Exercise. Prove or refute the following:
\\[(\forall t \in T)(\phi_s'(t) = \alpha(\phi^t(s)) \iff \phi_s'(0) = \alpha(\phi^0(s)) \iff \phi_s'(0) = \alpha(s)\\]
This says that to show that \\(\phi^t\\) is a global flow for \\(\alpha\\), it suffices to show that \\(\alpha\\) equals the time-derivative of \\(\phi^t\\) just at time t=0.
Hint: consider using the fact that \\(\phi^t\\) is an _action_ of \\(T\\) on \\(M\\).