Proposition 18.12.4625C5E3FF1B3

Proof. For $v\in \mathrm{PSH}(Y,\theta ')$, define

Over a sufficiently small ball in $X$, this is the finite maximum of the functions obtained by composing $v$ with the local inverse branches of $r$. Each of them is $\theta $-plurisubharmonic, so $v_{\max }\in \mathrm{PSH}(X,\theta )$ by lemma 5.24.1. If $v\preceq u'$, then $v_{\max }\preceq u$. If $v\sim _{\mathcal I}u'$, a local biholomorphic change of variables makes each branch multiplier-ideal equivalent to $u$, and the maximum rule in lemma 5.37.4 gives $v_{\max }\sim _{\mathcal I}u$. The obstacle bound $v\leq F\circ r$ gives $v_{\max }\leq F$.

Thus every competitor upstairs is dominated by the pullback of a corresponding competitor downstairs. Conversely, pullback of a competitor downstairs preserves the background inequality, the ordinary order or multiplier-ideal equivalence, and the obstacle bound. These two correspondences between the competitor families give the envelope identities; upper regularization commutes with $r$, since $r$ is locally biholomorphic.

Since $r$ is surjective, it reflects the ordinary order: $u'\preceq \varphi \circ r$ if and only if $u\preceq \varphi $. Fix $a\in (0,1)$. Pullback takes any stretching competitor $h$ with $a\varphi +(1-a)h\leq u$ to a stretching competitor upstairs. Conversely, if $v\in \mathrm{PSH}(Y,\theta ')$ satisfies

then taking the maximum over the fibers gives $a\varphi +(1-a)v_{\max }\leq u$; this also holds at poles, because the convex sums are defined there. Applying the definition of the relative class for every $a$ proves the last assertion. $\square$