Chapter 15: per-component centred contractions satisfy a uniform displacement budget

Let $m$ be finite, let $c_1,\dots,c_m\in\mathbb{R}^3$, let $R>0$, and let $a_1,\dots,a_m\in(0,1]$ satisfy $\sum_i (1-a_i)<\tfrac12$. Let $\chi_i:\mathbb{R}^3\to\mathbb{R}$ be continuous cut-offs which are exactly $1$ on the ball $B(c_i,R/2)$ and exactly $0$ outside $B(c_i,5R/2)$, and let $S_i$ be the centred contraction $S_i x = c_i + a_i (x-c_i)$. Then the displacement field $E(x)=\sum_i \chi_i(x)\,(S_i(x)-x)$ is Lipschitz on $\mathbb{R}^3$ with $\operatorname{Lip}(E)\le 2\sum_i(1-a_i)$, and this constant is strictly less than $1$.

This is the single quantitative estimate that the cut-off patching route to Chapter 15, Theorem 1 requires. The proved theorem BookSixth.bump_perturbation_is_homeomorph_v4 turns any displacement field that is $q$-Lipschitz with $q<1$ into a homeomorphism of $\mathbb{R}^3$ with continuous inverse, which is exactly the hypothesis that route is otherwise missing, and no theorem in the catalogue currently supplies it.

The two bounds that do exist, BookSixth.bump_perturbation_lipschitz_v3 and BookSixth.bump_perturbation_lipschitz, bound the field by a sum of per-component displacement radii and cut-off Lipschitz constants. That sum grows both with the size of the family and with the width of the cut-offs, so it cannot fall below $1$ for a large family or for wide cut-offs. Centring each contraction at its own component removes the dependence on both: the derivative of $S_i$ is the scalar $a_i$ and the constant term is absorbed by the centre $c_i$, so each component contributes exactly the scalar deficit $1-a_i$ and only the deficit sum survives. The content of the bound is therefore that the controlling invariant is $\sum_i(1-a_i)$ and not $\sum_i\|S_i-\mathrm{id}\|$.