 Solution

For $0<\kappa \le 4$, use the allowed probability-one event that the whole SLE trace lies in $\mathbb{H}$ at every positive time. At each deterministic rational $s>0$, part (b) makes the centered mapped future a fresh SLE with the same property. Consequently, almost surely, for every $t>0$,This inverse relation holds through the trace's boundary-extension construction and puts positive-time future points in the open surviving domain. It therefore excludes every contact with the old filled hull, not just contacts with a fixed old point.

Intersect these probability-one events over all positive rational $s$. If ${\gamma }_u={\gamma }_v$ for $0<u<v$, choose rational $s$ with $u<s<v$. The first occurrence belongs to $K_s$, whereas the second lies in $D_s$, a contradiction. A return to ${\gamma }_0=0$ is excluded by the original positive-imaginary-part assertion. This simplicity of SLE from boundary-avoiding restarts proves