Etale commutativity of the companion-ideal step

The trichotomy. If the pulled-back support is empty, every center has empty inverse image, so the induced sequence consists of isomorphisms and the assertion is immediate. Otherwise, along Step 2a we compare ${\mathrm{ord}}_N(I_{r_l})$ with ${\mathrm{ord}}_N({\phi }^{\ast }(I_{r_l}))$; by [F4] the pointwise residual orders agree; the global maximum on the image can be smaller, so the pullback may omit a companion pass. If the value at stage $r_l$ exceeds the value of the pullback, the centers of $(X_i{)}_{r_l\le i<r_{l+1}}$ lie in the locus where ${\mathrm{ord}}_N$ attains its maximal value, which does not meet the image of $\phi$, so the induced morphisms are isomorphisms; if the values agree, the companion ideals correspond, ${\phi }^{\ast }(O(I_{r_l}))=O({\phi }^{\ast }{I}_{r_l})$, and [F3] gives the commutativity of the maximal-order step together with equality of the invariants.

The monomial end and conclusion. If the residual maximum is zero, restrict to the open neighbourhood of the support where $N(I_{r_k})$ is a unit. There $I_{r_k}=M(I_{r_k})$ is monomial, as is its pullback by [F4], and both resolutions are controlled by the invariant $\rho$ on subsets of $E$; the ordered boundary labels and exponents identify the pointwise values of $\rho$ and $\nu$. If the image misses the current global maximum of $\rho$, the inverse center is empty and its blowup pulls back to an isomorphism. If it meets that maximum, the pulled-back center is precisely the maximal $\rho$ locus on the pullback. Iterating these two cases gives the same nonempty centers and invariant values, with isomorphism steps inserted where a larger maximum is missed. All centers lie in the support, so these local monomial sequences extend by the identity off it and agree on overlaps. Values on lower residual-order strata skipped by earlier companion passes are transferred through their unchanged lifts from the first later applicable pass, as in the proposition; the same transfers commute with étale pullback. Assembling the finitely many stages proves (1) and (2).