Lemma 54.9.6. In Situation 54.9.1 assume $A$ has a dualizing complex $\omega _ A^\bullet $. With $\omega _ X$ the dualizing module of $X$, the trace map $H^0(X, \omega _ X) \to \omega _ A$ is an isomorphism and consequently there is a canonical map $f^*\omega _ A \to \omega _ X$.
Proof. By Grauert-Riemenschneider (Proposition 54.7.8) we see that $Rf_*\omega _ X = f_*\omega _ X$. By duality we have a short exact sequence
\[ 0 \to f_*\omega _ X \to \omega _ A \to \mathop{\mathrm{Ext}}\nolimits ^2_ A(R^1f_*\mathcal{O}_ X, \omega _ A) \to 0 \]
(for example see proof of Lemma 54.8.8) and since $A$ defines a rational singularity we obtain $f_*\omega _ X = \omega _ A$. $\square$
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