A lemma from McDuff-Salamon says that $\psi:\mathbb{R}^{2n}\to\mathbb{R}^{2n}$ is symplectic iff $\{F,G\}\circ\psi=\{F\circ\psi,G\circ\psi\}$. I proved that. Then there is an exercise showing that $\{F,G\}=\omega(H_F,H_G)$, $H_F$ being the hamiltonian field associated to $F$. Prompted by this, I tried to generalize the lemma to manifolds. From a proposition in Hofer-Zehnder, I know that on a symplectic manifold $(M,\omega)$ there exist a Riemannian metric $\langle\cdot,\cdot\rangle$ and an almost complex structure $J$ with $J^2=-1$ such that $\omega(u,Jv)=\langle u,v\rangle$ for all $u,v$ in the appropriate tangent space. If $\nabla F$ denotes the gradient of $F$ w.r.t. the aforementioned Riemannian metric, then I proved $H_F(x)=J_x\nabla F(x)$.
With this in mind, I set out to extend the lemma to manifolds. I took local coordinates and with the hypothesis that $x,\psi(x)$ were in the same chart for any $x\in M$, I used some linear algebra to conclude the generalization.
The next step, of course, is generalizing this lemma to not have the hypothesis of those charts. But I seem to be unable to conclude. Is there a way to complete this generalization?
Symplecticity equates to $\psi^\ast\omega=\omega$, and to $\psi^\ast X_F=X_{F\circ\psi}$ for "enough" smooth functions, for example all smooth functions. With this, symplecticity equates to the two stars in the following being valid for all $F,G$:
\begin{align*} \{F,G\}\circ\psi(x)={}&\{F,G\}(\psi(x))=\omega_{\psi(x)}(X_F(\psi(x)),X_G(\psi(x)))={} \\ {}={}&[(\psi^{-1})^\ast\omega]_x((d\psi)^{-1}(X_F(\psi(x))),(d\psi)^{-1}(X_G(\psi(x))))={} \\ {}\stackrel\ast={}&\omega_x((d\psi)^{-1}(X_F(\psi(x))),(d\psi)^{-1}(X_G(\psi(x))))={} \\ {}={}&\omega_x(\psi^\ast X_F(x),\psi^\ast X_G(x))={} \\ {}\stackrel\ast={}&\omega_x(X_{F\circ\psi}(x),X_{G\circ\psi}(x))={} \\ {}={}&\{F\circ\psi,G\circ\psi\}(x). \end{align*}
What I had trouble seeing was, I guess, the converse, because thinking the proof out mentally I probably couldn't see the application of the "enough" thing.
Edit
No, that won't work. I used symplecticity twice in the above, I cannot say the equality implies symplecticity. Had I used symplecticity only once, the argument would have worked, but seen as two passages use it, I cannot say making both of them is equivalent to symplecticity.
However, I can go local, and compare $\psi^\ast\omega$ with $\omega$. I am basically saying the two forms in question have the same Poisson brackets. That is because:
\begin{align*} \{F\circ\psi,G\circ\psi\}_{\psi^\ast\omega}(x)={}&(\psi^\ast\omega)_x(\tilde X_{F\circ\psi}(x),\tilde X_{G\circ\psi}(x))={} \\ {}={}&\omega_{\psi(x)}(d\psi(\tilde X_{F\circ\psi}(x)),d\psi(\tilde X_{G\circ\psi}(x))), \end{align*}
where $\tilde X_F$ is the Hamiltonian field of $F$ w.r.t. $\psi^\ast\omega$. So if I prove this is like calculating $\omega_{\psi(x)}$ in $X_F(\psi(x)),X_G(\psi(x))$ I will have that is $\{F,G\}_\omega\circ\psi(x)$, which by hypothesis is $\{F\circ\psi,G\circ\psi\}_\omega(x)$. So I must prove $d\psi(\tilde X_{F\circ\psi}(x))=X_F(\psi(x))$, or in other words, taking $d\psi$ to the right by inverting, that $\psi^\ast X_F=\tilde X_{F\circ\psi}$. In other words, I now set out to prove:
$$\iota_{\psi^\ast X_F}\psi^\ast\omega=-d(F\circ\psi).$$
I know the left side equates to $\psi^\ast(\iota_X\omega)$. But $\iota_{X_F}\omega$ is by definition $-dF$. So the left side is $-\psi^\ast dF$. But $d$ and $\psi^\ast$ commute, meaning that is $-d(\psi^\ast F)=-d(F\circ\psi)$, as I wanted.
Now, the Poisson bracket can be expressed as $\omega^{k\ell}\partial_\ell F\partial_kG$, where $\omega^{k\ell}$ is the (antisymmetric) matrix locally representing $\omega$. I won't prove this, it boils down to finding expressions for the Hamiltonian fields and then applying $\omega$.
This means that if two forms have the same Poisson brackets on any two functions, they must coincide: one just need to feed in local coordinates and that's it. So symplecticity is indeed equivalent to the Poisson bracket condition.