An immersed disk in the four-ball cannot always be cleaned relative to its boundary

Change the middle positive crossing to a negative crossing, giving ${\sigma }_1{\sigma }_1^{-1}{\sigma }_1$. Cancel the first inverse pair by an explicit local ambient isotopy. In a braid cylinder $D^2\times I$, write its two strands as $(\pm r{e}^{i\theta (u)},u)$, where $\theta$ makes one half-turn and then its inverse and is zero near both cylinder ends. Choose a smooth cutoff $\eta$ of the squared radius, equal to one near $r^2$ and zero near the boundary value. The maps $(z,u)\mapsto (e^{-is\eta (\mathrm{\mid }z{\mathrm{\mid }}^2)\theta (u)}z,u)$ preserve radius, have inverse obtained by changing $s$ to $-s$, and are the identity near the cylinder boundary. They extend by the identity to ambient isotopies of $S^3$ and straighten these two strands at $s=1$. The remaining one-crossing closure bounds an explicit embedded disk in $S^3$: take the two disks spanning its oriented smoothing circles at separate heights and join them by the single narrow half-twisted crossing band. This surface is embedded, and two disks joined by one band connecting their components form a disk. Its boundary is exactly the one-crossing closure. Thus the knot movie from the trefoil reaches a disk-bounding knot with one crossing change and otherwise only the displayed ambient isotopy; no external Reidemeister theorem is used.

Put this movie into $S^3\times [0,\epsilon ]$ by sending a strand point at movie time $t$ to $(k_t(u),t)$. Away from the crossing-change time, each time slice is embedded and the time coordinate separates distinct slices. Near the event use spatial coordinates $(x,y,z)$ and time $t$, with the two sheets parametrized by $(u,t)\mapsto (u,0,t,t)$ and $(v,t)\mapsto (0,v,-t,t)$. They coincide only at $u=v=t=0$. Their tangent planes are spanned by $(1,0,0,0),(0,0,1,1)$ and by $(0,1,0,0),(0,0,-1,1)$, respectively; these four vectors are independent. Both branches are immersions and meet transversely at this single point. Patch this local movie to the stationary outside strands, and choose stationary time collars at both endpoints. This constructs an immersed annulus with exactly one transverse interior double point.

Use the embedded disk in the inner collar sphere constructed in step 1.1 as a cap. In a fresh inward collar write its graph as $(d(x),\epsilon +\tau (x))$, where $d:D^2\hookrightarrow S^3$ is that disk, $\tau$ vanishes on its boundary, is positive in its interior, and has positive inward derivative near its boundary. The graph is embedded because $d$ is, and its boundary is the inner movie knot. Its interior lies deeper than the movie annulus, so there are no new coincidences. Glue along their stationary boundary collars and round the corner. Annulus plus disk is a properly immersed disk whose only double point is the transverse crossing-change event of step 2.1.

The local trefoil nonsliceness lemma proves that no smooth proper embedded disk has boundary $K$: a putative slice disk would have a rationally acyclic branched double cover, but its trefoil boundary cover has first homology of order $3$, contradicting the locally proved square-order consequence of duality. Thus the immersed disk cannot be homotoped relative to its boundary to a proper embedding. This is a disk-cleaning obstruction; calling the disk a Whitney disk additionally requires sheet arcs and their boundary data, which are not part of this witness.