Proposition 9.11.7622B386B559D

Proof. The preceding theorem identifies the competitor families. Let $s$ be their pointwise supremum on $X$; its pullback is the pointwise supremum upstairs. The regularization of $s\circ \pi $ on $Y$ is $\pi ^*\theta $-plurisubharmonic and not identically $-\infty $, since there is a competitor below it. By the theorem, it descends to a $\theta $-plurisubharmonic function $q$. Since $q\geq s$ and $q$ is upper semicontinuous, $q\geq s^*$.

For the reverse inequality, fix $x\in X$ and $y\in \pi ^{-1}(x)$. For each neighborhood $U$ of $x$, the unregularized supremum upstairs is bounded by $\sup _ U s$ on the open neighborhood $\pi ^{-1}(U)$ of $y$. Therefore $q(x)\leq \sup _ U s$, and taking the infimum over $U$ gives $q(x)\leq s^*(x)$. This proves the identity and shows that $P_\theta (f)=s^*$ is $\theta $-plurisubharmonic. The proof for $P_\theta [u]$ is identical, because pullback preserves the ordinary singularity order together with its constants. $\square$