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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