Mapping property of $w=\sin z$.
Let $G=\{z: -\frac{\pi}{2}<\Re z<\frac{\pi}{2}, \Im z>0\}$.
This is one of pink-colored vertical stripe regions of Fig.1.
When $z$ moves along the half-line $l_1=\{z: z=-\frac{\pi}{2}+it, t:\infty \to 0\}\subset \partial G$,
$w=\sin (-\frac{\pi}{2}+it)=-\frac{1}{2}(e^{-t}+e^t)$ moves along $L_1=(-\infty,-1).$
Arrows indicate those directions of moving. When $z$ moves along the segment $l_2=[-\frac{\pi}{2}, \frac{\pi}{2}]\subset \partial G$, $w=\sin z $ moves along $L_2=[-1,1]$. Also when $z$ moves along the half-line $l_3=\{z: z=\frac{\pi}{2}+it, 00\}$ injectively.
The same argument shows that each pink-colored stripe region is mapped onto $H^+$ injectively by $w=\sin z$ and all $l_i$ correspond to $L_i$, $i=1,2,3$.
Similarly all white-colored stripe regions are mapped onto the lower half-plane $H^-=\{w:\Im w<0\}$.

Now we consider four curves $a,b,c$ and $d$ depicted in Fig.2. Of course $a\cdot b\cdot c\cdot d=u\cdot v$ and $c\cdot d\cdot a\cdot b=v\cdot u$.
The lifting of $u\cdot v.$
The curve $a$ starts at the origin, lies in $H^-$ and ends at a point $w=2\in L_3.$ Therefore it's lifting $\tilde{a}$ must start at the origin, lie in a white-colored region, and end at a point $P(=\sin^{-1}2)\in l_3$. Also the lifting of $b$, say $\tilde{b}$, starts at $P$, lies in a pink-colored region and ends at a point $Q(=\pi)\in l_2$, since $b$ starts at the end point of $a$, lies in a pink-colored region and ends at $w=0\in L_2$. These are illustrated in Fig.2.
We continue our arguments.
The lifting of $c$ must start at $Q$, lie in a pink-colored region and end at a point $R\in l_1$, since $c$ starts at the end point of $b$, lies in a pink-colored region and ends at $w=-2\in L_1$.
Finally the lifting of $d$ starts at $R$, lies in a white-colored region and ends at $z_0\in l_2$, since $d$ starts at the end point of $c\, (w=-2),$ lies in a white-colored region and ends at $w=0$.
Since $\sin z_0=0,\, \frac{3\pi}{2}
We only depict the lifting of $v\cdot u$ in Fig.2. We have $w_2(1)=z_1=-2\pi$.
