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Oct 17, 2020 at 11:08 comment added Tobias Diez Thanks for the explanation. That construction works indeed, so you just silently suppressed the fiber resealing in your answer.
Oct 16, 2020 at 12:01 comment added Dmitri Panov Tobias, you don't really need $t=1$ for this. The point is that if you consider a map from the cotangent bundle to itself that sends $\alpha\to t\alpha$, the canonical form $\omega$ pulls back to $t\omega$. So, any cotangent bundle with $\omega_{can}$ is simplectomorphic to one with $t\omega_{can}$. At the same time, this operation of scaling the fibers by $t$ doesn't affect the pullback of $\sigma$. Hopefully this answers your question.
Oct 16, 2020 at 11:10 comment added Tobias Diez Why does $t \omega + \sigma'$ give you the magnetic form $\omega_{\textrm{can}} + \sigma$ on $U$ (seen as a subset of $T^* \Sigma$)? Don't you need $t = 1$ for this?
Oct 16, 2020 at 1:57 vote accept Rohil Prasad
Oct 15, 2020 at 20:42 history edited Dmitri Panov CC BY-SA 4.0
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Oct 15, 2020 at 20:22 history answered Dmitri Panov CC BY-SA 4.0