A smooth vector field $X$ on $M$ is complete if for each $t\in \mathbb{R}$ we have $$\mathcal{D}_t = \{m\in M : t\in (a(m), b(m))\} = M,$$ where $(a(m),b(m))$ is the maximal interval of a integral curve $\sigma$ starting at $m$, that is
- $0\in (a(m), b(m))$ and $\sigma(0) = m$
- if $\gamma: (c,d) \rightarrow M$ is an integral curve of $X$ satisfies 1. above, then $(c,d)\subset (a(m), b(m))$ and $\gamma = \sigma|(c,d)$.
We can define a transformation $X_t$ with domain $\mathcal{D}_t$ by setting $X_t(m) = \sigma_m(t)$, and we have the following property: the domain of $X_s\circ X_t$ is contained in $\mathcal{D}_{s+t}$. If $s,t$ have the same sign, then the domain is $\mathcal{D}_{s+t}$ and $X_s\circ X_t = X_{s+t}$. I can also use the property that each $\mathcal D_t$ is open.
My proof: argue by contrapositive, suppose $\mathcal{D}_t \neq M$ for some $t\in \mathbb{R}$, then we show that $\{\mathcal{D}_{t/n}\}_{n\in \mathbb{N}}$ is an open cover of $M$ and it has no finite sub covers.
First to show it is an open cover, for each $m\in M$, I can use the result that $(a(m), b(m))$ is a non empty open interval around zero (this is from the theory of ode), then $m\in \mathcal{D}_{t/n^*}$ for some large $n^*$.
To show it does not have any finite subcover, note that if $\mathcal{D}_{t/2} = M$, then the domain of $X_{t/2}\circ X_{t/2}$ is $M$, but at the same time, the domain is also $\mathcal{D}_{t/2 + t/2} = \mathcal{D}_t$ which is impossible. By this argument $\mathcal{D}_{t/n} \neq M$ for each $n\in \mathbb{N}$. Also since $\mathcal{D}_{t/n}$ is monotone increasing in $n$, so there can not be any finite subcovers.