I've alrealdy asked this question, but now I have more clear ideas, so I'm going to ask again and see if I'll understand a bit more. It's about the trivialisation of the Moebius strip as a bundle on $S^1$ with fibre an open interval $I\subset \mathbb{R}$.
First of all, I write the strip as $E=[0,1]^2/\sim$, where $(0,y)\sim(1,1-y)$ and the other points of $[0,1]^2$ are equivalent to themselves. Then I write $S^1=[0,1]/\sim$, where $0\sim 1$. Now I define the projection $\pi: E\rightarrow S^1$, $\pi([(x,y)])=[x]$ (here with $[\cdot]$ I mean the equivalence class of $\cdot$). Now I want to find an open cover $U_1,U_2$ of $S^1$ (two arcs of $S^1$ substantially) such that there exist homeomorphisms $\phi_i:\pi^{-1}(U_i)\rightarrow U_i\times I$, $i=1,2$ and this homeomorphisms have the property that $\pi_1\circ \phi_i=\pi$, where $\pi_1$ is the canonical projection onto the first factor.
Now, my problem is write properly the homeomorphisms. Suppose that $S^1\ni[0]\notin U_1$, so $[0]\in U_2$. Then $\pi^{-1}(U_1)$ is exactly $U_1\times I$ and as homeomorphism I can choose the identity. But now my problem is $\pi^{-1}(U_2)$ which contains the gluing edges. Can anyone help me, please?
For example if $S^1\supset U_2=([0,2/5)\cup(3/5,1])/\sim$, then I could define $\phi_2([(x,y)])=(x,y)$ if $0\le x<2/5$ and $\phi_2([(x,y)])=(x,1-y)$ if $3/5<x<1$. But now the problem is prove that $\phi_2$ is a homeomorphism, I'm confused.