An LKB kernel braid fixes every standard adjacent edge up to isotopy

For $n=2$, [F5] forces a kernel braid to be the identity class, proving both edge fixing and label preservation. For $n=1$, the braid group is trivial and there are no edges. Hence assume $n\ge 3$. Fix $i$ and a standard fork with tine $E_i$. For each $j\mathrm{\notin }\{i,i+1\}$, choose its closing neighborhoods small enough to miss $M_j$; its compact replacement pairs to zero with $y_{M_j}$ because the tine is disjoint and the closing pieces also miss the crosscut. The kernel fixes that absolute class. The image replacement therefore still pairs to zero; [F3] makes $\sigma (E_i)$ individually disjoinable from each $M_j$. By [F4] isotope this one arc simultaneously off all those crosscuts. No previously arranged edge is invoked or trimmed.

Each $M_j$ separates a disk $V_j$ containing only $p_j$ from all other punctures. A connected arc disjoint from it whose two distinct ends are punctures cannot have an end $p_j$: otherwise the entire arc would lie in that component and there would be no possible second puncture end. Consequently $\{\sigma (p_i),\sigma (p_{i+1})\}=\{p_i,p_{i+1}\}\mathrm{.}$ The disjoined arc lies in the remaining closed disk $R_i$ obtained by removing the interiors of the caps $V_j$ for $j\mathrm{\notin }\{i,i+1\}$, with their crosscut boundaries retained. This disk has exactly two marked interior points $p_i,p_{i+1}$ and also contains $E_i$.

In a disk with exactly two marked interior points, every simple arc joining them is isotopic, as an unoriented image, to any other. Here is the relative construction rather than a simply-connectedness assertion about the punctured disk. Extend each such arc by two access arcs to distinct boundary points, yielding a crosscut and an outer-circle graph. Prescribe the target graph map to the corresponding graph for the original arc, fixing the outer boundary and taking the two ordered marked vertices to themselves; relative Schoenflies on the Jordan faces extends it to an orientation-preserving marked disk homeomorphism $h$. The two-point mapping-class theorem identifies $[h]$ with a power of its standard half twist, because $B_2\cong \mathbb{Z}$. That half twist has a representative supported around the target arc and preserves its image. Thus an isotopy from $h$ to that representative applied to the target arc takes the original image to the target image, with both marked points fixed during the isotopy whenever $h$ fixes them: its power is then even. Extend the isotopy of $R_i$ by identity over the removed caps, since it fixes $\mathrm{\partial }{R}_i$. This proves $\sigma (E_i)\simeq E_i$ relative to all of $P\cup \mathrm{\partial }D$.

Repeat the independent argument for every $i$, obtaining both its edge image class and its unordered endpoint pair. The pairs for $i=1,2$ intersect in the singleton $p_2$, so their preserved images force $\sigma (p_2)=p_2$ and then $\sigma (p_1)=p_1$, $\sigma (p_3)=p_3$. Each subsequent pair forces the next puncture fixed. Thus every label is preserved. Once this is known, the edge isotopies in step 2.1 may also be parametrized from $p_i$ to $p_{i+1}$: any final increasing interval reparametrization $f$ is corrected by $(1-s)f+s\mathrm{id}$. The lemma claims the individual edge classes; simultaneous pointwise spine fixing is proved by its separate boundary-twist consumer.