Lemma 5.43.150018DC5A7DD0

Proof. On a base disk choose a smooth function $a$ with $\mathrm{dd}^{\mathrm{c}}a=\beta $. Off the branch values, $m_ h+a$ is the maximum of finitely many subharmonic functions obtained from $h+a\circ r$ by inverse branches. It is subharmonic there and locally bounded above. The removable-singularity theorem extends it across the branch values. This extension has the stated value: upper semicontinuity and properness give the upper bound at the fiber, while the genuine upper limits of $h+a\circ r$ at its points give the reverse bound. Thus the extensions glue. The function is not identically $-\infty $, since $h$ is finite at some point. This proves the curvature inequality and $h\leq m_ h\circ r$.

Choose coordinates in the disjoint disks over $s$ so that the map on the disk centered at $x$ is $t=z^{e_ x}$. Then

Divide by $\log \delta ^2<0$ and use the supremum definition of Lelong number. Each quotient on the right tends to $\nu (h,x)/e_ x$, and the maximum becomes the minimum after division. Smooth local corrections do not affect these limits.

An ordinary bound $h\leq v\circ r+c$ gives $m_ h\leq v+c$ on every fiber. On a curve, the all-weight ideal order is equivalent to the pointwise Lelong inequalities, by theorem 5.37.3. The pullback formula 5.43.8 makes $h\preceq _{\mathcal I}v\circ r$ equivalent to $\nu (h,x)\geq e_ x\nu (v,r(x))$ at every $x$. The minimum formula gives the required base inequalities. $\square$