Consider the following transformation:
$$q=P^\alpha \cos(\beta Q)$$
$$p=P^\alpha \sin(\beta Q)$$
for $\alpha=1/2$ and $\beta=2$.
Now, by convention, one takes $(q,p)$ to be the old coordinates and $(Q, P)$ to be the new coordinates such that if for a given transformation $$\{Q,P\}_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$
then the given transformation is canonical transformation.
If now we need to check whether the above-given transformation is canonical one might try to make $Q$ and $P$ subjects in the above equations such that $Q=Q(q,p)$ and $P=P(q,p)$ and then just check the Poisson bracket.
As one can see making $Q$ and $P$ subjects in the above equations and then taking the partial derivatives is an ugly job.
My friend suggested to me that we can instead simply check $$\{q,p\}_{Q, P}=1$$ and if it is true then the transformation is canonical since the system doesn't care which way we make the transformation.
Though this suggestion sounds plausible to me, I am not entirely convinced of it since I do not see how the "system doesn't care which way we make the transformation" argument justifies this since as far as I can remember in some cases an inverse transformation is not possible. (zero Jacobian determinant)
Is my friend's argument correct?
If yes, is there a better way (so that my dumb brain understands) of seeing it?
