Cylinder and Möbius strip as fiber bundles: trivializations and cocycles

I have to show that the cylinder and the Möbius strip are fiber bundles over $S^1$ with fiber an open interval and I have to write down explicitly the transition functions and the cocycles.

Cylinder.

The cylinder is $E:=I\times S^1$ where $I$ is our open interval.

Let's consider $$\pi : I \times S^1 \to S^1$$ the projection on the second factor (by definition it's continuous and surjective).

Let's take $S^1$ as basis of the fiber bundle and let $\{U_\alpha\}_\alpha$ be an open covering of $S^1$.

$\pi_1:U_\alpha\times I\to U_\alpha$ is the projection on the first factor.

The trivialization is $$\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times I\\ \phi_\alpha:=id|_{U_\alpha\times I}$$ In this case the transition functions are such that for all $\alpha,\beta$ $$\phi_{\alpha\beta}=\phi_\alpha\circ\phi_\beta^{-1}=id|_{(U_\alpha\cap U_\beta)\times I}.$$ Hence the cocycles are just $$g_{\alpha\beta}:U_\alpha\cap U_\beta\to Aut(I)\\ p\mapsto id|_I$$

Möbius band Let's start with the closed interval $[0,1]$ and the map $$p:[0,1]\to S^1\\ x\mapsto e^{i2\pi x}$$ which identifies the point $0$ with the point $1$.

Let's consider $[0,1]\times I$, where $I$ is an open interval, let's say $(0,1)$. Let's define an equivalence relation on $[0,1]\times I$: $$(0,y)\sim (1,1-y)$$ and we obtain the Möbius band $E$.

Let's consider an open covering $\{U_\alpha\}_\alpha$ of sphere $S^1$ and let $$\pi:E\to S^1\\ [(x,y)]\mapsto p(x)$$ be continuous and surjective and $$\pi_1:U_\alpha\times I\to U_\alpha$$ be continuous and surjective.

I can't find a trivialization. I just know that I have to choose a proper open covering of $S^1$ like two open interval overlapped.

• Can you find the trivialization in the case that $U_\alpha \subset p(0,1)$? – Lee Mosher Jul 29 '15 at 13:52
• Yes! Let's define $U_1:=S^1-\{0\}=p((0,1))$. Then I want to construct a homeomorphism $\phi_1:\pi^{-1}(U_1)\to U_1\times I$. Let's see what $\pi^{-1}(U_1)$ is. $\pi^{-1}(U_1)=\{[(x,y)]\in E: p(x)\in p((0,1))\}=(0,1)\times I$. Hence $\phi_1:\pi^{-1}(U_1)\to U_1\times I$ maps $[(x,y)]$ to $(p(x),y)$. Right? – avati91 Jul 29 '15 at 14:16
• Yup. That knocks off one big case. Now all you have to do is figure out the complementary case $U_\alpha \not\subset p(0,1)$. – Lee Mosher Jul 29 '15 at 17:15
• I think I solved my doubts :) Thanks – avati91 Jul 29 '15 at 17:20
• The example of gluing a Mobius strip from a cocycle is covered in Jeffrey Lee's book in Example 6.16 in case you want a textbook source. – ಠ_ಠ Oct 29 '15 at 4:16