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I am trying to prove that the Whitney sum of the Mobius bundle is trivial. I have read here: Global Trivialization of $M\oplus M$ a way to do it, but I don't know nothing about sections.

I am trying to prove it by defining a vector bundle isomorphism between $M\oplus M$ and $M\times \mathbb{R}^2$.

The elements of $M \oplus M$ are of the form $(e^{2\pi ti},([y_{1},y_{2}],[z_{1},z_{2}]))$ where $t\in [0,1]$ and $([y_{1},y_{2}],[z_{1},z_{2}])\in M\times M$ with $y_{1}=y_{2}=t$ if $t\neq 0,1$ and $y_{1},y_{1}\in \{0,1\}$ if $t\in \{0,1\}$. Hence, if $x\neq (1,0)$, I can send that point to $(x,(y_{1},y_{2})$), but if $x=(1,0)$ I don't know how to define the map in order to be well-defined and continuous.

Can anyone help me, please? Thanks in advance.

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    Two suggestions: Think of $M$ as a quotient of $S^1\times\Bbb R$, and think of the bundle $M'$ over $\Bbb RP^1$ whose fibers are the orthogonal complements of the fibers of $M$ (in $\Bbb R^2$). Can you show $M\cong M'$? – Ted Shifrin Jan 06 '18 at 17:05
  • Sorry, I don't understand the definition of $M'$. My knowledge about differential geometry is really poor. –  Jan 06 '18 at 17:50
  • This isn't differential geometry! It's just $\Bbb R^2$: Rotate the line spanned by $x$ $90º$ and you get the orthogonal line. – Ted Shifrin Jan 06 '18 at 18:01
  • So in $M$, the fiber of $(-1,0)$ would be a vertical line (the line parallel to the axis OZ passing through the point $(-1,0)$, and in $M'$ it would be the line parallel to OX passing through $(-1,0)$. Am I understanding correctly the spaces? –  Jan 06 '18 at 18:15
  • No, you have them backwards. :) The fiber of the Möbius bundle over $[x]$ is the line spanned by $x$. – Ted Shifrin Jan 06 '18 at 18:16
  • Ok! So they would be isomorphic vector bundles by projecting a point of the fiber to the corresponding point of the orthogonal line. But how can this help me with the Whitney sum? –  Jan 06 '18 at 19:02
  • If $L$ is a one-dimensional subspace of $\Bbb R^2$ and $L^\perp$ is its orthogonal complement, what is $L\oplus L^\perp$? – Ted Shifrin Jan 06 '18 at 20:03
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    Ok! I have understood it! Thank you very much for your help and patience! –  Jan 06 '18 at 20:12

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