# 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.

• yes, this is a property of open sets in a metric space: each point of the set is the center of an $\epsilon$-ball which is contained in the open set Jul 29 '21 at 23:02
• For $t$ near $0$, $(t,p_0)$ is, by definition, contained in $D$. How about $(0,p)$ or, more generally, $(t,p)$? Jul 29 '21 at 23:18
• Yes. Maybe it would clarify things if you stated, or tried to formalize what you mean by "near" in the case of $(t,p)$ being near $(0,p_0)$ Jul 29 '21 at 23:22
• This is a basic fact about the product topology: every open neighborhood of $(x,y)\in X\times Y$ contains a subset of the form $U\times\{y\}$, where $U\subseteq X$ is an open neighborhood of $x\in X$. Jul 29 '21 at 23:38

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.