Symplectomorphisms preserves Hamiltonian equations Let $(M_1,\omega_1)$, $(M_2,\omega_2)$ be symplectic manifolds and let $\psi:(M_1,\omega_1)\rightarrow (M_2,\omega_2)$ be a symplectomorphism. Consider a Hamiltonian $H\in\mathcal{C}^\infty(M_2)$. Show that a curve $t\mapsto \gamma(t)\in M_1$ solves Hamilton equation for $\tilde{H}:=H\circ\psi$ $\iff$ the curve $t\mapsto \psi\circ\gamma(t)\in M_2$ solves them for $H$.
It will be great if it is obtained as a corollary of the fact that $X_{H\circ\psi}=\psi^*(X_H)$.
Thank you
 A: My attempt.
Proposition 1
Let $(M_1,\omega_1)$, $(M_2,\omega_2)$ be symplectic manifolds and let $\psi:(M_1,\omega_1)\rightarrow (M_2,\omega_2)$ be a symplectomorphism. Let, moreover $H\in\mathcal{C}^\infty(M_2)$ be a Hamiltonian, with Hamiltonian vector field $X_H\in\Gamma(TM_2)$.
Then $\psi^*(X_H)$ is the Hamiltonian vector field w.r.t. the Hamiltonian $\tilde{H}:=H\circ\psi\in\mathcal{C}^\infty(M_1)$, i.e.
\begin{equation}
\label{ciao}
    X_{H\circ\psi}=\psi^*(X_H).
\end{equation}
proof
\begin{equation*}
        d(H\circ\psi)=d(\psi^*H)=\psi^*dH=-\psi^*\left(i_{X_H}\omega_2\right)=-i_{\psi^*X_H}\psi^*\omega_2=-i_{\psi^*X_H}\omega_1,
\end{equation*}
so $\psi^*X_H$ is the unique Hamiltonian vector field w.r.t $H\circ\psi$.
Corollary 1
$X_{H\circ\psi}$ and $X_H$ are $\psi$-related
proof
We can explicit conclusion of Proposition 1, which means that $\forall p\in M_1$, $\forall h\in\mathcal{C}^\infty(M_1)$
\begin{equation}
\label{expanded correlation Hamiltonian vector fields}
    (X_{H\circ\psi})_p(h)=(\psi^*X_H)_p(h)=:(X_H)_{\psi(p)}(h\circ\psi^{-1}),
\end{equation}
where we have developed the pull-back of vector fields through diffeomorphisms. Now, take any $p\in M_1$ and $g\in\mathcal{C}^\infty(M_2)$, then
\begin{equation}
\label{first related vector fields}
    \left[T_p\psi((X_{H\circ\psi})_p)\right](g)=(X_{H\circ\psi})_p(g\circ\psi);
\end{equation}
apply the first equation with $h:=g\circ \psi$, then we have
\begin{equation*}
    \left[T_p\psi((X_{H\circ\psi})_p)\right](g)=(X_H)_{\psi(p)}(g\circ\psi\circ\psi^{-1})=(X_H)_{\psi(p)}(g).
\end{equation*}
Since this hold for every $g\in\mathcal{C}^\infty(M_2)$, we conclude that
\begin{equation*}
    \left[T_p\psi((X_{H\circ\psi})_p)\right]=(X_H)_{\psi(p)}(g\circ\psi\circ\psi^{-1})=(X_H)_{\psi(p)},
\end{equation*}
which means exactly that $X_{H\circ\psi}$ and $X_H$ are $\psi$-related.
Proposition 2
Let $F:M\rightarrow N$ be a smooth map between manifolds and suppose that $X\in\Gamma(TM)$, $Y\in\Gamma(TN)$ are $F$-related vector fields. Then $F$ takes integral curves of $X$ to integral curves of $Y$.
Proof
Let $\gamma:\mathcal{I}\rightarrow M$ be an integral curve of $X$, we have to show that $\sigma:=F\circ\gamma$ is an integral curve of $Y$:
\begin{equation*}
    \dot{\sigma}(t)=\frac{d}{dt}(F\circ\gamma)(t)=T_{\gamma(t)}F(\dot{\gamma}(t))=T_{\gamma(t)}F(X_{\gamma(t)})=Y_{F(\gamma(t))}=Y_{\sigma(t)}.
\end{equation*}
Conclusion
Symplectomorphisms preserve Hamilton's equations.
proof
Let $\psi$ be a symplectomorphism, then, thanks to Corollary 1, we see that Hamiltonian vector fields $X_{H\circ\psi}$ and $X_H$ are related through $\psi$. Moreover by Proposition 2, $\psi$ maps integral curves to integral curves of $\psi$-related generic vector fields. But integral curves of Hamiltonian vector fields are solutions of Hamilton's equations and so $\psi$ preserves Hamilton's equations.
A: Since you mentioned Abraham-Marsden as a source, here are a few comments which I think you'll find beneficial (the notation is very much identical to how they use it). Here is a more "streamlined approach" (atleast in my opinion) which is at the "mapping level" rather than the "pointwise level".

*

*I hope you realize that the conclusion of Proposition 1 can be written as $\psi^*(X_H) = X_{\psi^*H}$, which of course makes it very memorable. Similarly, by replacing $\psi$ by $\psi^{-1}$, and using the fact that $(\psi)_*:= (\psi^{-1})^*$ (i.e push-forward is same as pull-back by inverse (by definition)), we get $\psi_*(X_H) = X_{(\psi_*H)}$ (of course you have to redefine where everything is defined)


*Recall that if $F:M \to N$ and $X$ and $Y$ are vector field on $M$ and $N$ respectively, then we say $X$ and $Y$ are $F$-related if $TF \circ X = Y \circ F$, and we write $X\sim_F Y$; i.e the following diagram commutes
$\require{AMScd}$
\begin{CD}
TM @>TF>> TN\\
@A{X}AA @AA{Y}A \\
M @>>F> N
\end{CD}
Finally, recall the definition of the pull-back of a vector field (this requires $F$ to be a diffeomorphism): $F^*(Y):= TF^{-1}\circ Y \circ F$ (and note that $T(F^{-1}) = (TF)^{-1}$, so simply writing $TF^{-1}$ isn't ambiguous). With this, corollary 1 is simple to prove:
\begin{align}
T\psi \circ X_{\psi^*H} &= T\psi \circ (\psi^*X_H) \tag{by proposition $1$} \\
&= T\psi \circ (T\psi^{-1}\circ X_H \circ \psi) \tag{by definition} \\
&= X_H \circ \psi
\end{align}
This says exactly that $X_{\psi^*H} \sim_{\psi}X_H$ that the two vector fields are $\psi$-related.


*We can rewrite the proof of Proposition $2$ as follows:
\begin{align}
(F\circ \gamma)' &= T F \circ \gamma' \\
&= TF \circ (X\circ \gamma) \\
&= (Y\circ F) \circ \gamma \tag{since $X\sim_F Y$} \\
&= Y \circ (F\circ \gamma)
\end{align}
This says exactly that $F\circ \gamma$ is an integral curve of $Y$. Here, I use $\gamma'$ where you use $\dot{\gamma}$; this is a curve in the tangent bundle $I\subset \Bbb{R}\to TM$
