Local meridian conjugacy at a Wirtinger crossing

Take ambient coordinates $(x,y,z)$ with overstrand tangent $+{\mathrm{\partial }}_x$, viewing normal $+{\mathrm{\partial }}_z$, and positive projection basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$. Source has $y<0$ and target $y>0$, with the understrand below the overstrand. Parameterize the under-core from source to target by $u$, whether or not this agrees with the knot orientation. Choose its product tube $\mathrm{\Phi }(u,v)$ so that the oriented knot tangent followed by the ordered normal frame is ambient-positive; take the first normal-frame vector toward the positive-$z$ side of the understrand and the second to complete that orientation. Thus increasing angle from ${\theta }_0=0$ gives the positive meridian even when $u$ runs opposite to the knot orientation. Choose a small positive radius $r$. The path used here is the push-off $\beta (u)=\mathrm{\Phi }(u,r(\cos {\theta }_0,\sin {\theta }_0))$, from source to target; it never lies on the knot core. The two meridian loops are the boundaries ${\mu }_u(\tau )=\mathrm{\Phi }(u,r(\cos ({\theta }_0+2\pi \tau ),\sin ({\theta }_0+2\pi \tau )))$, based at the endpoints of $\beta$. For exterior paths take $r$ at least the removed tube radius; shrink that radius first if necessary. Choose the product tube disjoint from the overstrand and other arc pieces, and $r$ smaller than the crossing-chart separation bounds. Choose upper-collar paths ${\alpha }_s,{\alpha }_t$ to these actual boundary points. In the normal $(y,z)$ plane of the overstrand, their concatenation ${\alpha }_s\beta {\alpha }_t^{-1}$ follows a small perturbation of the polygon $(0,H),(-d,-h),(d,-h),(0,H)$, which has positive signed area. The polygon and a sufficiently small surrounding strip avoid the origin; the chosen radius makes the perturbation lie in that strip, so its winding is still $+1$. For explicit complement control, straighten the local under-core to $x=0,z=-h$ in the product chart and take its upper push-off at $z=-h+r$. The access triangle is then entirely above $z=-h$, while the over-core projects to the origin. Radial interpolation of its positive polar-distance function to a sufficiently small circle about the origin stays inside that triangle minus the origin, hence above the under-core as well. Retain the positive-radius angular access in the under-tube until reaching these straightened paths. Thus the resulting homotopy is to one counterclockwise over-meridian and avoids both strands throughout. Thus all paths and homotopies lie in the complement, the loop represents $m_o$, and ${\alpha }_t\simeq m_o^{-1}{\alpha }_s\beta$ rel endpoints there.

The positive-radius cylinder $(u,\tau )\mapsto \mathrm{\Phi }(u,r(\cos ({\theta }_0+2\pi \tau ),\sin ({\theta }_0+2\pi \tau )))$ transports the boundary meridians, with its basepoint edge equal to $\beta$. Its image misses the knot core and, by the separated tube choice, the overstrand and every other arc. Reading the boundary of this cylinder gives $\beta {\mu }_t{\beta }^{-1}\simeq {\mu }_s$ in the complement; its disk orientation is transported continuously even when the source-to-target direction opposes the knot orientation. Therefore $[{\alpha }_t{\mu }_t{\alpha }_t^{-1}]=m_o^{-1}[{\alpha }_s\beta {\mu }_t{\beta }^{-1}{\alpha }_s^{-1}]m_o=m_o^{-1}{m}_s{m}_o$. Equivalently the annular-core word $m_o^{-1}{m}_s{m}_o{m}_t^{-1}$ bounds the cap of [F1], consistently with [F3]. Reversing the understrand orientation inverts both its meridians, so inversion gives the same side-based equation, rather than a switch of the conjugating direction.

The under-tangent follows $n_o$ exactly when incoming is source and outgoing is target, giving the first incoming/outgoing equation. Otherwise solve the same equation for $m_s$ to obtain the second. A reversed meridian is the inverse loop by [F2]. If a common access path $\alpha$ is replaced by $\lambda \alpha$, the class $[\alpha \mu {\alpha }^{-1}]$ becomes $[\lambda ][\alpha \mu {\alpha }^{-1}][\lambda {]}^{-1}$; this holds for each of the three meridians. These observations prove all remaining assertions.