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

Question

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 $$g=g_{\theta,t}=\{x\in\mathbb R^2\ |\ <x,\theta>=t\}\ ,\theta\in S^1,t\in\mathbb R$$

@MartinBrandenburg Is $G$ the straight lines in $\mathbb{R}^2$ or the straight lines through zero in $\mathbb{R}^2$? — Neal, Nov 23 '13 at 22:21
Affine lines in $\mathbb R^2$ can be identified naturally with $\mathbb R P^2$ minus a point, which is the Möbius band. — Dustan Levenstein, Nov 23 '13 at 22:28

score 2 · Accepted Answer · answered Dec 20 '13 at 06:53

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.

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

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