I want to check my understanding on the difference between symplectomorphism and canonical transformation. This is a follow-up of my previous post.
(A) A map $(q,p)$ to $(Q,P)$ is called a symplectomorphism if it preserves the symplectic two-form: $dp\wedge dq=dP\wedge dQ$.
(B) A map $(q,p)$ to $(Q,P)$ is called a canonical transformation if it satisfies $(pdq-H(q,p,t)dt)-(PdQ-K(Q,P,t)dt)=dF$ for a function $F$.
Let $M=\mathbb{R}^2\setminus\{(0,0)\}$ be the phase space. As an example, let us consider the following map: $$Q(q,p)=q\sqrt{1+\frac{2\epsilon}{p^2+q^2}},\quad P(q,p)=p\sqrt{1+\frac{2\epsilon}{p^2+q^2}},$$ where $\epsilon\geq0$ is a parameter. The inverse transformation reads $$q(Q,P)=Q\sqrt{1-\frac{2\epsilon}{P^2+Q^2}},\quad p(Q,P)=P\sqrt{1-\frac{2\epsilon}{P^2+Q^2}}.$$
This map is a symplectomorphism because $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial P}{\partial q}\frac{\partial Q}{\partial p}=1,$$ implying that $dp\wedge dq=dP\wedge dQ$.
This map is not a canonical transformation because the following candidate of $F$ and $K$ satisfy $(pdq-H(q,p,t)dt)-(PdQ-K(Q,P,t)dt)=dF$, but this $F$ is not globally defined. $$F=-\epsilon\Big(\frac{pq}{q^2+p^2}+\text{arctan}(q/p)\Big),\quad K(Q,P,t)=H(q,p,t).$$
The infinitesimal version of this transformation reads $$Q(q,p)=q+\epsilon\frac{q}{p^2+q^2}+O(\epsilon^2),\quad P(q,p)=p+\epsilon\frac{p}{p^2+q^2}+O(\epsilon^2),$$ which is suggested in this post.
I thought this is a good example, but it has a problem: the inverse function is only defined for $P^2+Q^2\geq2\epsilon$. Can we improve this map in such a way that it maps $M\to M$?
