$[F,F]$ conjugates each of its elements into a prescribed dyadic interval

Fix dyadic rationals $a, b$ with $0 < a < b < 1$. Let $w \in F$ be trivial in a neighbourhood of $0$ and in a neighbourhood of $1$; by Theorem 4.1 these are exactly the elements of the commutator subgroup $[F,F]$.

Then there is an element $\varphi \in F$, itself trivial near $0$ and near $1$ (so $\varphi \in [F,F]$ as well), and reals $c, d$ with

such that the conjugate $\varphi w \varphi^{-1}$ is supported inside $[c,d]$; that is, $\varphi w \varphi^{-1}$ fixes every point of $[0,1]$ outside $[c,d]$. The conclusion is phrased pointwise as: whenever $\varphi\big(w(\varphi^{-1}(z))\big) \neq z$ one has $c \le z \le d$.

This is the transitivity statement that the simplicity argument needs. Because $w$ is trivial near the two endpoints, its support is contained in some interval $[s,t] \subset (0,1)$; enlarging to dyadic endpoints and using the transitivity of $F$ on dyadic partitions of $[0,1]$, one builds $\varphi \in F$ which is the identity near $0$ and near $1$ and carries $[s,t]$ into a compact subinterval of $(a,b)$. Conjugation transports supports, $\operatorname{supp}(\varphi w \varphi^{-1}) = \varphi(\operatorname{supp} w)$, which gives the conclusion. The requirement $c > a$ and $d < b$ — strict containment, not merely support inside $(a,b)$ — is what later makes $\varphi w \varphi^{-1}$ trivial near the endpoints of $[a,b]$.