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In Lectures on Symplectic Geometry by AC da Silva, we have the following:

$\textbf{Moser Theorem}$: Suppose that a manifold $M$ is compact, the 2-forms $\omega_0$ and $\omega_1$ on $M$ are such that $[\omega_0] = [\omega_1]$ and the 2-form $\omega_t$ where $\omega_t = (1-t)\omega_0 + t \omega_1$ is symplectic for each $t \in [0,1]$. Then $\exists$ an isotopy $\rho: M \times \mathbb{R} \rightarrow M$ such that $\rho_t^{*} \omega_t = \omega_0$ $\forall$ $t \in [0,1]$.

Why would we need $[\omega_0] = [\omega_1]$? In the proof, I get that it is used directly to say there exists a 1-form $\mu$ such that $\omega_0-\omega_1 = d\mu$ but, it still isn't intuitively clear why this condition on the cohomology class is so essential to the Moser Trick. Thank you all in advance for any pointers.

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  • $\begingroup$ Recall that homotopic smooth applications induce identical morphisms in cohomology. $\endgroup$ – Mindlack Jul 12 at 12:32
  • $\begingroup$ See my edited answer to your previous question. $\endgroup$ – TheGeekGreek Jul 12 at 21:44
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Maybe you should read the paragraph just before, I can't say more than her I think. If you require that there is a symplectomotphism homotopic to the identity then the equality of cohomology class is a necessary condition, and often one ask himself if it is a sufficient condition: it is not except with further hypothesis.

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