*\mathcal{O}_ X = \Delta '_*\mathcal{O}_{X'} \quad \text{and}\quad L(g' \times g')^*L\text{pr}_1^*K|_ W = L\text{pr}_1^*K'|_{W'}
The first equality holds because X and X' \times _{S'} X' are tor independent over X \times _ S X (see for example More on Morphisms, Lemma 37.69.1). The second holds by transitivity of derived pullback (Cohomology, Lemma 20.27.2). Thus \xi ' = L(g' \times g')^*\xi can be viewed as a map
\xi ' : \Delta '_*\mathcal{O}_{X'} \longrightarrow L\text{pr}_1^*K'|_{W'}
Having said this the proof of the lemma is straightforward. First, K' is S'-perfect by Derived Categories of Schemes, Lemma 36.35.6. To check that \xi ' induces an isomorphism of \Delta '_*\mathcal{O}_{X'} to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{W'}}( \Delta '_*\mathcal{O}_{X'}, L\text{pr}_1^*K'|_{W'}) we may work affine locally. By Lemma 48.28.2 we reduce to the corresponding statement in algebra which is proven in Dualizing Complexes, Lemma 47.27.4.
\square
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