# K-equivalence does not depend on the choice of the third variety

By K-equivalent of two smooth varieties $X,Y$, we mean there exist a smooth variety $Z$, and birational morphism $q: Z \to X,\quad p: Z \to Y$ , such that $q^* \omega_X \cong p^* \omega_Y$.

Suppose $X,Y$ are K-equivalent, if one has another smooth variety $W$, and birational morphism $t : W \to X$, $s: W \to Y$, would it be $t^* \omega_X \cong s^* \omega_Y$? And why?

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You can use the usual diagonal trick: consider a smooth variety mapping birationally to both $Z$ and $W$ (this exists by resolution of singularities in characteristic zero) and then use the fact that the map on Picard groups induced by pullback for a birational proper morphism of smooth varieties is injective. –  ulrich Apr 10 '13 at 4:03
Thank you! I see what you mean. –  Li Yutong Apr 10 '13 at 15:08