Raw slicing reverses geometric stacking products

The lower half is the first slice loop. For $0\le t\le \frac{1}{2}$, [L1] gives $S(\gamma \star \beta )(t)=[(z_1(2t),\dots ,z_n(2t))]=S(\beta )(2t),$ which is the first half of the concatenation $S(\beta )\ast S(\gamma )$ by [L3].

The upper half is the second slice loop. For $\frac{1}{2}\le t\le 1$, [L1] gives the ordered tuple $(w_{\pi (\beta )(1)}(2t-1),\dots ,w_{\pi (\beta )(n)}(2t-1))$. Since $\pi (\beta )$ is a permutation, this is a reordering of the coordinates of $\gamma$ at height $2t-1$; [L4] therefore gives $S(\gamma \star \beta )(t)=S(\gamma )(2t-1),$ the second half of $S(\beta )\ast S(\gamma )$.

The piecewise paths agree at the seam. At $t=\frac{1}{2}$, the first half has value $S(\beta )(1)=[Q]$ and the second has value $S(\gamma )(0)=[Q]$ by [L2]. The stacked slice is also this same orbit by [L1]. Thus the two formulas establish the pointwise identity of based loops $S(\gamma \star \beta )=S(\beta )\ast S(\gamma )$ on all of $I$, including the shared endpoint of the two closed halves.

Taking path-homotopy classes of this equality and using the product convention [L3] gives $[S(\gamma \star \beta )]=[S(\beta )][S(\gamma )]$ in the group ${\pi }_1(C_n(\mathrm{int}{D}^2),[Q])$ of [L5]. For $n=0$ all loops are the unique empty loop and the identity holds; for $n=1$, $\pi (\beta )$ is the identity and the same two-half calculation applies without collision conditions.