Existence and rigidity of bigradings

Non-closed curves admit bigradings. If $c$ is an arc, $c\setminus \mathrm{\Delta }$ is a contractible interval [L3], hence path-connected and locally path-connected with trivial fundamental group. Choose any point over $s_c(z_0)$; the subgroup condition in [L2] is then automatic, and the lifting criterion gives a continuous lift of $s_c$, which is a bigrading.

Simple closed curves do not admit bigradings. By Jordan's theorem an essential simple closed curve encloses $k\ge 1$ marks. The two circle coordinates of ${\delta }_P{s}_c$ have winding $\pm (2-2k,k)$ by [L4]. A continuous real-coordinate lift around $c$ would return to its initial value, forcing both windings to be zero, contrary to $k\ge 1$. Hence no bigrading exists.

Uniqueness up to the deck action. Suppose $c$ admits bigradings $\tilde{c}$ and ${\tilde{c}}^{\mathrm{\prime }}$. Both are lifts of the same section $s_c$ over the connected base $c\setminus \mathrm{\Delta }$ [L3], so ${\tilde{c}}^{\mathrm{\prime }}(z)=\chi (\delta (z))\tilde{c}(z)$ for a continuous function $\delta \text{\hspace{0.17em}⁣}:c\setminus \mathrm{\Delta }\to {\mathbb{Z}}^2$; since ${\mathbb{Z}}^2$ is discrete this function is locally constant, and since $c\setminus \mathrm{\Delta }$ is connected it is constant, say $\delta \equiv (r_1,r_2)$. Thus ${\tilde{c}}^{\mathrm{\prime }}=\chi (r_1,r_2)\tilde{c}$, and $(r_1,r_2)$ is unique because $\chi$ acts freely on each fibre.

Freeness on isotopy classes. Parametrize the given bigraded isotopy by smooth embedded arcs ${\gamma }_t:[0,1]\to D$, choosing the parametrizations so that ${\gamma }_1={\gamma }_0$; this is possible by interpolating the increasing reparametrization of the returned arc. The base and tangent-line traces at each unmarked parameter value are therefore closed loops. If the arc has a boundary endpoint, that endpoint is fixed, and its tangent line stays transverse to the boundary throughout the isotopy. Its projective tangent trace lies in $\mathbb{R}{P}^1$ minus the boundary tangent line, a contractible interval; its base trace is constant. Hence its cover monodromy is zero. If both endpoints are distinct marks $q_0,q_1$, the loops ${\kappa }_s(t)={\gamma }_t(s)$ for $0<s<1$ are freely homotopic as $s$ varies, and thus have the same winding about every mark. Near $q_0$ all windings except the one about $q_0$ vanish, while near $q_1$ all except the one about $q_1$ vanish. Comparing these tuples shows that every winding is zero. For small $s>0$, ${\kappa }_s(t)-q_0=s{\gamma }_t^{\mathrm{\prime }}(0)+o(s)$ uniformly in $t$, so the nonzero vector loop ${\gamma }_t^{\mathrm{\prime }}(0)$ also has winding zero; the tangent-line trace at ${\gamma }_t(s)$ converges to its projectivization, and thus has zero fibre winding. The two coordinates of the covering monodromy in [L4] are consequently zero. The deck shift is constant along the connected arc, so in both endpoint cases $(r_1,r_2)=0$. This is the endpoint comparison of Khovanov--Seidel Lemma bigrading-isotopy, printed p. 24, with the tangent contribution made explicit.

Conclusion and isotopy lifting. Steps 1.1 and 1.2 give the existence criterion, step 1.3 gives uniqueness up to a unique deck element, and step 1.4 gives freeness on isotopy classes. Parametrize an arc isotopy by a fixed interval with its marked ends removed. Its tangent sections give a homotopy into $P$, which lifts uniquely from a prescribed initial bigrading by [L2]; interpreting the lifted map on each moving arc gives the required bigraded isotopy. No choice principle is used.