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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