In a smooth manifold M every flow generates a vector field. How can I prove this.
if X is a tangent vector at (0,p) in $\mathbb{R}\times M$ then the differential of the flow map generates a tangent vector at p in M. In this way we can always assign a tangent vector at each point p. But how this assignment is smooth?...or any other way to prove?
If $\Phi\colon \Bbb R\times M \to M$ is a flow, we fix $x \in M$ and look at the curve $$\Bbb R \ni t \mapsto \Phi(t,x) \in M.$$For $t = 0$, we see that it starts at $\Phi(0,x) = x$. The velocity of this curve gives us a vector $$X_x= \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \Phi(t,x) \in T_xM.$$So $x \mapsto X_x$ defines $X \in \mathfrak{X}(M)$. It is smooth because $\Phi$ is (more precisely, you're taking derivatives of something you assume smooth, hence the result is smooth as well).