I am trying to prove the the escape lemma (Lee's Intro to smooth manifolds Lemma 9.19) The proof can be broken down in three parts

a) First prove this lemma:

Lemma: $X$ is a smooth vector field on a smooth manifold $M$ . Let $\gamma : J \to M$ a maximum integral curve of $X$ such that $b := $sup$(J)$ is finite. Let $t_0 \in J$, and $K \subseteq M$ compact. Suppose $\gamma([t_0, b)) \subseteq K$.

Suppose $U$ and $V$ are relatively compact open subsets of $M$ such that $K \subseteq U$ and $\bar U \subseteq V$ . Let $\psi \in C^\infty(M )$ such that $\psi|_ \bar U \equiv 1$ and supp$(\psi) ⊂ V$ .

Then there is a $\varepsilon > 0$ such that $(t_0 − \varepsilon, b) \subseteq J$ and $\gamma|_{(t_0−\varepsilon,b)}$ an is an integral curve of $\psi X$.

b) Let $\delta$ be the maximal integral curve of $\psi X$ so that $\delta(t_0)=\gamma(t_0)$. Let $J_U \subseteq \Bbb R$ the connected component of $\delta^{-1}(U)$ that contains $t_0$. Prove that $[t_0, b]\subseteq J_U$

c) Derive a contradiction

Suppose step a is proven, **How should I prove step b and derive a contradiction **

The following proposition can be used twice in the whole proof

Proposition Let $X$ be a smooth vector field on a smooth manifold $M$ . Let $p \in M$, and $\gamma_p : J_p \to M$ the maximum integral curve of $X$ with $\gamma_p(0) = p$. Let $\gamma : J \to M$ be another integral curve of $X$ with $\gamma(0) = p$. Then $J \subseteq J_p$ and $\gamma = \gamma_p|_J$ .

Note I am reposting this question

How do I prove the escape lemma?

in which I mistankingly made people think I wanted any proof of the escape lemma, while my question is HOW TO PROVE (b), using the lemma(a) and the proposition, not how to prove the escape lemma in whatever way ignoring the whole post

  • $\begingroup$ This is the definition of a maximal integral curve. $\endgroup$
    – Didier
    Dec 7, 2023 at 11:45
  • $\begingroup$ @Didier Well since this proposition was also given to me as an exercise I guess there was be some other equivalent definition that is being taken as the definition. By the way, is this definition or proposition valid more generally for a $t_0$ instead of 0? $\endgroup$
    – darkside
    Dec 7, 2023 at 17:12
  • $\begingroup$ @Didie In Lee's book: A maximal integral curve is one that cannot be extended to an integral curve on any larger open interval. On the other hand in the proposition we have a priori two different curves, with just a point in common $\endgroup$
    – darkside
    Dec 7, 2023 at 17:27

1 Answer 1


Step b) is just basic general topology and does not use much of the Lemma. $\delta^{-1}(U)\subset\mathbb R$ is open, hence a disjoint union of open intervals, one of which is $J_U$. By definition $t_0\in J_U$ and $(t_0-\epsilon,b]\subset \delta^{-1}(U)$ and since $ (t_0-\epsilon,b]$ is connected and contains $t_0$, we must have $(t_0-\epsilon,b]\subset J_U$. This of course implies $[t_0,b]\subset J_U$.


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