PomeLemma 19.5.10.32DCABF4D374 The convex body $\Delta _{Y_{\bullet }}(T)$ defined in definition 19.5.7 is independent of the choice of $\pi $.
Lemma 19.5.1032DCABF4D374
Proof. Take another map $\pi '\colon Y'\rightarrow X$ with the same properties. We want to show that $\pi $ and $\pi '$ define the same $\Delta _{Y_{\bullet }}(T)$. We may assume that $\pi '$ dominates $\pi $ through $p\colon Y'\rightarrow Y$, so that we have a commutative diagram
We take $D$ and $R$ as in (19.5.8). Then
\[ \pi '^*T=[p^*D]+p^*R. \]
Write $(Z_{\bullet },g)$ and $(Z_{\bullet }',g')$ for the liftings of $Y_{\bullet }$ to $Y$ and $Y'$ respectively. We need to prove that
\[ \Delta _{Z_{\bullet }}(\{ R\} )g^{-1}+\nu _{Z_{\bullet }}([D])g^{-1}=\Delta _{Z_{\bullet }'}(\{ p^*R\} )g'^{-1}+\nu _{Z_{\bullet }'}([p^*D])g'^{-1}. \]
By theorem 19.5.4, applied to the class $\{ R\} $, we have
\[ \Delta _{Z'_\bullet }(\{ p^*R\} )=\Delta _{Z_\bullet }(\{ R\} )\, \operatorname{cor}(Z_\bullet ,p). \]
By proposition 19.3.14, applied to the divisor current $[D]$, we also have
\[ \nu _{Z'_\bullet }([p^*D])=\nu _{Z_\bullet }([D])\, \operatorname{cor}(Z_\bullet ,p). \]
Finally, proposition 19.3.17 gives
\[ g'=g\, \operatorname{cor}(Z_\bullet ,p). \]
Multiplying by $(g')^{-1}$ gives the desired equality. $\square$
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