Proof of Reeb's theorem without using Morse Lemma

802 Views Asked by At

I'm trying to prove Reeb's theorem as stated in Milnor's Morse Theory. That is, suppose we have an $n$-manifold $M$ together with a smooth function $f$ with exactly two critical points (both non-degenerate), then $M$ is homeomorphic to $S^n$.

I understand the most crucial part. That is, let $f$ be normalized such that the critical points are mapped to $0$ and $1$ respectively. We can then prove that $f^{-1}[0,a]$ is diffeomorphic to $f^{-1}[0,b]$ for any $0<a,b<1$ by constructing a certain vector field out of the gradient of $f$ and then using the flow of this vector field as the required diffeomorphism. Then if $f^{-1}[0,\epsilon]$ and $f^{-1}[1-\epsilon,1]$ are homeomorphic to a disk, we have that $M$ must be the union of two disks glued along their boundary, making $M$ homeomorphic to $S^n$.

Now my problem is the part where there is an $\epsilon>0$ such that $f^{-1}[0,\epsilon]$ is homeomorphic to an ($n$-dimensional) disk. Milnor uses a lemma (Morse Lemma) which constructs quadratic coordinates in some open neigborhood of $f^{-1}(0)$, but this seems to far too strong. So I was wondering if it's possible to prove this without resorting to that lemma.

My idea is first of all to note that $f^{-1}[0,\epsilon]$ is contractible since we can simply consider the family $f^{-1}[0,t]$ for $t\in[0,\epsilon]$. So if we shrink $f^{-1}[0,\epsilon]$ such that it fits inside a single chart, we can consider it as an closed contractible subset of $\mathbb{R}^n$. However this is not enough to make it homeomorphic to a disk.

Another approach would be to prove that $f^{-1}[\epsilon,a]$ is homeomorphic to $S^{n-1}\times [0,1]$, and then argue that since for any $0<t<\epsilon$, $f^{-1}[t,a]$is a deformation retract of $f^{-1}[\epsilon,a]$, so $f^{-1}[0,a]$ would be homeomorphic to $S^n\times [0,1]$ with $S^n\times\{0\}$ identified to a point. This space is then homeomorphic to a disk as well.

Milnor also remarks that the non-degeneracy condition on $f$ is not essential. The way I would like to prove it doesn't use non-degeneracy either. He refers to two sources that supposedly give a proof that don't use the non-degeneracy condition, but I wasn't able to find a copy online and my university's library doesn't stock any copies.

Any thoughts?


Edit:

In the end I was able to find a book after all. In Saaty's Lectures on Modern Mathematics vol. 2 Milnor has written a chapter on differential topology. In this section he gives a proof that doesn't require the non-degeneracy condition.

He uses a lemma by Brown and Stalling that says that for a paracompact manifold $M$, if any compact subset is contained in an open set diffeomorphic to $\mathbb{R}^n$, then $M$ is itself diffeomorphic to $\mathbb{R}^n$.

Let $x_0$, $x_1$ be the critical points, and let $U$ be a neigborhood of $x_0$ diffeomorphic to $\mathbb{R}^n$. Then using the gradient we can always stretch $U$ such that it covers any compact subset of $M-x_1$. Hence $M-x_1$ is diffeomorphic to $\mathbb{R}^n$. Now $M$ is homeomorphic to the one point compactification of $\mathbb{R}^n$ which is in turn homeomorphic to $S^n$, completing the proof.