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, 0, $w=\sin\left(\frac{\pi}{2}+it\right)=\frac{1}{2}(e^{-t}+e^t)$ moves along $L_3=(1, \infty)$. Therefore it follows from Darboux theorem that $w=\sin z$ mapps $G$ onto the upper half-plane $H^+=\{w:\Im w>0\}$ 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 know that $w_1(1)=z_0=2\pi.$
We only depict the lifting of $v\cdot u$ in Fig.2. We have $w_2(1)=z_1=-2\pi$. 