I am following da Silva's lectures on symplectic geometry.
She defines the lift of a diffeomorphism as follows:
Let $X_1$ and $X_2$ be $n$-dimensional manifolds with cotangent bundles $M_1=T^*X_1$ and $M_2=T^*X_2$ and suppose $f:X_1 \rightarrow X_2$ is a diffeomorphism.
Then the lift $f_\#:M_1 \rightarrow M_2$ is defined
$$f_\#(p_1) = p_2 = (x_2, \xi_2) = (f(x_1), \xi_2) $$ where $\xi_1 = (f_{x_1})^*\xi_2 = \xi_{2}\circ df_{x_1}$ and $\xi_2 \in (T^*X_2)_{f(x_1)}$
$f_\#\rvert_{T^*X_1}$ is therefore the inverse map of $(f_{x_1})^*$
and so $\xi_2 = [(f_{x_1})^*]^{-1}\xi_1 = (f^{-1}_{x_1})^*$
Now let $(M, \omega)$ be the symplectic manifold obtained by equipping the tangent bundle $M=T^*N$ with the canonical 2-form $\omega = \sum_i dx^i \wedge d\xi^i$ and let the Lie Group G act on N: $$\psi: G \rightarrow \text{Diff(M)}, \quad g \rightarrow \psi_g \\ $$ I am trying to prove the contangent lift of the action is symplectic (and hamiltonian)
We must have $$(\psi_g)_\#(p) = (\psi_g)_\#(x, \xi) = (\psi_g(x), (\psi_g^{-1})^*(\xi)) = (\psi_g(x), (\psi_{g^{-1}})^*(\xi)) $$
This must preserve the symplectic form, i.e. $$((f_\#)^*\omega)_p (u, v) = \omega_{f_\#(p)}(df_p(u), df_p(v)) $$
I am unsure what to do from here. I can see the symplectic form takes in two coordinates that transform in seemingly inverse ways. Could someone point me in the right direction ? It would also help if anyone spots any errors in the way I state the problem, I have studied physics so far and am slightly out my depth with the mathematics.
EDIT: Solution using tautological one-form (this method is in da Silva's notes)
We want to show $(f_\#)^*\alpha_2 = \alpha_1$
The tautological form on $M_1$, $M_2$ is defined: $$(\alpha_1)_{p_1} = \pi^*_{p_1}\xi_1, \quad \quad (\alpha_2)_{p_2} = \pi^*_{p_2}\xi_2$$
So we have $$\begin{align}(f_\#)^*_{p_1}(\alpha_2)_{p_2} &= (f_\#)^*_{p_1}\pi^*_{p_2}(\xi_2)] \\ &= (\pi_2 \circ f_\#)^*_{p_1} \xi_2 \\ &= (f \circ \pi_1)^*_{p_1} \xi_2 \quad\quad \text{(The lift is constructed as such)} \\ &= (\pi_1)^*_{p_1}f^*_{x_1}\xi_2 \\ &= (\pi_1)^*_{p_1}\xi_1 \quad\quad \text{(By definition of the lift)}\\ &= (\alpha_1)_{p_1} \end{align} $$
Therefore $f_\#$ preserves the tautological form as well as $d\alpha_1$, the canonical 2-form which is symplectic on $M_1$
Therefore the lift of the diffeomorphism $\psi_g:X_1 \rightarrow X_2$ is:
$(\psi_\#)_g:M_1 \rightarrow M_2$
and it is a symplectomorphism.
The lift of the action $\psi_\#$ is therefore symplectic.
It can also be shown the action is hamiltonian. It seems obvious since we are preserving the tautological one-form, although I'm not sure on the proof.
My own attempt attempt at an answer after reading the comments:
Proving that the lift of the flow of a vector field $V$ on $X_1$ is a hamiltonian vector field.
The flow of a vector field $V$ on a manifold $M$ is an action of the real numbers on the manifold: $f: \mathbb{R} \times M \rightarrow M$
There is a bijection between symplectic actions of $\mathbb{R}$ on $M$ and complete vector fields on $M$ (given by the flow of the vector field).
We have shown the lift of the flow $f_\#$ is a symplectic action on $M_1$. Let its associated vector field be $V_\#$.
$V_\#$ is hamiltonian if $ \iota_{V_\#} \omega$ is exact.
$$\iota_{V_\#} \omega = -\iota_{V_\#} d\alpha $$ By Cartan's magic formula: $$\iota_{V_\#} \omega = d\iota_{V_\#}\alpha - \mathcal{L}_{V_\#}\alpha = d\iota_{V_\#}\alpha $$ Since $f_\#$ preserves the tautological form $\alpha$
$V_\#$ is therefore hamiltonian with hamiltonian function $H = \iota_{V_\#}\alpha$