Analysis of a smooth flow on a smooth manifold Let $M$ be a smooth manifold. An open subset $D$ of $\mathbb{R}\times M$ is called a flow domain for $M$ if $\forall p\in M$, the set
$$D^{(p)}=\{t\in\mathbb{R}:(t,p)\in D\}$$
is an open interval containing $0$. A smooth flow on $M$ is a smooth map $F$ from a flow domain $D$ to $M$ that satisfies: $\forall p\in M$,
$$F(0,p)=p,$$
and for all $s\in D^{(p)}$ and $t\in D^{(F(s,p))}$ such that $s+t\in D^{(p)}$,
$$F(t,F(s,p))=F(t+s,p).$$
Here comes the question. For any $p_0\in M$, why is $F(t,p)$ defined and smooth for all $(t,p)$ sufficiently close to $(0,p_0)$? This property is mentioned in Lee's ISM, and Lee said that it is because $D$ is open. But my knowledge of topology is unable to get me out of the question. Would you do me a favor? Thank you.

I think I have found the answer. It is all about a basis for a topology. According to Munkres's book, the product topology on $\mathbb{R}\times M$ is the topology having as basis the collection of all sets of the form $U\times V$, where $U$ and $V$ are open sets in $\mathbb{R}$ and $M$, respectively. Now $(0,p_0)$ is in the open set $D$, so there exist open sets $U\subseteq\mathbb{R}$ and $V\subseteq M$ such that $(0,p_0)\in U\times V\subseteq D$. The rest is history.
