 Solution

Explore the two stopped curves in the opposite order. Begin with an ${\mathrm{SLE}}_6$ from $1$ to $\infty$, stopped on leaving $B(1,1)$, and then, in its unbounded complementary component, draw an ${\mathrm{SLE}}_6$ from $-1$ to $\infty$, stopped on leaving $B(-1,1)$. Before their respective stopping times, each curve is separated from the neighborhood in which the other hull changes the domain. The Locality property of SLE therefore says that mapping out the other stopped hull does not change either stopped marginal law.

The two exploration orders consequently define the same joint law for the pair of stopped hulls. Disintegrating this joint law with respect to the second curve shows that, conditional on ${\gamma }_2{|}_{[0,{\tau }_2]}$, the first curve is an ${\mathrm{SLE}}_6$ in the unbounded component offrom $-1$ to $\infty$, stopped when it leaves $B(-1,1)$, as required.