A local Whitney move in Euclidean space

Use an increasing coordinate change to put the zeros at $u=-1,1$, and reflect $v$ if necessary so $f$ is negative between them and positive outside. The ratio $\lambda (u)=(u^2-1)\mathrm{/}f(u)$ extends smoothly and positively over the two zeros by their nonzero first derivatives. The map $(u,v)\mapsto (u,\lambda (u)v)$ is a plane diffeomorphism fixing the axis and taking the other arc to $v=u^2-1$. At its corners the determinant of the ordered tangent directions $(1,0),(1,2u)$ is $2u$, giving one negative and one positive intersection.

Apply the explicit local flow of the model definition: $g(u)=b(u)(u^2-1-\epsilon )$, with $b=1$ on $[-1,1]$, and use a compact vertical cutoff equal to one on the swept segments. The axis is taken to $v=g(u)$ at time one. For $\mathrm{\mid }u\mathrm{\mid }\le 1$, $g=u^2-1-\epsilon <u^2-1$. For $\mathrm{\mid }u\mathrm{\mid }>1$, $u^2-1>0$ and $g=b(u)(u^2-1)-b(u)\epsilon <u^2-1$, also where $b=0$. Thus the moved axis and the unchanged graph are disjoint. The vector field has compact support, so its auxiliary time maps are diffeomorphisms; the first arc is embedded throughout. Pull back by the plane normalization to obtain the asserted isotopy of the original first arc, with the second held fixed.

In complementary dimensions the sheet factors are $E={\mathbb{R}}^{a-1}$ and $H={\mathbb{R}}^{b-1}$, with sheets $\{v=0,h=0\}$ and $\{v=u^2-1,e=0\}$. Use the compact normal cutoff from the theorem; a possible intersection still forces $e=h=0$, where the preceding calculation applies. The normal factors are split sheet directions, not a simultaneous product action on both images. This verifies the exact local picture and the cancellation of the opposite-sign pair.