A blackboard push-off doubles each crossing into mixed crossings

The zero section of $\nu K$ is compact, so after shrinking the parameter interval there is a $\delta >0$ such that $\rho (t,\epsilon ):=\pi \mathrm{\Phi }(K(t),\epsilon [b_D(t)])$ is smooth on $S^1\times (-\delta ,\delta )$, with $\rho (t,0)=\gamma (t)$. The normalized-chart condition says that ${\mathrm{\partial }}_{\epsilon }\rho (t,0)$ represents $R{\widehat{T}}_{\gamma }(t)$ modulo the tangent line spanned by ${\gamma }^{\mathrm{\prime }}(t)$. Therefore ${\gamma }^{\mathrm{\prime }}(t)$ and ${\mathrm{\partial }}_{\epsilon }\rho (t,0)$ are linearly independent at every $t$. By [F4], $(t,\epsilon )\mapsto \rho (t,\epsilon )$ is locally injective near every $(t,0)$. Compactness of $S^1$ and [F7] give an open diagonal neighbourhood $V_{\mathrm{\Delta }}\subseteq S^1\times S^1$ and a uniform ${\epsilon }_0>0$ such that, for $0<\epsilon <{\epsilon }_0$, whenever $(s,t)\in V_{\mathrm{\Delta }}$, the points $\rho (s,0)$ and $\rho (t,\epsilon )$, and the points $\rho (s,\epsilon )$ and $\rho (t,\epsilon )$, lie in one common injectivity chart. Shrink $V_{\mathrm{\Delta }}$, if necessary, so that it contains no off-diagonal crossing pair; this is possible because there are finitely many such pairs. Hence there are no projected coincidences between the original and offset branches with close parameters, and $t\mapsto \rho (t,\epsilon )$ is locally injective. Choose disjoint small open neighbourhoods of the finitely many ordered crossing pairs of $\gamma$ outside $V_{\mathrm{\Delta }}$; by [F1], these are the only off-diagonal equalities $\gamma (s)=\gamma (t)$. The compact complement of $V_{\mathrm{\Delta }}$ and those crossing neighbourhoods contains no such equality, so $\parallel \gamma (s)-\gamma (t)\parallel$ has a positive minimum there. Smoothness gives uniform convergence $\rho (\cdot ,\epsilon )\to \gamma$, excluding original/offset and offset/offset projected coincidences on this compact complement for sufficiently small $\epsilon$. Thus every projected coincidence lies in one of the chosen crossing neighbourhoods.

Fix one crossing disk with disjoint parameter intervals around the over branch $\alpha$ and under branch $\beta$, and with no other branch entering. For $i,j\in \{0,1\}$, the map $G_{ij}(s,t,\epsilon ):=(\rho (s,i\epsilon )-\rho (t,j\epsilon ),\epsilon )$ has invertible derivative at the original crossing and $\epsilon =0$: its first two columns are ${\alpha }^{\mathrm{\prime }}(0)$ and $-{\beta }^{\mathrm{\prime }}(0)$, which are linearly independent by transversality, and its last coordinate is $\epsilon$. By [F4], each of the four branch pairs has exactly one transverse projected intersection near this crossing for every sufficiently small $\epsilon$. They are, respectively, the original crossing, the two mixed crossings, and the crossing of the two offset branches. The local injectivity from step 1.1 excludes intersections between an original branch and its own offset; it also implies that the four listed crossing points are distinct, since a coincidence between any two would place an original branch and its own offset at the same projected point. Thus there are no triple points in this crossing disk. The strict height order of $\alpha$ over $\beta$ at the original crossing persists for all four nearby crossings when $\epsilon$ is small. The projected tangent determinants are nonzero and vary continuously, so both mixed crossings have sign $\mathrm{sgn}(c)$ by [F2].

There are finitely many crossing disks, so choose $\epsilon$ below all local bounds from steps 1.1 and 2.1. By [F6], the spatial push-off is an embedded circle disjoint from $K$. The projection of $K\cup K_{\epsilon }$ has only the finitely many transverse double points described above and no triple points, so it is a regular diagram. Each original crossing disk contributes exactly two mixed crossings, both with the original sign, and step 1.1 excludes all other mixed crossings. Countable Choice is used only through [F6] for the tubular push-off; the finite projected-crossing argument uses no choice.