# Canonical Transformation of Hamilton Equation

I have a problem understanding the criteria of an canonical transformation. I am preparing for my exam and came across this question:

For which $A$, $B$, $C$, $D$ is: $Q = A q^2+B p^2, \quad P = Cq^2+Dp^2$ a canonical transformation of $(q,p) \to (Q,P)$?

I know that it must hold: $\{Q,P\} = 1$

My Calculation results in $AD-BC = \frac{1}{4}pq$. But now i wonder if this is enough. Intuitively i think it should further hold that the mapping $(q,p) \to (Q,P)$ is invertible which is not the case, due to the the four Solutions $q = \pm 2\sqrt{DQ-BP}$ and $p = \pm 2\sqrt{PA-QC}$.

I am currently looking thorugh my lecture notes and the books. Still i cannot find anything regarding if the transformation must be bijective in order to be a canonical transformation.

My Question: Which assumption(s) must $(q,p) \to (Q,P)$ fulfill in order to be a canonical transformation.

EDIT: I was just looking at my calculation. I did a stupid mistake. The calculation does not result in: $AD-BC = \frac{1}{4}$ it should be $$AD-BC = \frac{1}{4}pq$$ and therefore this transformation can't be canonical. I corrected it above. However this mistake made me question about the assumptions for $(q,p) \to (Q,P)$ in order to be canonical. I finally got my answer in the Book "Theoretische Physik" written by W. Nolting. It's basically the same as in the accepted answer, Nolting further provides a clear proof of criteria for canonical transformations. So, whoever has the same question as i did, might find an answer there.

One condition is just the one you wrote, the other three are that the remaining Poisson commutators vanish. These conditions also imply that the Jacobian determinant of the transformation is $\pm 1$ and thus, in particular, the transformation il locally a diffeomorphism. In the definition of canonical transformation there is the requirement that it must be a diffeomorphism between two open sets of the space of phases. So you must restrict the domain and/or the codomain in order to have a bijective map. In your case it means that you have to (arbitrarily) chose a sign in your equations. However you must also impose the further requirements I pointed out above.