# Question about the proof of Reeb's theorem in Milnor's Morse Theory

Theorem. (Reeb) If $$M$$ is a compact manifold and $$f$$ is a differentiable function on $$M$$ with only two critical points, both of which are nondegenerate, then $$M$$ is homeomorphic to a sphere.

Proof) The two critical points must be the minimum and maximum points. Say that $$f(p)=0$$ is the minimum and $$f(q)=1$$ is the maximum. If $$\epsilon$$ is small enough then the sets $$f^{-1}[0,\epsilon]$$ and $$f^{-1}[1-\epsilon,1]$$ are closed $$n$$-cells by the Morse lemma.

What I can't understand is the last sentence. The Morse lemma is stated below. Since $$f(p)=0$$ is the minimum, we must have $$f=(y^1)^2+\cdots+(y^n)^2$$ near $$p$$. For small enough $$\epsilon$$, we will have $$\{y\in U: (y^1)^2+\cdots+(y^n)^2\leq \epsilon \} \subset f^{-1}[0,\epsilon]$$, and the former set is the closed $$n$$-disk. But how can we show the reverse inclusion $$\{y\in U: (y^1)^2+\cdots+(y^n)^2\leq \epsilon \} \supset f^{-1}[0,\epsilon]$$ holds for small $$\epsilon$$?

Morse Lemma. Let $$p$$ be a nondegenerate critical point of $$f$$. Then there is a local coordinate system $$(y^1,\dots,y^n)$$ in a neighborhood $$U$$ of $$p$$ with $$y^i(p)=0$$ for all $$i$$ and such that the identity $$f=f(p)-(y^1)^2-\cdots-(y^\lambda)^2+(y^{\lambda+1})^2+\cdots+(y^n)^2$$ holds throughout $$U$$, where $$\lambda$$ is the index of $$f$$ at $$p$$.

Let us look closer at the implications the normal form for $$f$$ given by Morse Lemma. Centered at $$p$$, being the minimum point (in particular a local minimum), there is a chart $$(y,U)$$ with $$y(r)=\big(y^1(r),\ldots, y^n(r)\big)$$ for $$r\in U$$, such that $$f\circ y^{-1}\big(y^1(r),\ldots, y^n(r)\big)=y^1(r)^2+\ldots+y^n(r)^2,$$ for any $$r\in U$$. This is the precise meaning of $$f=(y^1)^2+\ldots+(y^n)^2$$ near $$p$$.
Now, take $$\epsilon>0$$ such that $$y(U)\supset \mathbb{D}^n_{\epsilon}$$, where $$\mathbb{D}^n_{\epsilon}$$ is the closed ball in $$\mathbb{R}^n$$ with center $$0_{\mathbb{R}^n}$$ and radius $$\epsilon^{\frac{1}{2}}$$. You can do this, since $$y(U)$$ is an open neighborhood of $$y(p)=0_{\mathbb{R}^n}$$. Clearly, \begin{align} f^{-1}[0, \epsilon]&\supset\{r\in U\ |\ 0\leq f(r)\leq\epsilon\}\\ &= \{r\in U\ |\ 0\leq f\circ y^{-1}\big(y(r)\big)\leq\epsilon\}\\ &= \{r\in U\ |\ 0\leq y^1(r)^2+\ldots+y^n(r)^2\leq\epsilon\}. \end{align} For the other inclusion, put a Riemannian metric $$\langle, \rangle$$ on $$M$$ and define the gradient of $$f$$, $$\nabla f$$, by $$\langle \nabla f_p, X_p \rangle=df_p(X_p).$$ Since $$M$$ is compact, $$\nabla f$$ is a complete vector field. Denote its flow by $$\phi:M\times\mathbb{R}\rightarrow M$$.
Assume there is $$r\in M\setminus U$$ such that $$f(r)=\epsilon_0\in [0,\epsilon]$$.
For any $$t\in\mathbb{R}$$, \begin{align} \frac{d}{dt}f\phi(t,r)&=df_{\phi(t,r)}\big(\frac{\partial}{\partial t}\phi(t,r)\big)\\ &=\langle \frac{\partial}{\partial t}\phi(t,r), \nabla f_{\phi(t,r)}\rangle\\ &=|\nabla f_{\phi(t,r)}|^2>0, \end{align} since $$r$$ cannot be in the orbits of $$p$$, $$q$$, being constant the integral curves of $$\nabla f$$ through these points. Hence, the map $$\mathbb{R}\rightarrow [0,1]$$, $$t\mapsto f\phi(t,r)$$ strictly increasing.
Let $$[0,\infty)\ni\delta=\inf_{(-\infty,0]} |\nabla f_{\phi(t,r)}|^2.$$ If $$\delta>0$$, then \begin{align} f\phi(0,r)-f\phi(t,r)&=\int^0_t \frac{d}{ds}f\phi(s,r)\\ &=\int^0_t |\nabla f_{\phi(s,r)}|^2 ds\\ &\geq \int^0_t \delta ds = -\delta t. \end{align} Making $$t\to -\infty$$ gives a contradiction, since the LHS is a number in $$[0,1]$$.
So, $$\delta=0$$ and we have a sequence $$\{t_n\}_n\subset (-\infty,0]$$ such that $$\lim_{n\to\infty}|\nabla f_{\phi(t_n,r)}|=0.$$ If $$\{t_n\}_n$$ had a bounded subsequence, we could rename it to $$\{t_n\}_n$$ and assume w.l.o.g that $$t_n\to t_0\in (-\infty,0]$$. Hence, $$\nabla f_{\phi(t_0,r)}=0$$, and this is absurd, as we said before.
Thus $$t_n\to -\infty$$. Again, since $$M$$ is compact we can assume w.l.o.g that $$\lim_{n\to\infty}\phi(t_n,r)=r_0\in M,$$ and then $$r_0=p$$. This shows that $$\lim_{t\to -\infty}\phi(t,r)=p.$$ In particular, the curve $$t\mapsto \phi(t,r)$$ intersects the $$(n-1)$$-dimensional sphere $$f^{-1}(\epsilon_0)\cap U$$, for some $$t<0$$. Of course this cannot happen, as $$f$$ strictly increases along the integral curve $$t\mapsto \phi(t,r)$$.
Thus, such an $$r$$ does not exist, and we get the other inclusion.
• I can't see why the statement "since $f:M\to \Bbb R$ is continuous, $p\in U$, $U$ is open,and $f(p)=0$, there is an $\epsilon>0$ such that $f^{-1}(-\epsilon,\epsilon)\subset U$. This is not true if $M=\Bbb R$, $f\equiv 0$, $p=0$ and $U=(-1,1)$ (which is an example that does not satisfying the assumptions of the theorem, though) Sep 25 '20 at 3:26