# Liftings of curves $u\cdot v$ and $v\cdot u$ with respect to the sine covering map.

I'm trying to work through the exercises in Otto Forster's book on Riemann Surfaces. While most of them seemed not that hard, this one gives me a headache:

Let $X=\mathbb{C}\setminus\{\pm1\}$ and $Y = \mathbb{C}\setminus\{\frac{\pi}{2}+k\pi~|~k\in\mathbb{Z}\}$. Then $\sin: Y\rightarrow X$ is a covering map. This is clear.

Now consider the following curves:

$$u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$$

and

$$v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$$

Let $w_1: [0,1]\rightarrow Y$ be the lifting of $u\cdot v$ and $w_2$ be the lifting of $v\cdot u$ with $w_1(0)=w_2(0)=0$.

Show that $w_1(1) = 2\pi$ while $w_2(1) = -2\pi$.

As of now, I don't have many ideas how to solve this. Of course, the inverse sine is a multi-valued function so the two liftings should use different branches. But I don't see how exactly they come into play.

I'm also not sure whether the solution can be done by calculations alone or if there is some more general principle underlying. Any suggestions would be welcome.

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