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\$$.