# Cohomology classes in the Moser Theorem

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.

• Recall that homotopic smooth applications induce identical morphisms in cohomology. – Mindlack Jul 12 at 12:32
• See my edited answer to your previous question. – TheGeekGreek Jul 12 at 21:44