Corollary 18.19.39E4426687FC0

Proof. On $B=Y\times C_0$, with projection $\pi $ to $Y$, use the background $\gamma _{\sigma ,f}$ of the proof of lemma 18.18.1. Sums of ordinary competitors in the separate coordinates give $P_{\gamma _{\sigma ,f}}[u\circ \pi ]\geq \sum _ iq_ i\circ \pi _ i\circ \pi $. For the reverse inequality, replace an ordinary competitor by its maximum with a sufficiently low translate of $u\circ \pi $. Its restriction to every section $Y\times \{ z\} $ is then not identically $-\infty $, nonpositive and at least as singular as $u$, so the separation for the zero obstacle bounds it by $\sum _ iq_ i\circ \pi _ i$. The reduction of proposition 18.14.1, applied to the divisor $B\times D$ with weight one, proves the ordinary identity upstairs.

By corollary 18.4.4, each $q_ i$ has the multiplier ideals of $u_ i$, so pullback and addition give $U\sim _{\mathcal I}\Phi _\sigma (q)$. These two potentials therefore have the same ideal envelope, which lemma 18.18.1 identifies with $\Psi +Q\circ p$.

If $\mu _ i=E_ i$ for every $i$, then $u_ i\in \mathcal E_{\mathrm{env}}(C_ i,\sigma _ i,r_ i)$ by corollary 13.29.2, and we choose stretching competitors in these curve classes. The sum of their pullbacks plus $\Psi $ is $\Theta _{\sigma ,f}$-plurisubharmonic: its current is the sum of the pulled-back positive curve currents, the form $(f\circ p)s^*\tau $ and the current of the divisor with weight one. Adding the stretching inequalities gives the one for $U$ relative to $\Psi +Q\circ p$, also at poles, and the ordinary order follows in the same way. Thus $U$ belongs to that relative class, and proposition 18.12.2 gives equality below all obstacles. Taking the zero obstacle and using the two identities above recovers $q_ i=r_ i$, hence $\mu _ i=E_ i$. This closes the chain of equivalences. $\square$