Understanding Mobius bundle. I'm trying to solve the following question:

Consider the vector bundle
$$E:= [0,1] \times \mathbb{R}/\sim \quad \to S^1$$ 
where $\sim$ is the equivalence rleation $(0,t) \sim (1,-t)$ for all
   $t \in \mathbb{R}$. Does there exist a section of $E$ that is nowhere
  vanishing?

How can I visualise $E$? It looks like some kind of infinite Möbius strip. Also, how is the map $E \to S^1$ explicitely defined? I cannot see what map is meant here, but there should be an obvious choice. 
 A: The map $E\to S^1$ is explicitly defined by $\overline{(x,t)}\mapsto e^{2i\pi x}.$ You can show that it is an homeomorphism when restricted to $[0,1]\times\{0\}/\sim$. 
Then, you can show that the (continuous) sections of this bundle are in 1-1 correspondance with continuous maps $\sigma:[0,1]\to\mathbb{R}$ such that $\sigma(0)=-\sigma(1)$ by sending $\sigma$ to $\tilde{\sigma}:S^1\to E$ defined by $\tilde{\sigma}(e^{2i\pi x})=\overline{(x,\sigma(x))}$, with $0\leq x\leq 1$. Finally, the existence of a nowhere vanishing section would imply the existence of a nowhere vanishing function $\sigma:[0,1]\to\mathbb{R}$ with $\sigma(0)=-\sigma(1)$. Since $\sigma(0)\neq 0$, then $\sigma(0)$ and $\sigma(1)$ are of opposed signs. By continuity of $\sigma$, it contradicts the intermediate value theorem.
Edit to answer your questions:
Sounds good to me! 
For $\sigma(0)=-\sigma(1)$, note that $\tilde{\sigma}:S^1\to E$ is a section, which means that for all $x\in[0,1]$ there is a $y_x$ such that $\tilde{\sigma}(e^{2i\pi x})=\overline{(x,y_x)}$. Note that $\tilde{\sigma}(1)=\tilde{\sigma}(e^{2i\pi 0})=\overline{(0,y_0)}$ and $\tilde{\sigma}(1)=\tilde{\sigma}(e^{2i\pi 1})=\overline{(1,y_1)}$, so by definition of the equivalence relation we get $y_0=-y_1$. Since with our definition we get $\sigma(0)=y_0$ and $\sigma(1)=y_1$, we get the wanted result.
A non-vanishing section $s:B\to E$ of a vector bundle is a section such that 
$$\forall x\in B,s(x)\neq 0_{E_x}$$ 
where $0_{E_x}$ is the $0$ of the vector space $E_x=\pi^{-1}(x)$.
