Show that $G:=\{g\ |\ g\text{ is a straight line in }\mathbb R^2 \}$ is homeomorphic to the Moebius strip 
Show that $G:=\{g\ |g\text{ is a straight line in }\mathbb R^2 \}$ is homeomorphic to the Moebius strip

I defined
$\bar f:\mathbb R\times S^1/\tilde{}$ where $(t,\theta)\tilde{} (-t,-\theta)$
with $(t,\theta)\mapsto\{x\in\mathbb R:<x,\theta>=t\}$
then we have $\pi:\mathbb R\times S^1\rightarrow\mathbb R\times S^1/\tilde{}$ (continuous by assumption)
here i was told to make use of $f$ which is $f=\bar f\circ\pi$. but,
Why is $f$ continuous ? and
How can we prove that $\bar f^{-1}$ is continuous?
edit: I think it must be the OPEN moebius strip.
G is the set, which contains all straight lines in $\mathbb R^2$ and as a hint i have that every straight line can be written in the form $$g=g_{\theta,t}=\{x\in\mathbb R^2\ |\ <x,\theta>=t\}\ ,\theta\in S^1,t\in\mathbb R$$

Starting with 
$$G=\{\{x\in\mathbb R^2\ |\ \langle x,\theta \rangle=t\}  : \theta\in S^1,t\in\mathbb R\}$$
let $C\subset S^1$ be a closed half-circle with endpoints $\alpha,\beta$. Observe that every element of $G$ can be written with $\theta\in C$. The product $C\times \mathbb R$ is topologically a rectangle which is open on two sides  ($t$ direction) and closed on two others ($\theta=\alpha$, $\theta=\beta$). It remains to perform the identification $(\alpha,t)\sim (\beta,-t)$, which is precisely what is done to glue a Möbius strip from a rectangle. 
