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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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    $\begingroup$ 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. $\endgroup$
    – naf
    Apr 10, 2013 at 4:03
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    $\begingroup$ What was behind of the mind of Kawamata for this conjecture ?: let $X$ and $Y$ be two smooth projective varieties. They are called D-equivalent if $D^b(Coh(X))$ and $D^b(Coh(Y))$ are equivalent as triangulated categories. The varieties $X$ and $Y$ are called K-equivalent if there is a third smooth projective variety $Z$ together with birational morphisms $f:Z→X$ and $g:Z→Y$ such that $f^∗K_X∼g^∗K_Y $(linear equivalence of divisors) or $X$ is a Fourier-Mukai partner of $Y$ $\endgroup$
    – user21574
    Oct 21, 2017 at 8:43
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    $\begingroup$ ....from Kawamata, we are expected K-equivalence be equivalent with D-equivalence: The main idea of the conjecture of K-equivalence<==>D-equivalence was that the Serre functor is invariant in the derived category of triangulated coherent sheaves and Serre functor is related to canonical divisor also, and so Kawamata gave such natural conjecture. $\endgroup$
    – user21574
    Oct 21, 2017 at 8:43
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    $\begingroup$ Uehara gave an counter example to this conjecture of Kawamata: Uehara, Hokuto: An example of Fourier-Mukai partners of minimal elliptic surfaces. Math. Res. Lett. 11 (2004), no. 2-3, 371–375. $\endgroup$
    – user21574
    Oct 21, 2017 at 8:49

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