Why isn't it important, after which coordinates the Variation of the action integral is done?

Question

I often read,that if the lagrangian $L=p\dot{q}-H$ of a pair of coordinates in phase space $(q,p)$ and $P\dot{Q}- K $, for some new pair of coordinates $(Q,P)$ only differ by a total time derivative $dF/dt$, the Variation of the action functional doesn't differ too, which implies that the canonical equations for the new coordinates hold.

My question now is: if the action functional $S(q,p,t)=S'(Q,P,t) + c$ where $c= \int (dF/dt )dt $ is the integrated total time derivative of $F$ (which disappears in the Variation); why then is the variation of both sides equal? I mean surely the Variation of the $(q,p)$ coordinates would be the same for both sides, but why should the Variation of the $(Q,P)$ coordinates on the right hand side be the same as the Variation of the $(q,p)$ coordinates on the left hand side?

Answer 1

It is important to stress that the $2n$ new variables $Z^J\equiv (Q^j,P_j)$ and the $2n$ old variables $z^I\equiv (q^i,p_i)$ are not $4n$ but only $2n$ independent variables.

We assume that the new variables $Z^J$ are a function of the old variables $z^I$ and time $t$ (and vice versa), and that the Jacobian matrix $\partial Z^J / \partial z^I$ is invertible.

In particular an infinitesimal variation $\delta z$ of $z$ induces a corresponding infinitesimal variation $\delta Z$ of $Z$.

See also this related Phys.SE post.