The Whitney move removes a cancelling pair of intersection points

Apply the adapted-tube lemma to the supplied smooth clean framed bigon and its compatible corner collars. It gives an open plane extension and a uniformly thin embedded tube whose sheet inverse images are precisely the two extended edges with their respective normal blocks. The affine-in-$v$ plane change $(u,v)\mapsto (u,(v-(1-u^2))\mathrm{/}2)$ sends the upper edge of $\mathcal{B}$ to $v=0$ and the lower edge to $v=u^2-1$; its determinant is $1\mathrm{/}2$ even at the two corners. Reparametrize the tube by this diffeomorphism, keeping the $E,H$ coordinates. The selected sheets are therefore exactly $\{v=0,h=0\}$ and $\{v=u^2-1,e=0\}$ on the adapted neighbourhood.

Choose $\epsilon >0$ small and the transition of $b$ close to the two corners. The swept plane set $S=\{(u,tg(u)):u\in \mathrm{supp}b,0\le t\le 1\}$ is compact, as the continuous image of a compact product; these choices put $S$ inside the open plane tube domain $U$, since the swept segments lie in the bigon plus an arbitrarily thin collar. Choose a compact neighbourhood $K_S$ of $S$ inside $U$ using finitely many sufficiently small closed balls, and a relatively compact open set $O$ with $K_S\subset O\subset \bar{O}\subset U$. The Euclidean bump supplier gives $c(u,v)=1$ on $K_S$ with support in $O$; its support is compact since it is closed and lies in the compact $\bar{O}$. Choose the smooth normal cutoff $\chi$ with compact support in the normal tube balls, equal to one near zero, and with $0\le \chi \le 1$. The model vector field $g(u)c(u,v)\chi (e,h){\mathrm{\partial }}_v$ is then smooth and compactly supported in the tube. On the first sheet it keeps $u,e,h$ fixed and takes $v=0$ to $v=tg(u)\chi (e,0)$: the entire trajectory belongs to the swept segment in $S$, where $c=1$. Its support avoids the tube boundary and selected patch boundary. Transport it and extend it by zero to a compactly supported smooth ambient vector field on $X$. Compact-support completeness gives its global smooth flow, and the flow time maps are diffeomorphisms with inverse the negative-time map. This is the asserted auxiliary isotopy.

At a possible intersection of the moved first sheet with the fixed second one must have $e=0$ and $h=0$. At time one the second coordinate of the moved sheet is $g(u)$, since $\chi (0,0)=1$. For $\mathrm{\mid }u\mathrm{\mid }\le 1$ it is $f(u)-\epsilon <f(u)$. Outside that interval $f(u)>0$, and $g(u)=b(u)f(u)-b(u)\epsilon <f(u)$, including where $b=0$. Hence no model intersection remains. The first sheet stays embedded because the auxiliary time map is a diffeomorphism. The tube can avoid every remaining sheet part outside its designated arc collars: the disk interior is clean, and the closed sheet parts outside smaller designated collar neighbourhoods are disjoint from the compact disk and can be excluded by shrinking its neighbourhood, while the product charts handle the endpoints. All intersections outside the tube are fixed.

For global embedded sheets, apply this auxiliary isotopy to $A$ and compare with the unchanged $B$, obtaining the asserted $A^{\mathrm{\prime }}$. For an immersion patch $P$, set $f_t=G_t\circ f$ on $P$ and $f_t=f$ outside $P$. Agreement on an open collar of $\mathrm{\partial }P$ makes these formulas smooth. On $P$, the derivative is $d{G}_t\circ df$ and remains injective; outside it is unchanged. Tube avoidance of all other source-image branches excludes extra coincidences. Intermediate intersections with the fixed second sheet may be tangent, while each source branch remains immersed. The cancellation formula at time one follows from step 3.1. The map $G_t\circ f$ on the whole source would preserve all coincidences and is not this construction.