Poincaré–Bendixson Theorem on a compact, connected, orientable, two-dimensional manifold I'm currently reading the article "A Generalization of a Poincaré–Bendixson Theorem to Closed Two-Dimensional Manifolds" by Arthur Shwartz. The paper first establishes a result for minimal sets, which are closed, non-empty subsets of $M$ which are invariant under the flow of some $C^2$ vector field, and contain no such proper subsets. In particular, the result can be summarized as follows:

Suppose that $M$ is a compact, connected, two-dimensional manifold and $\Sigma \subset M$ is a minimal set. Then $\Sigma$ is either:

*

*a fixed point,

*a periodic orbit (homeomorphic to $S^1$), or

*$M$ itself (which must be homeomorphic to $T^2$).


I was able to follow the proof of this theorem, but fail to understand the subsequent corollary, which may be stated as:

Suppose that $M$ is a compact, connected, orientable, two-dimensional manifold such that $M$ is not a minimal set and for some $x \in M$, the $\omega$-limit set of $x$—$\omega(x)$—contains no fixed points. Then $\omega(x)$ is homeomorphic to $S^1$ and the flow of $x$—$\varphi_{t}(x)$—winds towards $\omega(x)$ as $t \to \infty$.

The argument presented is that $\omega(x)$ must contain a minimal set (say $\Sigma$) which is homeomorphic to $S^1$ by the previous theorem. Now since $M$ is orientable, there exists a neighborhood $N$ of $\Sigma$ which is homeomorphic to $(-1, 1)\times S^1$ such that $\Sigma$ is mapped onto $\{0\}\times S^1$ and the line segments $(-1, 1)\times \{y\}$ are transversal arcs for $y \in S^1$. The setup is clear to me, but I don't follow the remaining steps of the proof: "It follows from the continuity of the flow, and the fact that $\Sigma \subset \omega(x)$, that there exist $0 < t_1 < t_2$, $q \in \omega(x)$, and a transversal $I_q$ through $q$ such that $\varphi_{t_2}(x)$ is between $\varphi_{t_1}(x)$ and $q$ on $I_q$. Thenceforth $\varphi_{t}(x)$ is "trapped" between the closed curve formed by $\varphi_{[t_1, t_2]}(x)$ and the portion of $I_q$ between $\varphi_{t_1}(x)$ and $\varphi_{t_2}(x)$ and $\Sigma$. Thus we find the next intersection of $\varphi_{t}(x)$ with $I_q$ between $\varphi_{t_2}(x)$ and $q$. Repetition of this argument yields the conclusion of the corollary."
I believe that we're aiming to show that $\omega(x) = \Sigma$ (in particular by a monotonicity-type argument on the surface of the cylinder analogous to that used in the proof of the Poincaré–Bendixson theorem in $\mathbb{R}^2$). But how can we be sure that $q \in N$, or that the flow remains within $N$ (and hence its image under the homeomorphism remains on the surface of the cylinder, which I understand to be what "traps" it)? Even if so, how would this ultimately show that $q \in \Sigma$?
 A: To address your two questions:

*

*Why is $q \in N$?

If I understand correctly, they could have (and should have) just started with $q \in \Sigma$, since for any $q \in \Sigma$ the orbit $\phi_t(x)$ approaches $q$ arbitrarily closely.
To be precise, because $\Sigma \subset \omega(x)$, there's a $t_1 > 0$ such that $\phi_{t_1}(x) \in I_q \subset N$, and there's a $t_2 > t_1$ such that $\phi_{t_2}(x) \in I_q$ and is even closer to $q$ than $\phi_{t_1}(x)$. By orientability of $M$, $\phi_{t_2}(x)$ is between $q$ and $\phi_{t_1}(x)$.


*Why does the flow stay in $N$?

The proof doesn't need the flow to stay within $N$. Instead, it requires something weaker. Consider the subsegment $T$ of the transversal $I_q$ between $q$ and $\phi_{t_1}(x)$. By continuity, if we choose $t_1$ so that $\phi_{t_1}(x)$ is sufficiently close to $q$, then the orbit of any point $T$ will stay within $N$ up until it hits $I_q$ again.
That is enough to rule out any "weird manifold topology" and conclude there is a "trapped" region bounded by $\phi_{[t_1, t_2]}$, $\Sigma$, and a piece of $I_q$, as in the Poincare-Bendixson proof.
