Consider the covering map $p: \mathbb{R} \times \mathbb{R}_+ \to \mathbb{R}^2-0 $ given by $p(x,t) = (t\cos(2\pi x), t\sin(2\pi x))$ Find liftings of the paths


*

*$f(t) = (2-s,0),$

*$g(t) = ((1+s)\cos(2 \pi s), (1+s)\sin(2 \pi s))$,

*$h(t) = f*g$.
By definition we must find functions $f_1$, $g_1$, $h_1 : X \to  \mathbb{R} \times \mathbb{R}_+$ such that $p \circ f_1 = f$, $p \circ g_1 = g$ and $p \circ h_1 = h$. But I don't know how to do that.
Any help?
 A: Let the covering map $p:\mathbb R \times \mathbb R_+\rightarrow \mathbb R^2-0$ be given by $$p(x,y)=(y\cos{2\pi x},y\sin{2\pi x})$$
Let the following $f,g,h:[0,1]\rightarrow \mathbb R^2-0$ be given: 
$$\begin{array}{cc}
f(t) = & (2-t,0)\\
g(t) = & ((1+t)\cos{2\pi t},(1+t)\sin{2\pi t})\\
h(t) = & f*g(t)
\end{array}$$
For respective lifts $f',g',h':[0,1]\rightarrow \mathbb R\times \mathbb R_+$ we must have $p \circ f' = f $, that is $p(f'_1,f'_2)=f$ (respectively $g,h$). So we solve:
$$\begin{array}{rl}
p(x,y)=& (2-t,0)\\
y\cos{2\pi x} = & 2-t\\
y\sin{2\pi x} = &0
\end{array}$$
We see this has a nice solution setting $$x= 0$$
$$y=2-t$$
So we let 
$$f'(t)= (0,2-t)$$
Which is easily verified. Again for $g'$ we solve: 
$$\begin{array}{cc}
p(x,y) = & ((1+t)\cos{2\pi t},(1+t)\sin{2\pi t})\\
y\cos{2\pi x} = & (1+t)\cos{2\pi t}\\
y\sin{2\pi x} = & (1+t)\sin{2\pi t}
\end{array}$$
And this has solution $$x=t$$ $$y= 1+t$$
So we let $$ g'(t)= (t,1+t)$$
The case of $h'$ should be simpler, set $h'=f' *g'$. This is well defined as $f'(1)= (0,1)$ and $g'(0)=(0,1)$. We verify: $$\begin{array}{rl}
p\circ h' = & p \circ (f'*g')\\
= & p\circ \left\{ \begin{array}{cc}
f'(2t) & t \in [0,\frac{1}{2}]\\
g'(2t-1) & t\in [\frac{1}{2}, 1]\\
\end{array}\right. \\
= &\left\{ \begin{array}{cc}
p \circ f'(2t) & t \in [0,\frac{1}{2}]\\
p\circ g'(2t-1) & t\in [\frac{1}{2}, 1]\\
\end{array}\right. \\
= & \left\{ \begin{array}{cc}
f(2t) & t \in [0,\frac{1}{2}]\\
g(2t-1) & t\in [\frac{1}{2}, 1]\\
\end{array}\right. \\
= & f*g\\
= & h
\end{array}$$
And we have all the desired lifts.
