If $(q,p)$ to $(Q,P)$ is a canonical transformation, then does this imply $(Q,P)$ to $(q,p)$ is also?

Question

If $(q,p)$ to $(Q,P)$ is a canonical transformation, then does this imply $(Q,P)$ to $(q,p)$ is also, assuming Hamilton's equations hold for the coordinates $(q,p)$?

This seems like it should be true from the derivation for a canonical transformation using Poisson brackets, but I was wondering if anyone knew any better (or had some specific counter examples).

Answer 1

A canonical transformation leaves the symplectic form on the cotangent bundle invariant. Hence its determinant is 1 globally, which means that your coordinate transformation can be inverted everywhere. The inverse transformation then preserves the symplectic form too, hence is a canonical transformation.