I was trying to prove, that for a transformation to be Canonical, one must have a relationship:
$$ \left\{ Q_a,P_i \right\} = \delta_{ai} $$
Where $Q_a = Q_a(p_i,q_i)$ and $P_a = P_a(p_i,q_i)$.
Now to do the proof I started with $\dot{Q_a}$:
Chain rule and Hamilton's equation for initial coordinates $q_i,p_i$ $$ \dot{Q_a} = \frac{\partial Q_a}{\partial q_j} \dot{q_j} + \frac{\partial Q_a}{\partial p_j} \dot{p_j} = \frac{\partial Q_a}{\partial q_j} \frac{\partial H_a}{\partial p_j} - \frac{\partial Q_a}{\partial p_j} \frac{\partial H_a}{\partial q_j} $$
Then I apply chain rule for the Hamiltonian derivatives: $$ \dot{Q_a} = \frac{\partial Q_a}{\partial q_j} \left( \frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial p_j} + \frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial p_j} \right) - \frac{\partial Q_a}{\partial p_j} \left( \frac{\partial H}{\partial Q_i} \frac{\partial Q_i}{\partial q_j} + \frac{\partial H}{\partial P_i} \frac{\partial P_i}{\partial q_j} \right) $$
Now reordering the terms yields us: $$ \dot{Q_a} = \frac{\partial H}{\partial Q_i} \left\{ Q_a,Q_i \right\} + \frac{\partial H}{\partial P_i} \left\{ Q_a,P_i \right\} $$
Now here the problem, for the transformation from a coordinate system $(q_i,p_i)$ to a coordinate system $(Q_a(q_i,p_i), P_a(q_i,p_i))$ to be canonical we require:
$$\left\{ Q_a,P_i \right\} = \delta_{ai}$$
But why we have as well the following requirement, or is it just too obvious or true because of some property of any coordinate transformation?
$$\left\{ Q_a,Q_i \right\} = 0$$
The problem I am having is as follows. I agree, that the following two are true (if I used the covariant notation correctly):
$$ \left\{ q^i,q_j \right\}_{q,p} = \frac{\partial q^i}{\partial q^k} \frac{\partial q_j}{\partial p_k} - \frac{\partial q^i}{\partial p^k} \frac{\partial q_j}{\partial q_k} = 0 $$
$$ \left\{ Q^i,Q_j \right\}_{Q,P} = 0 $$
But why its the case that the following is also true?
$$ \left\{ Q^i,Q_j \right\}_{q,p} = 0 $$