Global Trivialization of $M\oplus M$

Question

$\mathbb S^1$'s $\mathbb R^1$-bundle is $$\{\mathbb S^1\times\mathbb R^1\text{, open Möbius strip}\}$$ and its $\mathbb R^2$-bundle is $$\{\mathbb S^1\times\mathbb R^2\text{, open solid Klein bottle}\}$$

I am encounter with a question that

Let $M$ denote the open Möbius strip, and $\pi:M\to\mathbb S^1$ be the $\mathbb R^1$-bundle. Prove that Whintney sum $\pi:M\oplus M\to\mathbb S^1$ is the trivial $\mathbb R^2$-bundle.

We have $M=[0,1]\times\mathbb R^1/$~ and $\pi([t,x])=\exp(2\pi it)$. Now we should construct a homeomorphism $$h:M\oplus M\to\mathbb S^1\times\mathbb R^2$$

I have no idea. Any advice is helpful. Thank you.

Answer 1

In the hope that a picture is worth a thousand words:

The geometric point is, if $E \to [0, 2\pi]$ is the trivial plane bundle over an interval, and if $U_{1}$ and $U_{2}$ are the non-vanishing sections $$ U_{1}(t) = (\cos t/2, \sin t/2),\qquad U_{2}(t) = (-\sin t/2, \cos t/2), $$ then each $U_{i}$ spans a trivial line subbundle $L_{i}$ of $E$ whose fibre "rotates half a turn between $0$ and $2\pi$", and $L_{1} \oplus L_{2} = E$ as vector bundles over $[0, 2\pi]$. Now map $E$ to the trivial plane bundle over the circle by identifying the fibres over $0$ and $2\pi$, and observe that each $L_{i}$ descends to a non-trivial line bundle over the circle, i.e., to the line bundle $M$ whose total space is a Möbius strip. (The $U_{i}$ themselves do not descend to continuous sections of $M$, of course.)