I'm stuck on this question and tried to follow the partial answer of Neal. Erno's answer is fine too but it seems like I need to find the local trivializations of the Mobius bundle, which requires a bit more work. So here is what I got so far:

We have these two local trivializations of $F$:

$\Phi:\pi^{-1}(S^1-\{1\})\to S^1-\{1\}\times\Bbb{R}$ and $\Psi:\pi^{-1}(S^1-\{-1\})\to S^1-\{-1\}\times\Bbb{R}$. Now I need to construct a diffeomorphism between $F$ and $(I\times\Bbb{R})/\sim$. I can do this with the smooth version of gluing lemma but I need to be careful at the intersection and the transition function $\tau$ will take care of it, but how exactly?

  • $\begingroup$ Sir, how could you post the same title again? ;) $\endgroup$
    – Silent
    Dec 21, 2013 at 14:26
  • $\begingroup$ I just added the "is" there ;) $\endgroup$
    – figura
    Dec 21, 2013 at 14:26
  • $\begingroup$ HAHA, very nice, it was something hot on math.se in this winter! $\endgroup$
    – Silent
    Dec 21, 2013 at 14:28

1 Answer 1


You've got a map from $[0, 1] \times \mathbb R$ onto the Mobius bundle, so I'm going to give it a name -- $f$ -- and identify each point in the Mobius bundle as $f(x, t)$ for some $x$ and $t$. The only place that's ambiguous is that certain points have two names: $f(0, t)$ is the same point as $f(1, -t)$. But I'll make a rule: I'll always use the name that has a $0$ as its first argument. (IF you like, I've restricted the domain of $f$ to $[0, 1) \times \mathbb R$.) That means that $F$ provides a 1-1 correspondence between its domain and the Mobius bundle. (The correspondence is NOT continuous at the two "ends" of $[0, 1]$, but at least it gives me a unique way to refer to every point of the total space of the bundle.)

I'm going to define a map $H$ from a subset of the domain of $f$ to $ V \times \mathbb R$ (in the language of your assignment).

$$ H : (0, 1) \times \mathbb R \to V : (x, t) \mapsto (e^{2\pi i x}, t) $$

I'm going to define another map $K$ from a subset of the domain of $f$ to $U \times \mathbb R$:

$$ K : [0, \frac{1}{2}) \cup (\frac{1}{2}, 1) \to U \times \mathbb R: (x, t) \mapsto \begin{cases} (e^{2\pi i x}, t) & \text{for $x < \frac{1}{2}$} \\ (e^{2\pi i x}, -t) & \text{for $x > \frac{1}{2}$} \end{cases} $$

These two maps together map the total space of the first bundle to the total space of the second in a way that respects the gluing maps on each side.

I believe that this is what you need.

But you may be saying "Wait, that's a map, but how can I know it's a smooth isomorphism of bundles?" OK. Here goes.

First, I'm going to place two coordinate charts on the base space of the Mobius bundle. Let me call that bundle $M$, OK? Once again, I'll be using the map $f$ to identify points in $M$.

The first chart's domain is $$ A = (0, 1) . $$ The coordinate map $\phi_A: A \to \mathbb R$ is exactly $x \mapsto x$.

The second chart's domain is $$ B = [0, \frac{1}{2}) \cup (\frac{1}{2}, 1)] / \sim $$

And the coordinate map $\phi_B$ is slightly more complicated: $$ \phi_B( [x] ) = \begin{cases} x & \text{for $0 < x < \frac{1}{2}$} \\ 0 & \text{for $[x] = [0] = [1]$} \\ x-1 & \text{for $\frac{1}{2} < x < 1$} \end{cases} $$

The image of $\phi_B$ is the interval $(-\frac{1}{2}, \frac{1}{2})$.

And now I'm going to build coordinate charts on the bundle; I'll call these $\psi_A$ and $\psi_B$, because one of them will trivialize the portion of the bundle over $A$, and the other will trivialize the portion of the bundle over $B$. Once again, I'll use the labelling of points of the bundle provided by $f$. Here goes:

$$ \psi_A : \pi^{-1}(A) \to (0, 1) \times \mathbb R : f(x, t) \mapsto (x, t). $$

Clear? $\psi_A$ is just the inverse of the map $f$, with its domain somewhat restricted.

Now for $\psi_B$.

$$ \psi_B : \pi^{-1}(B) \to (-\frac{1}{2}, \frac{1}{2}) \times \mathbb R : f(x, t) \mapsto \begin{cases} (x, t) & \text{for $x < \frac{1}{2}$} \\ (x, -t) & \text{for $x > \frac{1}{2}$} \end{cases}. $$

Now let $AB$ denote the intersection of $A$ and $B$.

The transition function $\tau_{AB} : \psi_A(AB) \to \psi_B(AB)$ is once again defined by cases. I'll write it down, and then you have to check that it is, in fact, just $\psi_B \circ \psi_A^{-1}$ on the domain $\psi_A(AB)$.

What's $\psi_A(AB)$? It's $((0, \frac{1}{2}) \cup (\frac{1}{2}, 1)) \times \mathbb R$. What's $\psi_B(AB)$? It's $((-frac{1}{2}, 0) \cup (0, \frac{1}{2})) \times \mathbb R$.

Now let's write down the map between them. $$ \tau_{AB}(x, t) = \begin{cases} (x, t) & 0 < x < \frac{1}{2} \\ (x, -t) & \frac{1}{2} < x < 1 \end{cases} $$


Now you've got a similar structure on the other bundle: charts $U$ and $V$, coordinate functions $\phi_U$ and $\phi_V$. (I'd recommend $\phi_V(e^{2\pi ix}) = x$ for $ -\frac{1}{2} < x < \frac{1}{2}$, and something similar for $\phi_U$.) You can also write down the local trivializations $\psi_U$ and $\psi_V$, which are really pretty simple, and the transition function, which will look a lot like the $\tau_{AB}$ that I wrote down.

Then all that's left to do is to show that the function defined by $H$ and $K$ plays nice with these coordinate charts and transition functions (i.e., check the things that Erno suggested in his answer to your previous question).

If you write out $\phi_U, \phi_V,$ and $\tau_{UV}$ here, I'll go ahead and respond with something about checking those conditions on $H$ and $K$.

  • $\begingroup$ where do you use the transition function $\tau$? $\endgroup$
    – figura
    Dec 21, 2013 at 15:04
  • $\begingroup$ I didn't use it. But that's because I didn't prove that this was a smooth isomorphism -- I left that to you. I'll add some detail to my answer. $\endgroup$ Dec 21, 2013 at 15:20
  • $\begingroup$ Is $\psi_B$ continuous at $x=0$? Shouldn't local trivializations be continuous? $\endgroup$
    – Hamed
    Dec 8, 2019 at 20:09

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