Lemma 47.17.2. Let $A$ be a Noetherian ring and let $\omega _ A^\bullet $ be a dualizing complex. Let $A \to B$ be a surjective ring map and let $\omega _ B^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ be the dualizing complex for $B$ of Lemma 47.15.9. Then we have

\[ \delta _{\omega _ B^\bullet } = \delta _{\omega _ A^\bullet }|_{\mathop{\mathrm{Spec}}(B)} \]

**Proof.**
This follows from the definition of the functions and Lemma 47.16.2.
$\square$

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